4_01_3A_Vectors_in_R

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.01%3A_Vectors_in_R

Skip to main content

Outcomes

1. Find the position vector of a point in \\\mathbb{R}^n\\.

The notation \\\mathbb{R}^{n}\\ refers to the collection of ordered lists of \\n\\ real numbers, that is \\\mathbb{R}^{n} = \left\\ \left( x\_{1}\cdots x\_{n}\right) :x\_{j}\in \mathbb{R}\text{ for }j=1,\cdots ,n\right\\\nonumber \\ In this chapter, we take a closer look at vectors in \\\mathbb{R}^n\\. First, we will consider what \\\mathbb{R}^n\\ looks like in more detail. Recall that the point given by \\0=\left( 0, \cdots, 0 \right)\\ is called the origin.

Now, consider the case of \\\mathbb{R}^n\\ for \\n=1.\\ Then from the definition we can identify \\\mathbb{R}\\ with points in \\\mathbb{R}^{1}\\ as follows: \\\mathbb{R} = \mathbb{R}^{1}= \left\\ \left( x\_{1}\right) :x\_{1}\in \mathbb{R} \right\\\nonumber \\ Hence, \\\mathbb{R}\\ is defined as the set of all real numbers and geometrically, we can describe this as all the points on a line.

Now suppose \\n=2\\. Then, from the definition, \\\mathbb{R}^{2}= \left\\ \left(x\_{1}, x\_{2}\right) :x\_{j}\in \mathbb{R}\text{ for }j=1,2 \right\\\nonumber \\ Consider the familiar coordinate plane, with an \\x\\ axis and a \\y\\ axis. Any point within this coordinate plane is identified by where it is located along the \\x\\ axis, and also where it is located along the \\y\\ axis. Consider as an example the following diagram.

Figure $\PageIndex{1}$

Hence, every element in \\\mathbb{R}^2\\ is identified by two components, \\x\\ and \\y\\, in the usual manner. The coordinates \\x, y\\ (or \\x_1\\,\\x_2\$ uniquely determine a point in the plan. Note that while the definition uses \\x_1\\ and \\x_2\\ to label the coordinates and you may be used to \\x\\ and \\y\\, these notations are equivalent.

Now suppose \\n=3\\. You may have previously encountered the \\3\\-dimensional coordinate system, given by \\\mathbb{R}^{3}= \left\\ \left( x\_{1}, x\_{2}, x\_{3}\right) :x\_{j}\in \mathbb{R}\text{ for }j=1,2,3 \right\\\nonumber \\

Points in \\\mathbb{R}^3\\ will be determined by three coordinates, often written \\\left(x,y,z\right)\\ which correspond to the \\x\\, \\y\\, and \\z\\ axes. We can think as above that the first two coordinates determine a point in a plane. The third component determines the height above or below the plane, depending on whether this number is positive or negative, and all together this determines a point in space. You see that the ordered triples correspond to points in space just as the ordered pairs correspond to points in a plane and single real numbers correspond to points on a line.

The idea behind the more general \\\mathbb{R}^n\\ is that we can extend these ideas beyond \\n = 3.\\ This discussion regarding points in \\\mathbb{R}^n\\ leads into a study of vectors in \\\mathbb{R}^n\\. While we consider \\\mathbb{R}^n\\ for all \\n\\, we will largely focus on \\n=2,3\\ in this section.

Consider the following definition.

Definition \\\PageIndex{1}\\ THe Position Vector

Let \\P=\left( p\_{1},\cdots ,p\_{n}\right)\\ be the coordinates of a point in \\\mathbb{R}^{n}.\\ Then the vector \\\overrightarrow{0P}\\ with its tail at \\0=\left( 0,\cdots ,0\right)\\ and its tip at \\P\\ is called the position vector of the point \\P\\. We write \\\overrightarrow{0P} = \left $$ \begin{array}{c} p\_{1} \\ \vdots \\ p\_{n} \end{array} \right $$\nonumber \\

For this reason we may write both \\P=\left( p\_{1},\cdots ,p\_{n}\right) \in \mathbb{R}^{n}\\ and \\\overrightarrow{0P} = \left $$ p\_{1} \cdots p\_{n} \right $$^T \in \mathbb{R}^{n}\\.

This definition is illustrated in the following picture for the special case of \\\mathbb{R}^{3}\\.

Figure $\PageIndex{2}$

Thus every point \\P\\ in \\\mathbb{R}^{n}\\ determines its position vector \\\overrightarrow{0P}\\. Conversely, every such position vector \\\overrightarrow{0P}\\ which has its tail at \\0\\ and point at \\P\\ determines the point \\P\\ of \\\mathbb{R}^{n}\\.

Now suppose we are given two points, \\P,Q\\ whose coordinates are \\\left( p\_{1},\cdots ,p\_{n}\right)\\ and \\\left( q\_{1},\cdots ,q\_{n}\right)\\ respectively. We can also determine the position vector from \\P\\ to \\Q\\ (also called the vector from \\P\\ to \\Q\\) defined as follows. \\\overrightarrow{PQ} = \left $$ \begin{array}{c} q\_{1}-p\_{1}\\ \vdots \\ q\_{n}-p\_{n} \end{array} \right $$ = \overrightarrow{0Q} - \overrightarrow{0P}\nonumber \\

Now, imagine taking a vector in \\\mathbb{R}^n\\ and moving it around, always keeping it pointing in the same direction as shown in the following picture.

Figure $\PageIndex{3}$

After moving it around, it is regarded as the same vector. Each vector, \\\overrightarrow{0P}\\ and \\\overrightarrow{AB}\\ has the same length (or magnitude) and direction. Therefore, they are equal.

Consider now the general definition for a vector in \\\mathbb{R}^n\\.

Definition \\\PageIndex{2}\\ Vectors in \\\mathbb{R}^n\\

Let \\\mathbb{R}^{n} = \left\\ \left( x\_{1}, \cdots, x\_{n}\right) :x\_{j}\in \mathbb{R}\text{ for }j=1,\cdots ,n\right\\ .\\ Then, \\\vec{x} = \left $$ \begin{array}{c} x\_{1} \\ \vdots \\ x\_{n} \end{array} \right $$\nonumber \\ is called a vector. Vectors have both size (magnitude) and direction. The numbers \\x\_{j}\\ are called the components of \\\vec{x}\\.

Using this notation, we may use \\\vec{p}\\ to denote the position vector of point \\P\\. Notice that in this context, \\\vec{p} = \overrightarrow{0P}\\. These notations may be used interchangeably.

You can think of the components of a vector as directions for obtaining the vector. Consider \\n=3\\. Draw a vector with its tail at the point \\\left( 0,0,0\right)\\ and its tip at the point \\\left( a,b,c\right)\\. This vector it is obtained by starting at \\\left( 0,0,0\right)\\, moving parallel to the \\x\\ axis to \\\left( a,0,0\right)\\ and then from here, moving parallel to the \\y\\ axis to \\\left( a,b,0\right)\\ and finally parallel to the \\z\\ axis to \\\left( a,b,c\right).\\ Observe that the same vector would result if you began at the point \\\left( d,e,f \right)\\, moved parallel to the \\x\\ axis to \\\left( d+a,e,f\right) ,\\ then parallel to the \\y\\ axis to \\\left( d+a,e+b,f\right) ,\\ and finally parallel to the \\z\\ axis to \\\left( d+a,e+b,f+c\right)\\. Here, the vector would have its tail sitting at the point determined by \\A= \left( d,e,f\right)\\ and its point at \\B=\left( d+a,e+b,f+c\right) .\\ It is the same vector because it will point in the same direction and have the same length. It is like you took an actual arrow, and moved it from one location to another keeping it pointing the same direction.

We conclude this section with a brief discussion regarding notation. In previous sections, we have written vectors as columns, or \\n \times 1\\ matrices. For convenience in this chapter we may write vectors as the transpose of row vectors, or \\1 \times n\\ matrices. These are of course equivalent and we may move between both notations. Therefore, recognize that \\\left $$ \begin{array}{r} 2 \\ 3 \end{array} \right $$ = \left $$ \begin{array}{rr} 2 & 3 \end{array} \right $$^T\nonumber \\

Notice that two vectors \\\vec{u} = \left $$ u\_{1} \cdots u\_{n}\right $$^T\\ and \\\vec{v}=\left $$ v\_{1} \cdots v\_{n}\right $$^T\\ are equal if and only if all corresponding components are equal. Precisely, \\\begin{array}{c} \vec{u}=\vec{v} \\ \mbox{if and only if}\\ u\_{j}=v\_{j} \\ \mbox{for all}\\ j=1,\cdots ,n \end{array}\nonumber \\ Thus \\\left $$ \begin{array}{rrr} 1 & 2 & 4 \end{array} \right $$^T \in \mathbb{R}^{3}\\ and \\\left $$ \begin{array}{rrr} 2 & 1 & 4 \end{array} \right $$^T \in \mathbb{R}^{3}\\ but \\\left $$ \begin{array}{rrr} 1 & 2 & 4 \end{array} \right $$^T \neq \left $$ \begin{array}{rrr} 2 & 1 & 4 \end{array} \right $$^T\\ because, even though the same numbers are involved, the order of the numbers is different.

For the specific case of \\\mathbb{R}^3\\, there are three special vectors which we often use. They are given by \\\vec{i} = \left $$ \begin{array}{rrr} 1 & 0 & 0 \end{array} \right $$^T\nonumber \\ \\\vec{j} = \left $$ \begin{array}{rrr} 0 & 1 & 0 \end{array} \right $$^T\nonumber \\ \\\vec{k} = \left $$ \begin{array}{rrr} 0 & 0 & 1 \end{array} \right $$^T\nonumber \\ We can write any vector \\\vec{u} = \left $$ \begin{array}{rrr} u_1 & u_2 & u_3 \end{array} \right $$^T\\ as a linear combination of these vectors, written as \\\vec{u} = u_1 \vec{i} + u_2 \vec{j} + u_3 \vec{k}\\. This notation will be used throughout this chapter.

---

4_02_3A_Vector_Algebra

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.02%3A_Vector_Algebra

Skip to main content

##### Outcomes

1. Understand vector addition and scalar multiplication, algebraically.

2. Introduce the notion of linear combination of vectors.

Addition and scalar multiplication are two important algebraic operations done with vectors. Notice that these operations apply to vectors in \\\mathbb{R}^{n}\\, for any value of \\n\\. We will explore these operations in more detail in the following sections.

Addition of Vectors in \\\mathbb{R}^n\\

Addition of vectors in \\\mathbb{R}^n\\ is defined as follows.

##### Definition \\\PageIndex{1}\\: Addition of Vectors in \\\mathbb{R}^n\\

If \\\vec{u}=\left $$ \begin{array}{c} u\_{1} \\ \vdots \\ u\_{n} \end{array} \right $$,\\ \vec{v}= \left $$ \begin{array}{c} v\_{1} \\ \vdots \\ v\_{n} \end{array} \right $$ \in \mathbb{R}^{n}\\ then \\\vec{u}+\vec{v}\in \mathbb{R}^{n}\\ and is defined by

\\\begin{aligned} \vec{u}+\vec{v} &= \left $$ \begin{array}{c} u\_{1} \\ \vdots \\ u\_{n} \end{array} \right $$ + \left $$ \begin{array}{c} v\_{1} \\ \vdots \\ v\_{n} \end{array} \right $$\\ & = \left $$ \begin{array}{c} u\_{1}+v\_{1} \\ \vdots \\ u\_{n}+v\_{n} \end{array} \right $$\end{aligned}\\

To add vectors, we simply add corresponding components. Therefore, in order to add vectors, they must be the same size.

Addition of vectors satisfies some important properties which are outlined in the following theorem.

##### Theorem \\\PageIndex{1}\\: Properties of Vector Addition

The following properties hold for vectors \\\vec{u},\vec{v}, \vec{w} \in \mathbb{R}^{n}\\.

• The Commutative Law of Addition \\\vec{u}+\vec{v}=\vec{v}+\vec{u}\nonumber \\
• The Associative Law of Addition \\\left( \vec{u}+\vec{v}\right) +\vec{w}=\vec{u}+\left( \vec{v}+\vec{w}\right)\nonumber \\
• The Existence of an Additive Identity \\\vec{u}+\vec{0}=\vec{u} \label{vectoridentity}\\
• The Existence of an Additive Inverse \\\vec{u}+\left( -\vec{u}\right) =\vec{0}\nonumber \\

The additive identity shown in Equation \\\eqref{vectoridentity}\\ is also called the zero vector, the \\n \times 1\\ vector in which all components are equal to \\0\\. Further, \\-\vec{u}\\ is simply the vector with all components having same value as those of \\\vec{u}\\ but opposite sign; this is just \$-1)\vec{u}\\. This will be made more explicit in the next section when we explore scalar multiplication of vectors. Note that subtraction is defined as \\\vec{u}-\vec{v} = \vec{u}+\left( -\vec{v} \right)\\.

Scalar Multiplication of Vectors in \\\mathbb{R}^n\\

Scalar multiplication of vectors in \\\mathbb{R}^n\\ is defined as follows.

##### Definition \\\PageIndex{2}\\: Scalar Multiplication of Vectors in \\\mathbb{R}^n\\

If \\\vec{u}\in \mathbb{R}^{n}\\ and \\k\in \mathbb{R}\\ is a scalar, then \\k\vec{u}\in \mathbb{R}^{n}\\ is defined by \\k\vec{u}=k\left $$ \begin{array}{c} u\_{1} \\ \vdots \\ u\_{n} \end{array} \right $$ = \left $$ \begin{array}{c} ku\_{1} \\ \vdots \\ ku\_{n} \end{array} \right $$\nonumber \\

Just as with addition, scalar multiplication of vectors satisfies several important properties. These are outlined in the following theorem.

##### Theorem \\\PageIndex{2}\\: Properties of Scalar Multiplication

The following properties hold for vectors \\\vec{u},\vec{v}\in \mathbb{R}^{n}\\ and \\k,p\\ scalars.

• The Distributive Law over Vector Addition \\k \left( \vec{u}+\vec{v}\right) = k\vec{u}+ k\vec{v}\nonumber\\
• The Distributive Law over Scalar Addition \\\left( k + p \right)\vec{u} = k \vec{u}+p \vec{u}\nonumber\\
• The Associative Law for Scalar Multiplication \\k \left( p \vec{u}\right) = \left(k p \right)\vec{u}\nonumber\\
• Rule for Multiplication by \\1\\ \\1\vec{u}=\vec{u}\nonumber\\

Proof

We will show the proof of: \\k \left( \vec{u}+\vec{v}\right) = k \vec{u}+ k \vec{v}\nonumber\\ Note that: \\\begin{array}{ll} k \left( \vec{u}+\vec{v}\right) & =k \left $$ u\_{1}+v\_{1} \cdots u\_{n}+v\_{n}\right $$^T \\ & = \left $$ k \left( u\_{1}+v\_{1}\right) \cdots k \left( u\_{n}+v\_{n}\right) \right $$^T \\ & = \left $$ k u\_{1}+ k v\_{1} \cdots k u\_{n}+ k v\_{n}\right $$^T \\ & = \left $$ k u\_{1} \cdots k u\_{n} \right $$^T + \left $$ k v\_{1} \cdots k v\_{n} \right $$^T \\ & = k \vec{u}+k \vec{v} \\ \end{array}\nonumber\\

We now present a useful notion you may have seen earlier combining vector addition and scalar multiplication

##### Definition \\\PageIndex{3}\\: Linear Combination

A vector \\\vec{v}\\ is said to be a linear combination of the vectors \\\vec{u}\_1,\cdots , \vec{u}\_n\\ if there exist scalars, \\a\_{1},\cdots ,a\_{n}\\ such that \\\vec{v} = a_1 \vec{u}\_1 + \cdots + a_n \vec{u}\_n\nonumber \\

For example, \\3 \left $$ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right $$ + 2 \left $$ \begin{array}{r} -3 \\ 0\\ 1 \end{array} \right $$ = \left $$ \begin{array}{r} -18 \\ 3 \\ 2 \end{array} \right $$.\nonumber \\ Thus we can say that \\\vec{v}= \left $$ \begin{array}{r} -18 \\ 3 \\ 2 \end{array} \right $$\nonumber \\ is a linear combination of the vectors \\\vec{u}\_1 = \left $$ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right $$ \mbox{ and } \vec{u}\_2 = \left $$ \begin{array}{r} -3 \\ 0\\ 1 \end{array} \right $$\nonumber \\

---

4_03_3A_Geometric_Meaning_of_Vector_Addition

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.03%3A_Geometric_Meaning_of_Vector_Addition

Skip to main content

Learning Objectives

1. Understand vector addition, geometrically.

Recall that an element of \\\mathbb{R}^{n}\\ is an ordered list of numbers. For the specific case of \\n=2,3\\ this can be used to determine a point in two or three dimensional space. This point is specified relative to some coordinate axes.

Consider the case \\n=3\\. Recall that taking a vector and moving it around without changing its length or direction does not change the vector. This is important in the geometric representation of vector addition.

Suppose we have two vectors, \\\vec{u}\\ and \\\vec{v}\\ in \\\mathbb{R}^{3}\\. Each of these can be drawn geometrically by placing the tail of each vector at \\0\\ and its point at \\\left( u\_{1}, u\_{2}, u\_{3}\right)\\ and \\\left( v\_{1}, v\_{2}, v\_{3}\right)\\ respectively. Suppose we slide the vector \\\vec{v}\\ so that its tail sits at the point of \\\vec{u}\\. We know that this does not change the vector \\\vec{v}\\. Now, draw a new vector from the tail of \\\vec{u}\\ to the point of \\\vec{v}\\. This vector is \\\vec{u}+\vec{v}\\.

The geometric significance of vector addition in \\\mathbb{R}^n\\ for any \\n\\ is given in the following definition.

Definition \\\PageIndex{1}\\: Geometry of Vector Addition

Let \\\vec{u}\\ and \\\vec{v}\\ be two vectors. Slide \\\vec{v}\\ so that the tail of \\\vec{v}\\ is on the point of \\\vec{u}\\. Then draw the arrow which goes from the tail of \\\vec{u}\\ to the point of \\\vec{v}\\. This arrow represents the vector \\\vec{u}+\vec{v}\\.

Figure $\PageIndex{1}$

This definition is illustrated in the following picture in which \\\vec{u}+\vec{v}\\ is shown for the special case \\n=3\\.

Figure $\PageIndex{2}$

Notice the parallelogram created by \\\vec{u}\\ and \\\vec{v}\\ in the above diagram. Then \\\vec{u} + \vec{v}\\ is the directed diagonal of the parallelogram determined by the two vectors \\\vec{u}\\ and \\\vec{v}\\.

When you have a vector \\\vec{v}\\, its additive inverse \\-\vec{v}\\ will be the vector which has the same magnitude as \\\vec{v}\\ but the opposite direction. When one writes \\\vec{u}-\vec{v,}\\ the meaning is \\\vec{u} + \left( -\vec{v}\right)\\ as with real numbers. The following example illustrates these definitions and conventions.

Example \\\PageIndex{1}\\: Graphing Vector Addition

Consider the following picture of vectors \\\vec{u}\\ and \\\vec{v}\\.

Figure $\PageIndex{3}$

Sketch a picture of \\\vec{u}+\vec{v},\vec{u}-\vec{v}.\\

###### Solution

We will first sketch \\\vec{u}+\vec{v}.\\ Begin by drawing \\\vec{u}\\ and then at the point of \\\vec{u}\\, place the tail of \\\vec{v}\\ as shown. Then \\\vec{u}+\vec{v}\\ is the vector which results from drawing a vector from the tail of \\\vec{u}\\ to the tip of \\\vec{v}\\.

Figure $\PageIndex{4}$

Next consider \\\vec{u}-\vec{v}.\\ This means \\\vec{u}+\left( -\vec{v} \right) .\\ From the above geometric description of vector addition, \\-\vec{v}\\ is the vector which has the same length but which points in the opposite direction to \\\vec{v}\\. Here is a picture.

Figure $\PageIndex{5}$

---

4_04_3A_Length_of_a_Vector

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.04%3A_Length_of_a_Vector

Skip to main content

Learning Objectives

• Find the length of a vector and the distance between two points in \\\mathbb{R}^n\\.
• Find the corresponding unit vector to a vector in \\\mathbb{R}^n\\.

In this section, we explore what is meant by the length of a vector in \\\mathbb{R}^n\\. We develop this concept by first looking at the distance between two points in \\\mathbb{R}^n\\.

First, we will consider the concept of distance for \\\mathbb{R}\\, that is, for points in \\\mathbb{R}^1\\. Here, the distance between two points \\P\\ and \\Q\\ is given by the absolute value of their difference. We denote the distance between \\P\\ and \\Q\\ by \\d(P,Q)\\ which is defined as \\d(P,Q) = \sqrt{ \left( P-Q\right) ^{2}} \label{distance1}\\

Consider now the case for \\n=2\\, demonstrated by the following picture.

Figure $\PageIndex{1}$

There are two points \\P =\left( p\_{1},p\_{2}\right)\\ and \\Q = \left(q\_{1},q\_{2}\right)\\ in the plane. The distance between these points is shown in the picture as a solid line. Notice that this line is the hypotenuse of a right triangle which is half of the rectangle shown in dotted lines. We want to find the length of this hypotenuse which will give the distance between the two points. Note the lengths of the sides of this triangle are \\\left\| p\_{1}-q\_{1}\right\|\\ and \\\left\| p\_{2}-q\_{2}\right\|\\, the absolute value of the difference in these values. Therefore, the Pythagorean Theorem implies the length of the hypotenuse (and thus the distance between \\P\\ and \\Q\$ equals \\\left( \left\| p\_{1}-q\_{1}\right\| ^{2}+\left\| p\_{2}-q\_{2}\right\| ^{2}\right) ^{1/2}=\left( \left( p\_{1}-q\_{1}\right) ^{2}+\left( p\_{2}-q\_{2}\right) ^{2}\right) ^{1/2} \label{distance2}\\

Now suppose \\n=3\\ and let \\P = \left( p\_{1},p\_{2},p\_{3}\right)\\ and \\Q = \left( q\_{1},q\_{2},q\_{3}\right)\\ be two points in \\\mathbb{R}^{3}.\\ Consider the following picture in which the solid line joins the two points and a dotted line joins the points \\\left( q\_{1},q\_{2},q\_{3}\right)\\ and \\\left( p\_{1},p\_{2},q\_{3}\right) .\\

Figure $\PageIndex{2}$

Here, we need to use Pythagorean Theorem twice in order to find the length of the solid line. First, by the Pythagorean Theorem, the length of the dotted line joining \\\left( q\_{1},q\_{2},q\_{3}\right)\\ and \\\left( p\_{1},p\_{2},q\_{3}\right)\\ equals \\\left( \left( p\_{1}-q\_{1}\right) ^{2}+\left( p\_{2}-q\_{2}\right) ^{2}\right) ^{1/2}\nonumber \\ while the length of the line joining \\\left( p\_{1},p\_{2},q\_{3}\right)\\ to \\\left( p\_{1},p\_{2},p\_{3}\right)\\ is just \\\left\| p\_{3}-q\_{3}\right\| .\\ Therefore, by the Pythagorean Theorem again, the length of the line joining the points \\P = \left( p\_{1},p\_{2},p\_{3}\right)\\ and \\Q = \left( q\_{1},q\_{2},q\_{3}\right)\\ equals \\\left( \left( \left( \left( p\_{1}-q\_{1}\right) ^{2}+\left( p\_{2}-q\_{2}\right) ^{2}\right) ^{1/2}\right) ^{2}+\left( p\_{3}-q\_{3}\right) ^{2}\right) ^{1/2}\nonumber \\ \\=\left( \left( p\_{1}-q\_{1}\right) ^{2}+\left( p\_{2}-q\_{2}\right) ^{2}+\left( p\_{3}-q\_{3}\right) ^{2}\right) ^{1/2} \label{distance3}\\

This discussion motivates the following definition for the distance between points in \\\mathbb{R}^n\\.

Definition \\\PageIndex{1}\\: Distance Between Points

Let \\P=\left( p\_{1},\cdots ,p\_{n}\right)\\ and \\Q=\left( q\_{1},\cdots ,q\_{n}\right)\\ be two points in \\\mathbb{R}^{n}\\. Then the distance between these points is defined as \\\text{ distance between }P\text{ and } Q\text{ } = d( P, Q ) = \left( \sum\_{k=1}^{n}\left\vert p\_{k}-q\_{k}\right\vert ^{2}\right) ^{1/2}\nonumber \\ This is called the distance formula. We may also write \\\left\vert P - Q \right\vert\\ as the distance between \\P\\ and \\Q\\.

From the above discussion, you can see that Definition $\PageIndex{1}$ holds for the special cases \\n=1,2,3\\, as in Equations \\\eqref{distance1}\\, \\\eqref{distance2}\\, \\\eqref{distance3}\\. In the following example, we use Definition $\PageIndex{1}$ to find the distance between two points in \\\mathbb{R}^4\\.

Example \\\PageIndex{1}\\: Distance Between Points

Find the distance between the points \\P\\ and \\Q\\ in \\\mathbb{R}^{4}\\, where \\P\\ and \\Q\\ are given by \\P=\left( 1,2,-4,6\right)\nonumber \\ and \\Q=\left( 2,3,-1,0\right)\nonumber \\

###### Solution

We will use the formula given in Definition $\PageIndex{1}$ to find the distance between \\P\\ and \\Q\\. Use the distance formula and write \\d(P,Q)= \left( \left( 1-2\right) ^{2}+\left( 2-3\right) ^{2}+\left( -4-\left( -1\right) \right) ^{2}+\left( 6-0\right)^{2}\right) ^{\frac{1}{2}} = 47\nonumber \\

Therefore, \\d( P,Q) = \sqrt{47}.\\

There are certain properties of the distance between points which are important in our study. These are outlined in the following theorem.

Theorem \\\PageIndex{1}\\: Properties of Distance

Let \\P\\ and \\Q\\ be points in \\\mathbb{R}^n\\, and let the distance between them, \\d( P, Q)\\, be given as in Definition $\PageIndex{1}$. Then, the following properties hold.

• \\d( P, Q) = d( Q, P)\\
• \\d( P, Q) \geq 0\\, and equals 0 exactly when \\P = Q.\\

There are many applications of the concept of distance. For instance, given two points, we can ask what collection of points are all the same distance between the given points. This is explored in the following example.

Example \\\PageIndex{2}\\: The Plane Between Two Points

Describe the points in \\\mathbb{R}^3\\ which are at the same distance between \\\left( 1,2,3\right)\\ and \\\left( 0,1,2\right) .\\

###### Solution

Let \\P = \left( p_1 , p_2, p_3\right)\\ be such a point. Therefore, \\P\\ is the same distance from \\\left( 1,2,3\right)\\ and \\\left( 0,1,2\right) .\\ Then byDefinition $\PageIndex{1}$, \\\sqrt{\left( p_1 -1\right) ^{2}+\left( p_2 -2\right) ^{2}+\left( p_3-3\right) ^{2}}= \sqrt{\left( p_1 - 0 \right)^{2}+\left( p_2-1\right) ^{2}+\left( p_3-2\right) ^{2}}\nonumber \\ Squaring both sides we obtain \\\left( p_1 -1\right) ^{2}+\left( p_2 -2\right) ^{2}+\left( p_3 -3\right) ^{2}=p_1^{2}+\left( p_2-1\right) ^{2}+\left( p_3 -2\right) ^{2}\nonumber \\ and so \\ \\ p_1^{2}-2p_1+14+p_2^{2}-4p_2+p_3^{2}-6p_3=p_1^{2}+p_2^{2}-2p_2+5+p_3^{2}-4p_3\nonumber \\ Simplifying, this becomes \\-2p_1+14-4p_2-6p_3=-2p_2+5-4p_3\nonumber \\ which can be written as \\2p_1+2p_2+2p_3=-9 \label{distanceplane}\\ Therefore, the points \\P = \left( p_1,p_2,p_3\right)\\ which are the same distance from each of the given points form a plane whose equation is given by \\\eqref{distanceplane}\\.

We can now use our understanding of the distance between two points to define what is meant by the length of a vector. Consider the following definition.

Definition \\\PageIndex{2}\\: Length of a Vector

Let \\\vec{u} = \left$$ u\_{1} \cdots u\_{n} \right$$^T\\ be a vector in \\\mathbb{R}^n\\. Then, the length of \\\vec{u}\\, written \\\\ \vec{u} \\\\ is given by \\\\ \vec{u} \\ = \sqrt{ u\_{1}^2 + \cdots + u\_{n}^2}\nonumber \\

This definition corresponds to Definition $\PageIndex{1}$, if you consider the vector \\\vec{u}\\ to have its tail at the point \\0 = \left( 0, \cdots ,0 \right)\\ and its tip at the point \\U = \left(u_1, \cdots, u_n \right)\\. Then the length of \\\vec{u}\\ is equal to the distance between \\0\\ and \\U\\, \\d(0,U)\\. In general, \\d(P,Q)=\|\|\vec{PQ}\|\|\\.

ConsiderExample $\PageIndex{1}$. ByDefinition $\PageIndex{2}$, we could also find the distance between \\P\\ and \\Q\\ as the length of the vector connecting them. Hence, if we were to draw a vector \\\overrightarrow{PQ}\\ with its tail at \\P\\ and its point at \\Q\\, this vector would have length equal to \\\sqrt{47}\\.

We conclude this section with a new definition for the special case of vectors of length \\1\\.

Definition \\\PageIndex{3}\\: Unit Vector

Let \\\vec{u}\\ be a vector in \\\mathbb{R}^{n}\\. Then, we call \\\vec{u}\\ a unit vector if it has length 1, that is if \\\\ \vec{u} \\ = 1\nonumber \\

Let \\\vec{v}\\ be a vector in \\\mathbb{R}^{n}\\. Then, the vector \\\vec{u}\\ which has the same direction as \\\vec{v}\\ but length equal to \\1\\ is the corresponding unit vector of \\\vec{v}\\. This vector is given by \\\vec{u} = \frac{1}{\\ \vec{v} \\} \vec{v}\nonumber \\

We often use the term normalize to refer to this process. When we normalize a vector, we find the corresponding unit vector of length \\1\\. Consider the following example.

Example \\\PageIndex{3}\\: Finding a Unit Vector

Let \\\vec{v}\\ be given by \\\vec{v} = \left$$ \begin{array}{rrr} 1 & -3 & 4 \end{array} \right$$^T\nonumber \\ Find the unit vector \\\vec{u}\\ which has the same direction as \\\vec{v}\\.

###### Solution

We will use Definition $\PageIndex{3}$ to solve this. Therefore, we need to find the length of \\\vec{v}\\ which, by Definition $\PageIndex{2}$ is given by \\\\ \vec{v} \\ = \sqrt{ v\_{1}^2 + v\_{2}^2+ v\_{3}^2}\nonumber \\ Using the corresponding values we find that \\\begin{aligned} \\ \vec{v} \\ &= \sqrt{ 1^2 + \left(-3 \right)^2 + 4^2} \\ &= \sqrt{ 1 + 9 + 16} \\ &= \sqrt{26} \end{aligned}\\ In order to find \\\vec{u}\\, we divide \\\vec{v}\\ by \\\sqrt{26}\\. The result is \\\begin{aligned} \vec{u} &= \frac{1}{\\ \vec{v} \\} \vec{v} \\ &= \frac{1}{\sqrt{26}} \left$$ \begin{array}{rrr} 1 & -3 & 4 \end{array} \right$$^T \\ &= \left$$ \begin{array}{rrr} \frac{1}{\sqrt{26}} & -\frac{3}{\sqrt{26}} & \frac{4}{\sqrt{26}} \end{array} \right$$^T\end{aligned}\\

You can verify using the Definition $\PageIndex{1}$ that \\\\ \vec{u} \\ = 1\\.

---

4_05_3A_Geometric_Meaning_of_Scalar_Multiplication

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.05%3A_Geometric_Meaning_of_Scalar_Multiplication

Skip to main content

Outcomes

1. Understand scalar multiplication, geometrically.

Recall that the point \\P=\left( p\_{1},p\_{2},p\_{3}\right)\\ determines a vector \\\vec{p}\\ from \\0\\ to \\P\\. The length of \\\vec{p}\\, denoted \\\\ \vec{p} \\\\, is equal to \\\sqrt{p\_{1}^{2}+p\_{2}^{2}+p\_{3}^{2}}\\ by Definition 4.4.1.

Now suppose we have a vector \\\vec{u} = \left$$ \begin{array}{lll} u_1 & u_2 & u_3 \end{array} \right$$^T\\ and we multiply \\\vec{u}\\ by a scalar \\k\\. By Definition 4.2.2, \\k\vec{u} = \left$$ \begin{array}{rrr} ku\_{1} & ku\_{2} & ku\_{3} \end{array} \right$$^T\\. Then, by using Definition 4.4.1, the length of this vector is given by \\\sqrt{\left( \left( k u\_{1}\right) ^{2}+\left( k u\_{2}\right) ^{2}+\left( k u\_{3}\right) ^{2}\right) }=\left\vert k \right\vert \sqrt{u\_{1}^{2}+u\_{2}^{2}+u\_{3}^{2}}\nonumber \\ Thus the following holds. \\\\ k \vec{u} \\ =\left\vert k \right\vert \\ \vec{u} \\\nonumber \\ In other words, multiplication by a scalar magnifies or shrinks the length of the vector by a factor of \\\left\vert k \right\vert\\. If \\\left\vert k \right\vert \> 1\\, the length of the resulting vector will be magnified. If \\\left\vert k \right\vert \<1\\, the length of the resulting vector will shrink. Remember that by the definition of the absolute value, \\\left\vert k \right\vert \>0\\.

What about the direction? Draw a picture of \\\vec{u}\\ and \\k\vec{u}\\ where \\k\\ is negative. Notice that this causes the resulting vector to point in the opposite direction while if \\k \>0\\ it preserves the direction the vector points. Therefore the direction can either reverse, if \\k \< 0\\, or remain preserved, if \\k \> 0\\.

Consider the following example.

Example \\\PageIndex{1}\\: Graphing Scalar Multiplication

Consider the vectors \\\vec{u}\\ and \\\vec{v}\\ drawn below.

Figure $\PageIndex{1}$

Draw \\-\vec{u}\\, \\2\vec{v}\\, and \\-\frac{1}{2}\vec{v}\\.

###### Solution

In order to find \\-\vec{u}\\, we preserve the length of \\\vec{u}\\ and simply reverse the direction. For \\2\vec{v}\\, we double the length of \\\vec{v}\\, while preserving the direction. Finally \\-\frac{1}{2}\vec{v}\\ is found by taking half the length of \\\vec{v}\\ and reversing the direction. These vectors are shown in the following diagram.

Figure $\PageIndex{2}$

Now that we have studied both vector addition and scalar multiplication, we can combine the two actions. Recall Definition 9.2.2 of linear combinations of column matrices. We can apply this definition to vectors in \\\mathbb{R}^n\\. A linear combination of vectors in \\\mathbb{R}^n\\ is a sum of vectors multiplied by scalars.

In the following example, we examine the geometric meaning of this concept.

Example \\\PageIndex{2}\\: Graphing a Linear Combination of Vectors

Consider the following picture of the vectors \\\vec{u}\\ and \\\vec{v}\\

Figure $\PageIndex{3}$

Sketch a picture of \\\vec{u}+2\vec{v},\vec{u}-\frac{1}{2}\vec{v}.\\

###### Solution

The two vectors are shown below.

Figure $\PageIndex{4}$

---

4_06_3A_Parametric_Lines

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.06%3A_Parametric_Lines

Skip to main content

Outcomes

1. Find the vector and parametric equations of a line.

We can use the concept of vectors and points to find equations for arbitrary lines in \\\mathbb{R}^n\\, although in this section the focus will be on lines in \\\mathbb{R}^3\\.

To begin, consider the case \\n=1\\ so we have \\\mathbb{R}^{1}=\mathbb{R}\\. There is only one line here which is the familiar number line, that is \\\mathbb{R}\\ itself. Therefore it is not necessary to explore the case of \\n=1\\ further.

Now consider the case where \\n=2\\, in other words \\\mathbb{R}^2\\. Let \\P\\ and \\P_0\\ be two different points in \\\mathbb{R}^{2}\\ which are contained in a line \\L\\. Let \\\vec{p}\\ and \\\vec{p_0}\\ be the position vectors for the points \\P\\ and \\P_0\\ respectively. Suppose that \\Q\\ is an arbitrary point on \\L\\. Consider the following diagram.

Figure $\PageIndex{1}$

Our goal is to be able to define \\Q\\ in terms of \\P\\ and \\P_0\\. Consider the vector \\\overrightarrow{P_0P} = \vec{p} - \vec{p_0}\\ which has its tail at \\P_0\\ and point at \\P\\. If we add \\\vec{p} - \vec{p_0}\\ to the position vector \\\vec{p_0}\\ for \\P_0\\, the sum would be a vector with its point at \\P\\. In other words, \\\vec{p} = \vec{p_0} + (\vec{p} - \vec{p_0})\nonumber \\

Now suppose we were to add \\t(\vec{p} - \vec{p_0})\\ to \\\vec{p}\\ where \\t\\ is some scalar. You can see that by doing so, we could find a vector with its point at \\Q\\. In other words, we can find \\t\\ such that \\\vec{q} = \vec{p_0} + t \left( \vec{p}- \vec{p_0}\right)\nonumber \\

This equation determines the line \\L\\ in \\\mathbb{R}^2\\. In fact, it determines a line \\L\\ in \\\mathbb{R}^n\\. Consider the following definition.

Definition \\\PageIndex{1}\\: Vector Equation of a Line

Suppose a line \\L\\ in \\\mathbb{R}^{n}\\ contains the two different points \\P\\ and \\P_0\\. Let \\\vec{p}\\ and \\\vec{p_0}\\ be the position vectors of these two points, respectively. Then, \\L\\ is the collection of points \\Q\\ which have the position vector \\\vec{q}\\ given by \\\vec{q}=\vec{p_0}+t\left( \vec{p}-\vec{p_0}\right)\nonumber \\ where \\t\in \mathbb{R}\\.

Let \\\vec{d} = \vec{p} - \vec{p_0}\\. Then \\\vec{d}\\ is the direction vector for \\L\\ and the vector equation for \\L\\ is given by \\\vec{p}=\vec{p_0}+t\vec{d}, t\in\mathbb{R}\nonumber \\

Note that this definition agrees with the usual notion of a line in two dimensions and so this is consistent with earlier concepts. Consider now points in \\\mathbb{R}^3\\. If a point \\P \in \mathbb{R}^3\\ is given by \\P = \left( x,y,z \right)\\, \\P_0 \in \mathbb{R}^3\\ by \\P_0 = \left( x_0, y_0, z_0 \right)\\, then we can write \\\left$$ \begin{array}{c} x \\ y \\ z \end{array} \right$$ = \left$$ \begin{array}{c} x_0 \\ y_0 \\ z_0 \end{array} \right$$ + t \left$$ \begin{array}{c} a \\ b \\ c \end{array} \right$$ \nonumber \\ where \\\vec{d} = \left$$ \begin{array}{c} a \\ b \\ c \end{array} \right$$\\. This is the vector equation of \\L\\ written in component form .

The following theorem claims that such an equation is in fact a line.

Proposition \\\PageIndex{1}\\: Algebraic Description of a Straight Line

Let \\\vec{a},\vec{b}\in \mathbb{R}^{n}\\ with \\\vec{b}\neq \vec{0}\\. Then \\\vec{x}=\vec{a}+t\vec{b},\\ t\in \mathbb{R}\\, is a line.

Proof

Let \\\vec{x\_{1}}, \vec{x\_{2}} \in \mathbb{R}^n\\. Define \\\vec{x\_{1}}=\vec{a}\\ and let \\\vec{x\_{2}}-\vec{x\_{1}}=\vec{b}\\. Since \\\vec{b} \neq \vec{0}\\, it follows that \\\vec{x\_{2}}\neq \vec{x\_{1}}.\\ Then \\\vec{a}+t\vec{b}=\vec{x\_{1}} + t\left( \vec{x\_{2}}-\vec{x\_{1}}\right)\\. It follows that \\\vec{x}=\vec{a}+t\vec{b}\\ is a line containing the two different points \\X_1\\ and \\X_2\\ whose position vectors are given by \\\vec{x}\_1\\ and \\\vec{x}\_2\\ respectively.

We can use the above discussion to find the equation of a line when given two distinct points. Consider the following example.

Example \\\PageIndex{1}\\: A Line From Two Points

Find a vector equation for the line through the points \\P_0 = \left( 1,2,0\right)\\ and \\P = \left( 2,-4,6\right).\\

###### Solution

We will use the definition of a line given above in Definition $\PageIndex{1}$ to write this line in the form

\\\vec{q}=\vec{p_0}+t\left( \vec{p}-\vec{p_0}\right)\nonumber \\

Let \\\vec{q} = \left$$ \begin{array}{c} x \\ y \\ z \end{array} \right$$B\\. Then, we can find \\\vec{p}\\ and \\\vec{p_0}\\ by taking the position vectors of points \\P\\ and \\P_0\\ respectively. Then,

\\\vec{q}=\vec{p_0}+t\left( \vec{p}-\vec{p_0}\right)\nonumber \\ can be written as

\\\left$$ \begin{array}{c} x \\ y \\ z \\ \end{array} \right$$B = \left$$ \begin{array}{c} 1 \\ 2 \\ 0 \end{array} \right$$B + t \left$$ \begin{array}{r} 1 \\ -6 \\ 6 \end{array} \right$$B, \\t\in \mathbb{R}\nonumber \\

Here, the direction vector \\\left$$ \begin{array}{r} 1 \\ -6 \\ 6 \end{array} \right$$B\\ is obtained by \\\vec{p} - \vec{p_0} = \left$$ \begin{array}{r} 2 \\ -4 \\ 6 \end{array} \right$$B - \left$$ \begin{array}{r} 1 \\ 2 \\ 0 \end{array} \right$$B\\ as indicated above in Definition $\PageIndex{1}$.

Notice that in the above example we said that we found “a” vector equation for the line, not “the” equation. The reason for this terminology is that there are infinitely many different vector equations for the same line. To see this, replace \\t\\ with another parameter, say \\3s.\\ Then you obtain a different vector equation for the same line because the same set of points is obtained.

In Example $\PageIndex{1}$, the vector given by \\\left$$ \begin{array}{r} 1 \\ -6 \\ 6 \end{array} \right$$B\\ is the direction vector defined in Definition $\PageIndex{1}$. If we know the direction vector of a line, as well as a point on the line, we can find the vector equation.

Consider the following example.

Example \\\PageIndex{2}\\: A Line From a Point and a Direction Vector

Find a vector equation for the line which contains the point \\P_0 = \left( 1,2,0\right)\\ and has direction vector \\\vec{d} = \left$$ \begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right$$B\\

###### Solution

We will use Definition $\PageIndex{1}$ to write this line in the form \\\vec{p}=\vec{p_0}+t\vec{d},\\ t\in \mathbb{R}\\. We are given the direction vector \\\vec{d}\\. In order to find \\\vec{p_0}\\, we can use the position vector of the point \\P_0\\. This is given by \\\left$$ \begin{array}{c} 1 \\ 2 \\ 0 \end{array} \right$$B.\\ Letting \\\vec{p} = \left$$ \begin{array}{c} x \\ y \\ z \end{array} \right$$B\\, the equation for the line is given by \\\left$$ \begin{array}{c} x \\ y \\ z \end{array} \right$$B = \left$$ \begin{array}{c} 1 \\ 2 \\ 0 \end{array} \right$$B + t \left$$ \begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right$$B, \\t\in \mathbb{R} \label{vectoreqn}\\

We sometimes elect to write a line such as the one given in \\\eqref{vectoreqn}\\ in the form \\\begin{array}{ll} \left. \begin{array}{l} x=1+t \\ y=2+2t \\ z=t \end{array} \right\\ & \mbox{where} \\ t\in \mathbb{R} \end{array} \label{parameqn}\\ This set of equations give the same information as \\\eqref{vectoreqn}\\, and is called the parametric equation of the line.

Consider the following definition.

Definition \\\PageIndex{2}\\: Parametric Equation of a Line

Let \\L\\ be a line in \\\mathbb{R}^3\\ which has direction vector \\\vec{d} = \left$$ \begin{array}{c} a \\ b \\ c \end{array} \right$$B\\ and goes through the point \\P_0 = \left( x_0, y_0, z_0 \right)\\. Then, letting \\t\\ be a parameter, we can write \\L\\ as \\\begin{array}{ll} \left. \begin{array}{c} x = x_0 + ta \\ y = y_0 + tb \\ z = z_0 + tc \end{array} \right\\ & \mbox{where} \\ t\in \mathbb{R} \end{array}\nonumber \\ This is called a parametric equation of the line \\L\\.

You can verify that the form discussed following Example $\PageIndex{2}$ in equation \\\eqref{parameqn}\\ is of the form given in Definition $\PageIndex{2}$.

There is one other form for a line which is useful, which is the symmetric form. Consider the line given by \\\eqref{parameqn}\\. You can solve for the parameter \\t\\ to write \\\begin{array}{l} t=x-1 \\ t=\frac{y-2}{2} \\ t=z \end{array}\nonumber \\ Therefore, \\x-1=\frac{y-2}{2}=z\nonumber \\ This is the symmetric form of the line.

In the following example, we look at how to take the equation of a line from symmetric form to parametric form.

Example \\\PageIndex{3}\\: Change Symmetric Form to Parametric Form

Suppose the symmetric form of a line is \\\frac{x-2}{3}=\frac{y-1}{2}=z+3\nonumber \\ Write the line in parametric form as well as vector form.

###### Solution

We want to write this line in the form given by Definition $\PageIndex{2}$. This is of the form \\\begin{array}{ll} \left. \begin{array}{c} x = x_0 + ta \\ y = y_0 + tb \\ z = z_0 + tc \end{array} \right\\ & \mbox{where} \\ t\in \mathbb{R} \end{array}\nonumber \\

Let \\t=\frac{x-2}{3},t=\frac{y-1}{2}\\ and \\t=z+3\\, as given in the symmetric form of the line. Then solving for \\x,y,z,\\ yields \\\begin{array}{ll} \left. \begin{array}{c} x=2 + 3t \\ y=1 + 2t \\ z=-3 + t \end{array} \right\\ & \mbox{with} \\t\in \mathbb{R} \end{array}\nonumber \\

This is the parametric equation for this line.

Now, we want to write this line in the form given by Definition $\PageIndex{1}$. This is the form \\\vec{p}=\vec{p_0}+t\vec{d}\nonumber\\ where \\t\in \mathbb{R}\\. This equation becomes \\\left$$ \begin{array}{c} x \\ y \\ z \end{array} \right$$B = \left$$ \begin{array}{r} 2 \\ 1 \\ -3 \end{array} \right$$B + t \left$$ \begin{array}{r} 3 \\ 2 \\ 1 \end{array} \right$$B, \\t\in \mathbb{R}\nonumber \\

---

4_07_3A_The_Dot_Product

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.07%3A_The_Dot_Product

Skip to main content

##### Outcomes

1. Compute the dot product of vectors, and use this to compute vector projections.

The Dot Product

There are two ways of multiplying vectors which are of great importance in applications. The first of these is called the dot product. When we take the dot product of vectors, the result is a scalar. For this reason, the dot product is also called the scalar product and sometimes the inner product. The definition is as follows.

##### Definition \\\PageIndex{1}\\: Dot Product

Let \\\vec{u},\vec{v}\\ be two vectors in \\\mathbb{R}^{n}\\. Then we define the dot product \\\vec{u}\bullet \vec{v}\\ as \\\vec{u}\bullet \vec{v} = \sum\_{k=1}^{n}u\_{k}v\_{k}\nonumber \\

The dot product \\\vec{u}\bullet \vec{v}\\ is sometimes denoted as \$\vec{u},\vec{v})\\ where a comma replaces \\\bullet\\. It can also be written as \\\left\langle \vec{u},\vec{v}\right\rangle\\. If we write the vectors as column or row matrices, it is equal to the matrix product \\\vec{v}\vec{w}^{T}\\.

Consider the following example.

##### Example \\\PageIndex{1}\\: Compute a Dot Product

Find \\\vec{u} \bullet \vec{v}\\ for \\\vec{u} = \left$$ \begin{array}{r} 1 \\ 2 \\ 0 \\ -1 \end{array} \right$$, \vec{v} = \left$$ \begin{array}{r} 0 \\ 1 \\ 2 \\ 3 \end{array} \right$$\nonumber \\

###### Solution

By Definition $\PageIndex{1}$, we must compute \\\vec{u}\bullet \vec{v} = \sum\_{k=1}^{4}u\_{k}v\_{k}\nonumber \\

This is given by \\\begin{aligned} \vec{u} \bullet \vec{v} &= (1)(0) + (2)(1) + (0)(2) + (-1)(3) \\ &= 0 + 2 + 0 + -3 \\ &= -1\end{aligned}\\

With this definition, there are several important properties satisfied by the dot product.

##### Proposition \\\PageIndex{1}\\: Properties of the Dot Product

Let \\k\\ and \\p\\ denote scalars and \\\vec{u},\vec{v},\vec{w}\\ denote vectors. Then the dot product \\\vec{u} \bullet \vec{v}\\ satisfies the following properties.

• \\\vec{u}\bullet \vec{v}= \vec{v}\bullet \vec{u}\\
• \\\vec{u}\bullet \vec{u}\geq 0 \text{ and equals zero if and only if }\vec{u}=\vec{0}\\
• \\\left( k\vec{u}+p\vec{v}\right) \bullet \vec{w}=k\left( \vec{u}\bullet \vec{w}\right) +p\left( \vec{v}\bullet \vec{w}\right)\\
• \\\vec{u}\bullet\left( k\vec{v}+p\vec{w}\right) =k\left( \vec{u}\bullet \vec{v}\right) +p\left( \vec{u}\bullet \vec{w}\right)\\
• \\\\ \vec{u}\\ ^{2}=\vec{u}\bullet \vec{u}\\

Proof

The proof is left as an exercise.

This proposition tells us that we can also use the dot product to find the length of a vector.

##### Example \\\PageIndex{2}\\: Length of a Vector

Find the length of \\\vec{u} = \left$$ \begin{array}{r} 2 \\ 1 \\ 4 \\ 2 \end{array} \right$$\nonumber \\ That is, find \\\\ \vec{u} \\ .\\

###### Solution

By Proposition $\PageIndex{1}$, \\\\ \vec{u} \\ ^{2} = \vec{u} \bullet \vec{u}\\. Therefore, \\\\ \vec{u} \\ = \sqrt {\vec{u} \bullet \vec{u}}\\. First, compute \\\vec{u} \bullet \vec{u}\\.

This is given by \\\begin{aligned} \vec{u} \bullet \vec{u} &= (2)(2) + (1)(1) + (4)(4) + (2)(2) \\ &= 4 + 1 + 16 + 4 \\ &= 25\end{aligned}\\

Then, \\\begin{aligned} \\ \vec{u} \\ &= \sqrt {\vec{u} \bullet \vec{u}} \\ &= \sqrt{25} \\ &= 5\end{aligned}\\

You may wish to compare this to our previous definition of length, given in Definition 4.4.2.

The Cauchy Schwarz inequality is a fundamental inequality satisfied by the dot product. It is given in the following theorem.

##### Theorem \\\PageIndex{1}\\: Cauchy Schwarz Inequality

The dot product satisfies the inequality \\\left\vert \vec{u}\bullet \vec{v}\right\vert \leq \\ \vec{u}\\ \\ \vec{v}\\ \label{cauchy}\\ Furthermore equality is obtained if and only if one of \\\vec{u}\\ or \\\vec{v}\\ is a scalar multiple of the other.

Proof

First note that if \\\vec{v}=\vec{0}\\ both sides of \\\eqref{cauchy}\\ equal zero and so the inequality holds in this case. Therefore, it will be assumed in what follows that \\\vec{v}\neq \vec{0}\\.

Define a function of \\t\in \mathbb{R}\\ by \\f\left( t\right) =\left( \vec{u}+t\vec{v}\right) \bullet \left( \vec{u}+ t\vec{v}\right)\nonumber \\ Then by Proposition $\PageIndex{1}$, \\f\left( t\right) \geq 0\\ for all \\t\in \mathbb{R}\\. Also from Proposition $\PageIndex{1}$ \\\begin{aligned} f\left( t\right) &=\vec{u}\bullet \left( \vec{u}+t\vec{v}\right) + t\vec{v}\bullet \left( \vec{u}+t\vec{v}\right) \\ &=\vec{u}\bullet \vec{u}+t\left( \vec{u}\bullet \vec{v}\right) + t \vec{v}\bullet \vec{u}+ t^{2}\vec{v}\bullet \vec{v} \\ &=\\ \vec{u}\\ ^{2}+2t\left( \vec{u}\bullet \vec{v}\right) +\\ \vec{v}\\ ^{2}t^{2}\end{aligned}\\

Now this means the graph of \\y=f\left( t\right)\\ is a parabola which opens up and either its vertex touches the \\t\\ axis or else the entire graph is above the \\t\\ axis. In the first case, there exists some \\t\\ where \\f\left( t\right) =0\\ and this requires \\\vec{u}+t\vec{v}=\vec{0}\\ so one vector is a multiple of the other. Then clearly equality holds in \\\eqref{cauchy}\\. In the case where \\\vec{v}\\ is not a multiple of \\\vec{u}\\, it follows \\f\left( t\right) \>0\\ for all \\t\\ which says \\f\left( t\right)\\ has no real zeros and so from the quadratic formula, \\\left( 2\left( \vec{u}\bullet \vec{v}\right) \right) ^{2}-4\\ \vec{u} \\ ^{2}\\ \vec{v}\\ ^{2}\<0\nonumber \\ which is equivalent to \\\left\vert \vec{u}\bullet \vec{v} \right\vert \<\\ \vec{u}\\ \\ \vec{v}\\\\.

Notice that this proof was based only on the properties of the dot product listed in Proposition $\PageIndex{1}$. This means that whenever an operation satisfies these properties, the Cauchy Schwarz inequality holds. There are many other instances of these properties besides vectors in \\\mathbb{R}^{n}\\.

The Cauchy Schwarz inequality provides another proof of the triangle inequality for distances in \\\mathbb{R}^{n}\\.

##### Theorem \\\PageIndex{2}\\: Triangle Inequality

For \\\vec{u},\vec{v}\in \mathbb{R}^{n}\\ \\\\ \vec{u}+\vec{v}\\ \leq \\ \vec{u}\\ +\\ \vec{v} \\ \label{triangleineq1}\\ and equality holds if and only if one of the vectors is a non-negative scalar multiple of the other.

Also \\\\ \\ \vec{u}\\ -\\ \vec{v}\\ \\ \leq \\ \vec{u}-\vec{v}\\ \label{triangleineq2}\\

Proof

By properties of the dot product and the Cauchy Schwarz inequality, \\\begin{aligned} \\ \vec{u}+\vec{v}\\ ^{2} &= \left( \vec{u}+\vec{v}\right) \bullet \left( \vec{u}+\vec{v}\right) \\ & =\left( \vec{u}\bullet \vec{u}\right) +\left( \vec{u}\bullet \vec{v}\right) +\left(\vec{v}\bullet \vec{u}\right) +\left( \vec{v}\bullet \vec{v}\right) \\ &=\\ \vec{u}\\ ^{2}+2\left( \vec{u}\bullet \vec{v}\right)+\\ \vec{v}\\ ^{2} \\ &\leq \\ \vec{u}\\ ^{2}+2\left\vert \vec{u}\bullet \vec{v}\right\vert +\\ \vec{v}\\ ^{2} \\ &\leq \\ \vec{u}\\ ^{2}+2\\ \vec{u}\\ \\ \vec{v}\\ +\\ \vec{v}\\ ^{2} =\left( \\ \vec{u}\\ +\\ \vec{v}\\\right) ^{2}\end{aligned}\\ Hence, \\\\ \vec{u}+\vec{v}\\ ^{2} \leq \left( \\ \vec{u}\\ +\\ \vec{v}\\ \right) ^{2}\nonumber \\ Taking square roots of both sides you obtain \\\eqref{triangleineq1}\\.

It remains to consider when equality occurs. Suppose \\\vec{u} = \vec{0}\\. Then, \\\vec{u} = 0 \vec{v}\\ and the claim about when equality occurs is verified. The same argument holds if \\\vec{v} = \vec{0}\\. Therefore, it can be assumed both vectors are nonzero. To get equality in \\\eqref{triangleineq1}\\ above, Theorem $\PageIndex{1}$ implies one of the vectors must be a multiple of the other. Say \\\vec{v}= k \vec{u}\\. If \\k \<0\\ then equality cannot occur in \\\eqref{triangleineq1}\\ because in this case \\\vec{u}\bullet \vec{v} =k \\ \vec{u}\\ ^{2}\<0\<\left\| k \right\| \\ \vec{u}\\ ^{2}=\left\| \vec{u}\bullet \vec{v}\right\|\nonumber \\ Therefore, \\k \geq 0.\\

To get the other form of the triangle inequality write \\\vec{u}=\vec{u}-\vec{v}+\vec{v}\nonumber \\ so \\\begin{aligned} \\ \vec{u}\\ & =\\ \vec{u}-\vec{v}+\vec{v}\\ \\ & \leq \\ \vec{u}-\vec{v}\\ +\\ \vec{v}\\ \end{aligned}\\ Therefore, \\\\ \vec{u}\\ -\\ \vec{v}\\ \leq \\ \vec{u}-\vec{v} \\ \label{triangleineq3}\\ Similarly, \\\\ \vec{v}\\ -\\ \vec{u}\\ \leq \\ \vec{v}-\vec{u} \\ =\\ \vec{u}-\vec{v}\\ \label{triangleineq4}\\ It follows from \\\eqref{triangleineq3}\\ and \\\eqref{triangleineq4}\\ that \\\eqref{triangleineq2}\\ holds. This is because \\\left\| \\ \vec{u}\\ -\\ \vec{v}\\ \right\|\\ equals the left side of either \\\eqref{triangleineq3}\\ or \\\eqref{triangleineq4}\\ and either way, \\\left\| \\ \vec{u}\\ -\\ \vec{v}\\ \right\| \leq \\ \vec{u}-\vec{v}\\\\.

The Geometric Significance of the Dot Product

Given two vectors, \\\vec{u}\\ and \\\vec{v}\\, the included angle is the angle between these two vectors which is given by \\\theta\\ such that \\0 \leq \theta \leq \pi\\. The dot product can be used to determine the included angle between two vectors. Consider the following picture where \\\theta\\ gives the included angle.

Figure $\PageIndex{1}$

##### Proposition \\\PageIndex{2}\\: The Dot Product and the Included Angle

Let \\\vec{u}\\ and \\\vec{v}\\ be two vectors in \\\mathbb{R}^n\\, and let \\\theta\\ be the included angle. Then the following equation holds. \\\vec{u}\bullet \vec{v}=\\ \vec{u}\\ \\ \vec{v} \\ \cos \theta\nonumber \\

In words, the dot product of two vectors equals the product of the magnitude (or length) of the two vectors multiplied by the cosine of the included angle. Note this gives a geometric description of the dot product which does not depend explicitly on the coordinates of the vectors.

Consider the following example.

##### Example \\\PageIndex{3}\\: Find the Angle Between Two Vectors

Find the angle between the vectors given by \\\vec{u} = \left$$ \begin{array}{r} 2 \\ 1 \\ -1 \end{array} \right$$, \vec{v} = \left$$ \begin{array}{r} 3 \\ 4 \\ 1 \end{array} \right$$\nonumber \\

###### Solution

By Proposition $\PageIndex{2}$, \\\vec{u}\bullet \vec{v}=\\ \vec{u}\\ \\ \vec{v} \\ \cos \theta\nonumber \\ Hence, \\\cos \theta =\frac{\vec{u}\bullet \vec{v}}{\\ \vec{u}\\ \\ \vec{v} \\}\nonumber \\

First, we can compute \\\vec{u}\bullet \vec{v}\\. By Definition $\PageIndex{1}$, this equals \\\vec{u}\bullet \vec{v} = (2)(3) + (1)(4)+(-1)(1) = 9\nonumber \\

Then, \\\begin{array}{c} \\ \vec{u} \\ = \sqrt{(2)(2)+(1)(1)+(1)(1)}=\sqrt{6}\\ \\ \vec{v} \\ = \sqrt{(3)(3)+(4)(4)+(1)(1)}=\sqrt{26} \end{array}\nonumber \\ Therefore, the cosine of the included angle equals \\\cos \theta =\frac{9}{\sqrt{26}\sqrt{6}}=0.7205766...\nonumber \\

With the cosine known, the angle can be determined by computing the inverse cosine of that angle, giving approximately \\\theta =0.76616\\ radians.

Another application of the geometric description of the dot product is in finding the angle between two lines. Typically one would assume that the lines intersect. In some situations, however, it may make sense to ask this question when the lines do not intersect, such as the angle between two object trajectories. In any case we understand it to mean the smallest angle between (any of) their direction vectors. The only subtlety here is that if \\\vec{u}\\ is a direction vector for a line, then so is any multiple \\k\vec{u}\\, and thus we will find complementary angles among all angles between direction vectors for two lines, and we simply take the smaller of the two.

##### Example \\\PageIndex{4}\\: Find the Angle Between Two Lines

Find the angle between the two lines \\L_1: \\ \left$$ \begin{array}{r} x \\ y \\ z \end{array} \right$$ = \left$$ \begin{array}{r} 1 \\ 2 \\ 0 \end{array} \right$$ +t\left$$ \begin{array}{r} -1 \\ 1 \\ 2 \end{array} \right$$\nonumber \\ and \\L_2: \\ \left$$ \begin{array}{r} x \\ y \\ z \end{array} \right$$ = \left$$ \begin{array}{r} 0 \\ 4 \\ -3 \end{array} \right$$ +s\left$$ \begin{array}{r} 2 \\ 1 \\ -1 \end{array} \right$$\nonumber \\

###### Solution

You can verify that these lines do not intersect, but as discussed above this does not matter and we simply find the smallest angle between any directions vectors for these lines.

To do so we first find the angle between the direction vectors given above: \\\vec{u}=\left$$ \begin{array}{r} -1 \\ 1 \\ 2 \end{array} \right$$,\\ \vec{v}=\left$$ \begin{array}{r} 2 \\ 1 \\ -1 \end{array} \right$$\nonumber \\

In order to find the angle, we solve the following equation for \\\theta\\ \\\vec{u}\bullet \vec{v}=\\ \vec{u}\\ \\ \vec{v} \\ \cos \theta\nonumber \\ to obtain \\\cos \theta = -\frac{1}{2}\\ and since we choose included angles between \\0\\ and \\\pi\\ we obtain \\\theta = \frac{2 \pi}{3}\\.

Now the angles between any two direction vectors for these lines will either be \\\frac{2 \pi}{3}\\ or its complement \\\phi = \pi - \frac{2 \pi}{3} = \frac{\pi}{3}\\. We choose the smaller angle, and therefore conclude that the angle between the two lines is \\\frac{\pi}{3}\\.

We can also use Proposition $\PageIndex{2}$ to compute the dot product of two vectors.

##### Example \\\PageIndex{5}\\: Using Geometric Description to Find a Dot Product

Let \\\vec{u},\vec{v}\\ be vectors with \\\\ \vec{u} \\ = 3\\ and \\\\ \vec{v} \\ = 4\\. Suppose the angle between \\\vec{u}\\ and \\\vec{v}\\ is \\\pi / 3\\. Find \\\vec{u}\bullet \vec{v}\\.

###### Solution

From the geometric description of the dot product in Proposition $\PageIndex{2}$ \\\vec{u}\bullet \vec{v}=(3)(4) \cos \left( \pi / 3\right) =3\times 4\times 1/2=6\nonumber \\

Two nonzero vectors are said to be perpendicular, sometimes also called orthogonal, if the included angle is \\\pi /2\\ radians (\\90^{\circ }).\\

Consider the following proposition.

##### Proposition \\\PageIndex{3}\\: Perpendicular Vectors

Let \\\vec{u}\\ and \\\vec{v}\\ be nonzero vectors in \\\mathbb{R}^n\\. Then, \\\vec{u}\\ and \\\vec{v}\\ are said to be perpendicular exactly when \\\vec{u} \bullet \vec{v} = 0\nonumber \\

Proof

This follows directly from Proposition $\PageIndex{2}$. First if the dot product of two nonzero vectors is equal to \\0\\, this tells us that \\\cos \theta =0\\ (this is where we need nonzero vectors). Thus \\\theta = \pi /2\\ and the vectors are perpendicular.

If on the other hand \\\vec{v}\\ is perpendicular to \\\vec{u}\\, then the included angle is \\\pi /2\\ radians. Hence \\\cos \theta =0\\ and \\\vec{u} \bullet \vec{v} = 0\\.

Consider the following example.

##### Example \\\PageIndex{6}\\: Determine if Two Vectors are Perpendicular

Determine whether the two vectors, \\\vec{u}= \left$$ \begin{array}{r} 2 \\ 1 \\ -1 \end{array} \right$$, \vec{v} = \left$$ \begin{array}{r} 1 \\ 3 \\ 5 \end{array} \right$$\nonumber \\ are perpendicular.

###### Solution

In order to determine if these two vectors are perpendicular, we compute the dot product. This is given by \\\vec{u} \bullet \vec{v} = (2)(1) + (1)(3) + (-1)(5) = 0\nonumber \\ Therefore, by Proposition $\PageIndex{3}$ these two vectors are perpendicular.

Projections

In some applications, we wish to write a vector as a sum of two related vectors. Through the concept of projections, we can find these two vectors. First, we explore an important theorem. The result of this theorem will provide our definition of a vector projection.

##### Theorem \\\PageIndex{3}\\: Vector Projections

Let \\\vec{v}\\ and \\\vec{u}\\ be nonzero vectors. Then there exist unique vectors \\\vec{v}\_{\|\|}\\ and \\\vec{v}\_{\bot }\\ such that \\\vec{v}=\vec{v}\_{\|\|}+\vec{v}\_{\bot } \label{projection}\\ where \\\vec{v}\_{\|\|}\\ is a scalar multiple of \\\vec{u}\\, and \\\vec{v}\_{\bot}\\ is perpendicular to \\\vec{u}\\.

Proof

Suppose \\\eqref{projection}\\ holds and \\\vec{v}\_{\|\|}= k \vec{u}\\. Taking the dot product of both sides of \\\eqref{projection}\\ with \\\vec{u}\\ and using \\\vec{v}\_{\bot }\bullet \vec{u}=0,\\ this yields \\\begin{array}{ll} \vec{v}\bullet \vec{u} & = ( \vec{v}\_{\|\|}+\vec{v}\_{\bot }) \bullet \vec{u} \\ & = k\vec{u} \bullet \vec{u} + \vec{v}\_{\bot} \bullet \vec{u} \\ & = k \\ \vec{u}\\ ^{2} \end{array}\nonumber \\ which requires \\k =\vec{v}\bullet \vec{u} / \\ \vec{u}\\ ^{2}.\\ Thus there can be no more than one vector \\\vec{v}\_{\|\|}\\. It follows \\\vec{v}\_{\bot }\\ must equal \\\vec{v}-\vec{v}\_{\|\|}.\\ This verifies there can be no more than one choice for both \\\vec{v}\_{\|\|}\\ and \\\vec{v}\_{\bot }\\ and proves their uniqueness.

Now let \\\vec{v}\_{\|\|} = \frac{\vec{v}\bullet \vec{u}}{\\ \vec{u}\\ ^{2}}\vec{u}\nonumber \\ and let \\\vec{v}\_{\bot }=\vec{v}-\vec{v}\_{\|\|}=\vec{v}-\frac{\vec{v}\bullet \vec{u}} {\\ \vec{u}\\ ^{2}}\vec{u}\nonumber \\ Then \\\vec{v}\_{\|\|}= k\vec{u}\\ where \\k =\frac{\vec{v}\bullet \vec{u}}{\\ \vec{u}\\ ^{2}}\\. It only remains to verify \\\vec{v}\_{\bot }\bullet \vec{u}=0.\\ But \\\begin{aligned} \vec{v}\_{\bot }\bullet \vec{u} &= \vec{v}\bullet \vec{u}-\frac{\vec{v}\bullet \vec{u}}{\\ \vec{u}\\ ^{2}}\vec{u}\bullet \vec{u} \\ &= \vec{v}\bullet\vec{u}-\vec{v}\bullet \vec{u}\\ &= 0 \end{aligned}\\

The vector \\\vec{v}\_{\|\|}\\ in Theorem $\PageIndex{3}$ is called the projection of \\\vec{v}\\ onto \\\vec{u}\\ and is denoted by \\\vec{v}\_{\|\|} = \mathrm{proj}\_{\vec{u}}\left( \vec{v}\right)\nonumber \\

We now make a formal definition of the vector projection.

##### Definition \\\PageIndex{2}\\: Vector Projection

Let \\\vec{u}\\ and \\\vec{v}\\ be vectors. Then, the projection of \\\vec{v}\\ onto \\\vec{u}\\ is given by \\\mathrm{proj}\_{\vec{u}}\left( \vec{v}\right) =\left( \frac{\vec{v}\bullet \vec{u}}{\vec{u}\bullet \vec{u}}\right) \vec{u} = \frac{\vec{v}\bullet \vec{u}}{\\ \vec{u}\\ ^{2}}\vec{u}\nonumber \\

Consider the following example of a projection.

##### Example \\\PageIndex{7}\\: Find the Projection of One Vector Onto Another

Find \\\mathrm{proj}\_{\vec{u}}\left( \vec{v}\right)\\ if \\\vec{u}= \left$$ \begin{array}{r} 2 \\ 3 \\ -4 \end{array} \right$$, \vec{v}= \left$$ \begin{array}{r} 1 \\ -2 \\ 1 \end{array} \right$$\nonumber \\

###### Solution

We can use the formula provided in Definition $\PageIndex{2}$ to find \\\mathrm{proj}\_{\vec{u}}\left( \vec{v}\right)\\. First, compute \\\vec{v} \bullet \vec{u}\\. This is given by \\\begin{aligned} \left$$ \begin{array}{r} 1 \\ -2 \\ 1 \end{array} \right$$ \bullet \left$$ \begin{array}{r} 2 \\ 3 \\ -4 \end{array} \right$$ &= (2)(1) + (3)(-2) + (-4)(1) \\ &= 2 - 6 - 4 \\ &= -8\end{aligned}\\ Similarly, \\\vec{u} \bullet \vec{u}\\ is given by \\\begin{aligned} \left$$ \begin{array}{r} 2 \\ 3 \\ -4 \end{array} \right$$ \bullet \left$$ \begin{array}{r} 2 \\ 3 \\ -4 \end{array} \right$$ &= (2)(2) + (3)(3) + (-4)(-4) \\ &= 4 + 9 + 16 \\ &= 29\end{aligned}\\

Therefore, the projection is equal to \\\begin{aligned} \mathrm{proj}\_{\vec{u}}\left( \vec{v}\right) &=-\frac{8}{29} \left$$ \begin{array}{r} 2 \\ 3 \\ -4 \end{array} \right$$ \\ &= \left$$ \begin{array}{r} - \frac{16}{29} \\ - \frac{24}{29} \\ \frac{32}{29} \end{array} \right$$\end{aligned}\\

We will conclude this section with an important application of projections. Suppose a line \\L\\ and a point \\P\\ are given such that \\P\\ is not contained in \\L\\. Through the use of projections, we can determine the shortest distance from \\P\\ to \\L\\.

##### Example \\\PageIndex{8}\\: Shortest Distance from a Point to a Line

Let \\P = (1,3,5)\\ be a point in \\\mathbb{R}^3\\, and let \\L\\ be the line which goes through point \\P_0 = (0,4,-2)\\ with direction vector \\\vec{d} = \left$$ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right$$\\. Find the shortest distance from \\P\\ to the line \\L\\, and find the point \\Q\\ on \\L\\ that is closest to \\P\\.

###### Solution

In order to determine the shortest distance from \\P\\ to \\L\\, we will first find the vector \\\overrightarrow{P_0P}\\ and then find the projection of this vector onto \\L\\. The vector \\\overrightarrow{P_0P}\\ is given by \\\left$$ \begin{array}{r} 1 \\ 3 \\ 5 \end{array} \right$$ - \left$$ \begin{array}{r} 0 \\ 4 \\ -2 \end{array} \right$$ = \left$$ \begin{array}{r} 1 \\ -1 \\ 7 \end{array} \right$$\nonumber \\

Then, if \\Q\\ is the point on \\L\\ closest to \\P\\, it follows that \\\begin{aligned} \overrightarrow{P_0Q} &= \mathrm{proj}\_{\vec{d}}\overrightarrow{P_0P} \\ &= \left( \frac{ \overrightarrow{P_0P}\bullet \vec{d}}{\\\vec{d}\\^2}\right) \vec{d} \\ &= \frac{15}{9} \left$$ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right$$ \\ &= \frac{5}{3} \left$$ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right$$\end{aligned}\\

Now, the distance from \\P\\ to \\L\\ is given by \\\\ \overrightarrow{QP} \\ = \\ \overrightarrow{P_0P} - \overrightarrow{P_0Q}\\ = \sqrt{26}\nonumber \\

The point \\Q\\ is found by adding the vector \\\overrightarrow{P_0Q}\\ to the position vector \\\overrightarrow{0P_0}\\ for \\P_0\\ as follows \\\begin{aligned} \left$$ \begin{array}{r} 0 \\ 4 \\ -2 \end{array} \right$$ + \frac{5}{3} \left$$ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right$$ &= \left$$ \begin{array}{r} \frac{10}{3} \\ \frac{17}{3} \\ \frac{4}{3} \end{array} \right$$\end{aligned}\\

Therefore, \\Q = (\frac{10}{3}, \frac{17}{3}, \frac{4}{3})\\.

---

4_08_3A_Planes_in_R

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.08%3A_Planes_in_R

Skip to main content

Outcomes

1. Find the vector and scalar equations of a plane.

Much like the above discussion with lines, vectors can be used to determine planes in \\\mathbb{R}^n\\. Given a vector \\\vec{n}\\ in \\\mathbb{R}^n\\ and a point \\P_0\\, it is possible to find a unique plane which contains \\P_0\\ and is perpendicular to the given vector.

Definition \\\PageIndex{1}\\: Normal Vector

Let \\\vec{n}\\ be a nonzero vector in \\\mathbb{R}^n\\. Then \\\vec{n}\\ is called a normal vector to a plane if and only if \\\vec{n} \bullet \vec{v} = 0\nonumber \\ for every vector \\\vec{v}\\ in the plane.

In other words, we say that \\\vec{n}\\ is orthogonal (perpendicular) to every vector in the plane.

Consider now a plane with normal vector given by \\\vec{n}\\, and containing a point \\P_0\\. Notice that this plane is unique. If \\P\\ is an arbitrary point on this plane, then by definition the normal vector is orthogonal to the vector between \\P_0\\ and \\P\\. Letting \\\overrightarrow{0P}\\ and \\\overrightarrow{0P_0}\\ be the position vectors of points \\P\\ and \\P_0\\ respectively, it follows that \\\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber \\ or \\\vec{n} \bullet \overrightarrow{P_0P} = 0\nonumber \\

The first of these equations gives the vector equation of the plane.

Definition \\\PageIndex{2}\\: Vector Equation of a Plane

Let \\\vec{n}\\ be the normal vector for a plane which contains a point \\P_0\\. If \\P\\ is an arbitrary point on this plane, then the vector equation of the plane is given by \\\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber \\

Notice that this equation can be used to determine if a point \\P\\ is contained in a certain plane.

Example \\\PageIndex{1}\\: A Point in a Plane

Let \\\vec{n} = \left$$ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right$$\\ be the normal vector for a plane which contains the point \\P_0 = \left( 2, 1, 4 \right)\\. Determine if the point \\P = \left( 5, 4, 1 \right)\\ is contained in this plane.

###### Solution

By Definition $\PageIndex{2}$, \\P\\ is a point in the plane if it satisfies the equation \\\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber \\

Given the above \\\vec{n}\\, \\P_0\\, and \\P\\, this equation becomes \\\begin{aligned} \left$$ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right$$ \bullet \left( \left$$ \begin{array}{r} 5 \\ 4 \\ 1 \end{array} \right$$ - \left$$ \begin{array}{r} 2 \\ 1 \\ 4 \end{array} \right$$ \right) &= \left$$ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right$$ \bullet \left( \left$$ \begin{array}{r} 3 \\ 3 \\ -3 \end{array} \right$$ \right) \\ &= 3 + 6 - 9 = 0\end{aligned}\\

Therefore \\P = ( 5, 4, 1)\\ is contained in the plane.

Suppose \\\vec{n} = \left$$ \begin{array}{c} a \\ b \\ c \end{array} \right$$\\, \\P = \left( x,y,z\right)\\ and \\P_0 = (x_0, y_0, z_0 )\\.

Then \\\begin{aligned} \vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) &= 0 \\ \left$$ \begin{array}{c} a \\ b \\ c \end{array} \right$$ \bullet \left( \left$$ \begin{array}{c} x \\ y \\ z \end{array} \right$$ - \left$$ \begin{array}{c} x_0 \\ y_0 \\ z_0 \end{array} \right$$ \right) &= 0 \\ \left$$ \begin{array}{c} a \\ b \\ c \end{array} \right$$ \bullet \left$$ \begin{array}{c} x - x_0 \\ y - y_0 \\ z - z_0 \end{array} \right$$ &= 0 \\ a(x - x_0) + b (y - y_0) + c (z-z_0) &= 0 \end{aligned}\\

We can also write this equation as \\ax + by + cz = ax_0 + by_0 + cz_0\nonumber \\

Notice that since \\P_0\\ is given, \\ax_0+by_0+cz_0\\ is a known scalar, which we can call \\d\\. This equation becomes \\ax + by + cz = d\nonumber \\

Definition \\\PageIndex{3}\\: Scalar Equation of a Plane

Let \\\vec{n} = \left$$ \begin{array}{c} a \\ b \\ c \end{array} \right$$\\ be the normal vector for a plane which contains the point \\P_0 = (x_0, y_0, z_0)\\.Then if \\P=(x,y,z)\\ is an arbitrary point on the plane, the scalar equation of the plane is given by \\ax + by + cz = d\nonumber \\ where \\a,b,c,d \in \mathbb{R}\\ and \\d = ax_0 + by_0 + cz_0\\.

Consider the following equation.

Example \\\PageIndex{2}\\: Finding the Equation of a Plane

Find an equation of the plane containing \\P_0 = (3, -2, 5)\\ and orthogonal to \\\vec{n} = \left$$ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right$$\\.

###### Solution

The above vector \\\vec{n}\\ is the normal vector for this plane. Using Definition $\PageIndex{2}$, we can determine the vector equation for this plane. \\\begin{aligned} \vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) &= 0 \\ \left$$ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right$$ \bullet \left(\left$$ \begin{array}{c} x \\ y \\ z \end{array} \right$$ - \left$$ \begin{array}{r} 3 \\ -2 \\ 5 \end{array} \right$$ \right) &= 0 \\ \left$$ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right$$ \bullet \left$$ \begin{array}{c} x - 3 \\ y + 2 \\ z - 5 \end{array} \right$$ &= 0 \end{aligned}\\

Using Definition $\PageIndex{3}$, we can determine the scalar equation of the plane. \\-2x + 4y + 1z = -2(3) + 4(-2) + 1(5) = -9\nonumber \\

Hence, the vector equation of the plane is \\\left$$ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right$$ \bullet \left$$ \begin{array}{c} x - 3 \\ y + 2 \\ z - 5 \end{array} \right$$ = 0\nonumber \\ and the scalar equation is \\-2x + 4y + 1z = -9\nonumber \\

Suppose a point \\P\\ is not contained in a given plane. We are then interested in the shortest distance from that point \\P\\ to the given plane. Consider the following example.

Example \\\PageIndex{3}\\: Shortest Distance From a Point to a Plane

Find the shortest distance from the point \\P = (3,2,3)\\ to the plane given by

\\2x + y + 2z = 2\\, and find the point \\Q\\ on the plane that is closest to \\P\\.

###### Solution

Pick an arbitrary point \\P_0\\ on the plane. Then, it follows that \\\overrightarrow{QP} = proj\_{\vec{n}}\overrightarrow{P_0P}\nonumber \\ and \\\\ \overrightarrow{QP} \\\\ is the shortest distance from \\P\\ to the plane. Further, the vector \\\overrightarrow{0Q} = \overrightarrow{0P} - \overrightarrow{QP}\\ gives the necessary point \\Q\\.

From the above scalar equation, we have that \\\vec{n} = \left$$ \begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right$$\\. Now, choose \\P_0 = (1, 0, 0)\\ so that \\\vec{n} \bullet \overrightarrow{0P} = 2 = d\\. Then, \\\overrightarrow{P_0P} = \left$$ \begin{array}{c} 3 \\ 2 \\ 3 \end{array} \right$$ - \left$$ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right$$ = \left$$ \begin{array}{c} 2 \\ 2 \\ 3 \end{array} \right$$\\.

Next, compute \\\overrightarrow{QP} = proj\_{\vec{n}}\overrightarrow{P_0P}\\. \\\begin{aligned} \overrightarrow{QP} &= proj\_{\vec{n}}\overrightarrow{P_0P} \\ &= \left( \frac{ \overrightarrow{P_0P} \bullet \vec{n}}{\\ \vec{n} \\ ^2}\right)\vec{n} \\ &= \frac{12}{9} \left$$ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right$$ \\ &= \frac{4}{3} \left$$ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right$$ \end{aligned}\\

Then, \\\\ \overrightarrow{QP} \\ = 4\\ so the shortest distance from \\P\\ to the plane is \\4\\.

Next, to find the point \\Q\\ on the plane which is closest to \\P\\ we have \\\begin{aligned} \overrightarrow{0Q} &= \overrightarrow{0P} - \overrightarrow{QP} \\ &= \left$$ \begin{array}{r} 3 \\ 2 \\ 3 \end{array} \right$$ - \frac{4}{3} \left$$ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right$$ \\ &= \frac{1}{3} \left$$ \begin{array}{r} 1 \\ 2 \\ 1 \end{array} \right$$\end{aligned}\\

Therefore, \\Q = (\frac{1}{3}, \frac{2}{3}, \frac{1}{3} )\\.

---

4_09_3A_The_Cross_Product

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.09%3A_The_Cross_Product

Skip to main content

Outcomes

1. Compute the cross product and box product of vectors in \\\mathbb{R}^3\\.

Recall that the dot product is one of two important products for vectors. The second type of product for vectors is called the cross product. It is important to note that the cross product is only defined in \\\mathbb{R}^{3}.\\ First we discuss the geometric meaning and then a description in terms of coordinates is given, both of which are important. The geometric description is essential in order to understand the applications to physics and geometry while the coordinate description is necessary to compute the cross product.

Consider the following definition.

Definition \\\PageIndex{1}\\: Right Hand System of Vectors

Three vectors, \\\vec{u},\vec{v},\vec{w}\\ form a right hand system if when you extend the fingers of your right hand along the direction of vector \\\vec{u}\\ and close them in the direction of \\\vec{v}\\, the thumb points roughly in the direction of \\\vec{w}\\.

For an example of a right handed system of vectors, see the following picture.

Figure $\PageIndex{1}$

In this picture the vector \\\vec{w}\\ points upwards from the plane determined by the other two vectors. Point the fingers of your right hand along \\\vec{u}\\, and close them in the direction of \\\vec{v}\\. Notice that if you extend the thumb on your right hand, it points in the direction of \\\vec{w}\\.

You should consider how a right hand system would differ from a left hand system. Try using your left hand and you will see that the vector \\\vec{w}\\ would need to point in the opposite direction.

Notice that the special vectors, \\\vec{i},\vec{j},\vec{k}\\ will always form a right handed system. If you extend the fingers of your right hand along \\\vec{i}\\ and close them in the direction \\\vec{j}\\, the thumb points in the direction of \\\vec{k}\\.

Figure $\PageIndex{2}$

The following is the geometric description of the cross product. Recall that the dot product of two vectors results in a scalar. In contrast, the cross product results in a vector, as the product gives a direction as well as magnitude.

Definition \\\PageIndex{2}\\: Geometric Definition of Cross Product

Let \\\vec{u}\\ and \\\vec{v}\\ be two vectors in \\\mathbb{R}^{3}.\\ Then the cross product, written \\\vec{u}\times \vec{v}\\, is defined by the following two rules.

1. Its length is \\\\ \vec{u}\times \vec{v}\\ =\\ \vec{u}\\ \\ \vec{v}\\ \sin \theta, \nonumber \\ where \\\theta\\ is the included angle between \\\vec{u}\\ and \\\vec{v}\\.

2. It is perpendicular to both \\\vec{u}\\ and \\\vec{v}\\, that is \\\left( \vec{u}\times \vec{v} \right) \cdot \vec{u}=0, \\\\\left( \vec{u}\times \vec{v} \right) \cdot \vec{v}=0, \nonumber\\ and \\\vec{u},\vec{v},\vec{u}\times \vec{v} \nonumber\\ form a right hand system.

The cross product of the special vectors \\\vec{i}, \vec{j}, \vec{k}\\ is as follows. \\\begin{array}{cc} \vec{i}\times \vec{j}=\vec{k} & \vec{j}\times \vec{i}=-\vec{k} \\ \vec{k}\times \vec{i}=\vec{j} & \vec{i}\times \vec{k}=-\vec{j} \\ \vec{j}\times \vec{k}=\vec{i} & \vec{k}\times \vec{j}=-\vec{i} \end{array}\nonumber \\ With this information, the following gives the coordinate description of the cross product.

Recall that the vector \\\vec{u}= \left$$ \begin{array}{ccc} u_1 & u_2 & u_3 \end{array} \right$$^T\\ can be written in terms of \\\vec{i}, \vec{j}, \vec{k}\\ as \\\vec{u}=u\_{1}\vec{i}+u\_{2}\vec{j}+u\_{3}\vec{k}\\.

Theorem \\\PageIndex{1}\\: Coordinate Description of Cross Product

Let \\\vec{u}=u\_{1}\vec{i}+u\_{2}\vec{j}+u\_{3}\vec{k}\\ and \\\vec{v}=v\_{1}\vec{i}+v\_{2}\vec{j}+v\_{3}\vec{k}\\ be two vectors. Then

\\\begin{array}{c} \vec{u}\times \vec{v} =\left( u\_{2}v\_{3}-u\_{3}v\_{2}\right) \vec{i}-\left( u\_{1}v\_{3} - u\_{3}v\_{1}\right) \vec{j}+ \left( u\_{1}v\_{2}-u\_{2}v\_{1}\right) \vec{k} \label{crossprod1} \end{array}\\

Writing \\\vec{u} \times \vec{v}\\ in the usual way, it is given by

\\\vec{u} \times \vec{v} = \left$$ \begin{array}{r} u\_{2}v\_{3}-u\_{3}v\_{2} \\ -(u\_{1}v\_{3}-u\_{3}v\_{1}) \\ u\_{1}v\_{2}-u\_{2}v\_{1} \end{array} \right$$\nonumber \\

We now prove this proposition.

Proof

From the above table and the properties of the cross product listed, \\\begin{aligned} \vec{u} \times \vec{v} &= \left( u\_{1}\vec{i}+u\_{2}\vec{j}+u\_{3}\vec{k}\right) \times \left( v\_{1}\vec{i}+v\_{2}\vec{j}+v\_{3}\vec{k}\right) \\ &= u\_{1}v\_{2}\vec{i}\times \vec{j}+u\_{1}v\_{3}\vec{i}\times \vec{k}+u\_{2}v\_{1}\vec{j}\times \vec{i}+ u\_{2}v\_{3}\vec{j}\times \vec{k}+ +u\_{3}v\_{1}\vec{k}\times \vec{i}+u\_{3}v\_{2}\vec{k}\times \vec{j} \\ &=u\_{1}v\_{2}\vec{k}-u\_{1}v\_{3}\vec{j}-u\_{2}v\_{1}\vec{k}+u\_{2}v\_{3} \vec{i}+u\_{3}v\_{1}\vec{j}-u\_{3}v\_{2}\vec{i} \\ &=\left( u\_{2}v\_{3}-u\_{3}v\_{2}\right) \vec{i}+\left( u\_{3}v\_{1}-u\_{1}v\_{3}\right) \vec{j}+\left( u\_{1}v\_{2}-u\_{2}v\_{1}\right) \vec{k} \end{aligned}\\ \\\label{crossprod2}\\

There is another version of \\\eqref{crossprod1}\\ which may be easier to remember. We can express the cross product as the determinant of a matrix, as follows.

\\\vec{u}\times \vec{v} = \left\vert \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ u\_{1} & u\_{2} & u\_{3} \\ v\_{1} & v\_{2} & v\_{3} \end{array} \right\vert \label{crossprod3}\\ Expanding the determinant along the top row yields \\\vec{i}\left( -1\right) ^{1+1}\left\vert \begin{array}{cc} u\_{2} & u\_{3} \\ v\_{2} & v\_{3} \end{array} \right\vert +\vec{j}\left( -1\right) ^{2+1}\left\vert \begin{array}{cc} u\_{1} & u\_{3} \\ v\_{1} & v\_{3} \end{array} \right\vert +\vec{k}\left( -1\right) ^{3+1}\left\vert \begin{array}{cc} u\_{1} & u\_{2} \\ v\_{1} & v\_{2} \end{array} \right\vert\nonumber \\\\

\\=\vec{i}\left\vert \begin{array}{cc} u\_{2} & u\_{3} \\ v\_{2} & v\_{3} \end{array} \right\vert -\vec{j}\left\vert \begin{array}{cc} u\_{1} & u\_{3} \\ v\_{1} & v\_{3} \end{array} \right\vert +\vec{k}\left\vert \begin{array}{cc} u\_{1} & u\_{2} \\ v\_{1} & v\_{2} \end{array} \right\vert\nonumber \\

Expanding these determinants leads to \\\left( u\_{2}v\_{3}-u\_{3}v\_{2}\right) \vec{i}-\left( u\_{1}v\_{3}-u\_{3}v\_{1}\right) \vec{j}+\left( u\_{1}v\_{2}-u\_{2}v\_{1}\right) \vec{k} \nonumber \\ which is the same as \\\eqref{crossprod2}\\.

The cross product satisfies the following properties.

Proposition \\\PageIndex{1}\\: Properties of the Cross Product

Let \\\vec{u}, \vec{v}, \vec{w}\\ be vectors in \\\mathbb{R}^3\\, and \\k\\ a scalar. Then, the following properties of the cross product hold.

1. \\\vec{u}\times \vec{v}= -\left( \vec{v}\times \vec{u}\right), \mbox{and} \\ \vec{u}\times \vec{u}=\vec{0}\\

2. \\\left( k \vec{u}\right)\times \vec{v}= k \left( \vec{u}\times \vec{v}\right) =\vec{u}\times \left( k \vec{v}\right)\\

3. \\\vec{u}\times \left( \vec{v}+\vec{w}\right) =\vec{u}\times \vec{v}+\vec{u}\times \vec{w}\\

4. \\\left( \vec{v}+\vec{w}\right) \times \vec{u}=\vec{v} \times \vec{u}+\vec{w}\times \vec{u}\\

Proof

Formula \\1.\\ follows immediately from the definition. The vectors \\\vec{u}\times \vec{v}\\ and \\\vec{v}\times \vec{u}\\ have the same magnitude, \\\left\vert \vec{u}\right\vert \left\vert \vec{v}\right\vert \sin \theta ,\\ and an application of the right hand rule shows they have opposite direction.

Formula \\2.\\ is proven as follows. If \\k\\ is a non-negative scalar, the direction of \\\left( k \vec{u}\right) \times \vec{v}\\ is the same as the direction of \\\vec{u}\times \vec{v}, k \left( \vec{u}\times \vec{v}\right)\\ and \\\vec{u}\times \left( k \vec{v}\right)\\. The magnitude is \\k\\ times the magnitude of \\\vec{u}\times \vec{v}\\ which is the same as the magnitude of \\k \left( \vec{u}\times \vec{v}\right)\\ and \\\vec{u}\times \left( k \vec{v}\right) .\\ Using this yields equality in \\2\\. In the case where \\k \<0,\\ everything works the same way except the vectors are all pointing in the opposite direction and you must multiply by \\\left\vert k \right\vert\\ when comparing their magnitudes.

The distributive laws, \\3.\\ and \\4.\\, are much harder to establish. For now, it suffices to notice that if we know that \\3.\\ is true, \\4.\\ follows. Thus, assuming \\3.\\, and using \\1.\\, \\\begin{aligned} \left( \vec{v}+\vec{w}\right) \times \vec{u}& =-\vec{u}\times \left( \vec{v}+\vec{w}\right) \\ & =-\left( \vec{u}\times \vec{v}+\vec{u}\times \vec{w}\right) \\ & =\vec{v}\times \vec{u}+\vec{w}\times \vec{u}\end{aligned}\\

We will now look at an example of how to compute a cross product.

Example \\\PageIndex{1}\\: Find a Cross Product

Find \\\vec{u} \times \vec{v}\\ for the following vectors

\\\vec{u} = \left$$ \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right$$, \vec{v} = \left$$ \begin{array}{r} 3 \\ -2 \\ 1 \end{array} \right$$\nonumber \\

###### Solution

Note that we can write \\\vec{u}, \vec{v}\\ in terms of the special vectors \\\vec{i}, \vec{j}, \vec{k}\\ as

\\\begin{array}{c} \vec{u} = \vec{i}-\vec{j}+2\vec{k} \\ \vec{v} = 3\vec{i}-2\vec{j}+\vec{k} \end{array}\nonumber \\

We will use the equation given by \\\eqref{crossprod3}\\ to compute the cross product.

\\\vec{u} \times \vec{v} = \left\vert \begin{array}{rrr} \vec{i} & \vec{j} & \vec{k} \\ 1 & -1 & 2 \\ 3 & -2 & 1 \end{array} \right\vert =\left\vert \begin{array}{rr} -1 & 2 \\ -2 & 1 \end{array} \right\vert \vec{i}-\left\vert \begin{array}{rr} 1 & 2 \\ 3 & 1 \end{array} \right\vert \vec{j}+\left\vert \begin{array}{rr} 1 & -1 \\ 3 & -2 \end{array} \right\vert \vec{k}=3\vec{i}+5\vec{j}+\vec{k}\nonumber \\

We can write this result in the usual way, as \\\vec{u} \times \vec{v} = \left$$ \begin{array}{r} 3 \\ 5 \\ 1 \end{array} \right$$\nonumber \\

An important geometrical application of the cross product is as follows. The size of the cross product, \\\\ \vec{u}\times \vec{v}\\\\, is the area of the parallelogram determined by \\\vec{u}\\ and \\\vec{v}\\, as shown in the following picture.

Figure $\PageIndex{3}$

We examine this concept in the following example.

Example \\\PageIndex{2}\\: Area of a Parallelogram

Find the area of the parallelogram determined by the vectors \\\vec{u}\\ and \\\vec{v}\\ given by

\\\vec{u} = \left$$ \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right$$, \vec{v} = \left$$ \begin{array}{r} 3 \\ -2 \\ 1 \end{array} \right$$\nonumber \\

###### Solution

Notice that these vectors are the same as the ones given in Example $\PageIndex{1}$. Recall from the geometric description of the cross product, that the area of the parallelogram is simply the magnitude of \\\vec{u} \times \vec{v}\\. From Example $\PageIndex{1}$, \\\vec{u} \times \vec{v} = 3\vec{i}+5\vec{j}+\vec{k}\\. We can also write this as

\\\vec{u} \times \vec{v} = \left$$ \begin{array}{r} 3 \\ 5 \\ 1 \end{array} \right$$\nonumber \\

Thus the area of the parallelogram is

\\\\ \vec{u} \times \vec{v} \\ = \sqrt{(3)(3) + (5)(5) + (1)(1)} = \sqrt{9+25+1}=\sqrt{35}\nonumber \\

We can also use this concept to find the area of a triangle. Consider the following example.

Example \\\PageIndex{3}\\: Area of Triangle

Find the area of the triangle determined by the points \\\left(1, 2, 3 \right) , \left( 0,2,5\right), \left( 5,1, 2 \right)\\

###### Solution

This triangle is obtained by connecting the three points with lines. Picking \\\left( 1,2,3\right)\\ as a starting point, there are two displacement vectors, \\\left$$ \begin{array}{rrr} -1 & 0 & 2 \end{array} \right$$^T\\ and \\\left$$ \begin{array}{rrr} 4 & -1 & -1 \end{array} \right$$^T\\. Notice that if we add either of these vectors to the position vector of the starting point, the result is the position vectors of the other two points. Now, the area of the triangle is half the area of the parallelogram determined by \\\left$$ \begin{array}{rrr} -1 & 0 & 2 \end{array} \right$$^T\\ and \\\left$$ \begin{array}{rrr} 4 & -1 & -1 \end{array} \right$$^T.\\ The required cross product is given by

\\\left$$ \begin{array}{r} -1 \\ 0 \\ 2 \end{array} \right$$ \times \left$$ \begin{array}{r} 4 \\ -1 \\ -1 \end{array} \right$$ = \left$$ \begin{array}{rrr} 2 & 7 & 1 \end{array} \right$$\nonumber \\

Taking the size of this vector gives the area of the parallelogram, given by

\\\sqrt{(2)(2) + (7)(7) + (1)(1)} = \sqrt{4+49+1} = \sqrt{54}\nonumber \\ Hence the area of the triangle is \\\frac{1}{2}\sqrt{54}= \frac{3}{2}\sqrt{6}.\\

In general, if you have three points in \\\mathbb{R}^{3}, P,Q,R\\, the area of the triangle is given by \\\frac{1}{2}\\ \vec{PQ} \times \vec{PR} \\\nonumber \\

Recall that \\\vec{PQ}\\ is the vector running from point \\P\\ to point \\Q\\.

Figure $\PageIndex{4}$

In the next section, we explore another application of the cross product.

The Box Product

Recall that we can use the cross product to find the the area of a parallelogram. It follows that we can use the cross product together with the dot product to find the volume of a parallelepiped. We begin with a definition.

Definition \\\PageIndex{3}\\: Parallelepiped

A parallelepiped determined by the three vectors, \\\vec{u},\vec{v}\\, and \\\vec{w}\\ consists of \\\left\\ r\vec{u}+s\vec{v}+t\vec{w}:r,s,t\in \left$$ 0,1\right$$ \right\\\nonumber \\

That is, if you pick three numbers, \\r,s,\\ and \\t\\ each in \\\left$$ 0,1\right$$\\ and form \\r\vec{u}+s\vec{v}+t\vec{w}\\ then the collection of all such points makes up the parallelepiped determined by these three vectors.

The following is an example of a parallelepiped.

Figure $\PageIndex{5}$

Notice that the base of the parallelepiped is the parallelogram determined by the vectors \\\vec{u}\\ and \\\vec{v}\\. Therefore, its area is equal to \\\\ \vec{u}\times \vec{v} \\\\. The height of the parallelepiped is \\\\ \vec{w}\\ \cos \theta\\ where \\\theta\\ is the angle shown in the picture between \\\vec{w}\\ and \\\vec{u}\times \vec{v}\\. The volume of this parallelepiped is the area of the base times the height which is just \\\\ \vec{u}\times \vec{v}\\ \\ \vec{w}\\ \cos \theta = \left( \vec{u}\times\vec{v}\right) \cdot \vec{w}\nonumber \\ This expression is known as the box product and is sometimes written as \\\left$$ \vec{u},\vec{v},\vec{w}\right$$ .\\ You should consider what happens if you interchange the \\\vec{v}\\ with the \\\vec{w}\\ or the \\\vec{u}\\ with the \\\vec{w}\\. You can see geometrically from drawing pictures that this merely introduces a minus sign. In any case the box product of three vectors always equals either the volume of the parallelepiped determined by the three vectors or else \\-1\\ times this volume.

Proposition \\\PageIndex{2}\\: The Box Product

Let \\\vec{u}, \vec{v}, \vec{w}\\ be three vectors in \\\mathbb{R}^n\\ that define a parallelepiped. Then the volume of the parallelepiped is the absolute value of the box product, given by \\\left\| \left(\vec{u}\times\vec{v}\right) \cdot \vec{w} \right\|\nonumber \\

Consider an example of this concept.

Example \\\PageIndex{4}\\: Volume of a Parallelepiped

Find the volume of the parallelepiped determined by the vectors

\\\vec{u} = \left$$ \begin{array}{r} 1 \\ 2 \\ -5 \end{array} \right$$, \vec{v} = \left$$ \begin{array}{r} 1 \\ 3 \\ -6 \end{array} \right$$, \vec{w} = \left$$ \begin{array}{r} 3 \\ 2 \\ 3 \end{array} \right$$\nonumber \\

###### Solution

According to the above discussion, pick any two of these vectors, take the cross product and then take the dot product of this with the third of these vectors. The result will be either the desired volume or \\-1\\ times the desired volume. Therefore by taking the absolute value of the result, we obtain the volume.

We will take the cross product of \\\vec{u}\\ and \\\vec{v}\\. This is given by

\\\vec{u} \times \vec{v} = \left$$ \begin{array}{r} 1 \\ 2 \\ -5 \end{array} \right$$ \times \left$$ \begin{array}{r} 1 \\ 3 \\ -6 \end{array} \right$$\nonumber \\ \\=\left\vert \begin{array}{rrr} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & -5 \\ 1 & 3 & -6 \end{array} \right\vert = 3\vec{i}+\vec{j}+\vec{k} = \left$$ \begin{array}{r} 3 \\ 1 \\ 1 \end{array} \right$$\nonumber \\

Now take the dot product of this vector with \\\vec{w}\\ which yields \\\begin{aligned} (\vec{u} \times \vec{v}) \cdot \vec{w} &= \left$$ \begin{array}{r} 3 \\ 1 \\ 1 \end{array} \right$$ \cdot \left$$ \begin{array}{r} 3 \\ 2 \\ 3 \end{array} \right$$ \\ &=\left( 3\vec{i}+\vec{j}+\vec{k}\right) \cdot \left( 3\vec{i}+2\vec{j}+3\vec{k}\right) \\ &=9+2+3 \\ &=14\end{aligned}\\

This shows the volume of this parallelepiped is 14 cubic units.

There is a fundamental observation which comes directly from the geometric definitions of the cross product and the dot product.

Proposition \\\PageIndex{3}\\: Order of the Product

Let \\\vec{u},\vec{v}\\, and \\\vec{w}\\ be vectors. Then \\\left( \vec{u}\times \vec{v}\right) \cdot \vec{w}=\vec{u}\cdot \left( \vec{v}\times \vec{w} \right) .\\

Proof

This follows from observing that either \\\left( \vec{u}\times \vec{v}\right) \cdot \vec{w}\\ and \\\vec{u}\cdot \left( \vec{v}\times \vec{w}\right)\\ both give the volume of the parallelepiped or they both give \\-1\\ times the volume.

Recall that we can express the cross product as the determinant of a particular matrix. It turns out that the same can be done for the box product. Suppose you have three vectors, \\\vec{u}=\left$$ \begin{array}{rrr} a & b & c \end{array} \right$$^T ,\vec{v}=\left$$ \begin{array}{rrr} d & e & f \end{array} \right$$^T ,\\ and \\\vec{w}=\left$$ \begin{array}{rrr} g & h & i \end{array} \right$$^T .\\ Then the box product \\\vec{u}\cdot \left(\vec{v}\times \vec{w}\right)\\ is given by the following. \\\begin{aligned} \vec{u}\cdot \left(\vec{v}\times \vec{w}\right) &= \left$$ \begin{array}{r} a \\ b \\ c \end{array} \right$$ \cdot \left\| \begin{array}{rrr} \vec{i} & \vec{j} & \vec{k} \\ d & e & f \\ g & h & i \end{array} \right\| \\ &=a\left\| \begin{array}{rr} e & f \\ h & i \end{array} \right\| -b\left\| \begin{array}{rr} d & f \\ g & i \end{array} \right\| +c\left\| \begin{array}{rr} d & e \\ g & h \end{array} \right\| \\ &= \det \left$$ \begin{array}{rrr} a & b & c \\ d & e & f \\ g & h & i \end{array} \right$$ \end{aligned}\\

To take the box product, you can simply take the determinant of the matrix which results by letting the rows be the components of the given vectors in the order in which they occur in the box product.

This follows directly from the definition of the cross product given above and the way we expand determinants. Thus the volume of a parallelepiped determined by the vectors \\\vec{u},\vec{v},\vec{w}\\ is just the absolute value of the above determinant.

---

4_10_3A_Spanning_Linear_Independence_and_Basis_in_R

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.10%3A_Spanning_Linear_Independence_and_Basis_in_R

Skip to main content

##### Outcomes

1. Determine the span of a set of vectors, and determine if a vector is contained in a specified span.

2. Determine if a set of vectors is linearly independent.

3. Understand the concepts of subspace, basis, and dimension.

4. Find the row space, column space, and null space of a matrix.

By generating all linear combinations of a set of vectors one can obtain various subsets of \\\mathbb{R}^{n}\\ which we call subspaces. For example what set of vectors in \\\mathbb{R}^{3}\\ generate the \\XY\\-plane? What is the smallest such set of vectors can you find? The tools of spanning, linear independence and basis are exactly what is needed to answer these and similar questions and are the focus of this section. The following definition is essential.

##### Definition \\\PageIndex{1}\\: Subset

Let \\U\\ and \\W\\ be sets of vectors in \\\mathbb{R}^n\\. If all vectors in \\U\\ are also in \\W\\, we say that \\U\\ is a subset of \\W\\, denoted \\U \subseteq W\nonumber \\

Spanning Set of Vectors

We begin this section with a definition.

##### Definition \\\PageIndex{2}\\: Span of a Set of Vectors

The collection of all linear combinations of a set of vectors \\\\ \vec{u}\_1, \cdots ,\vec{u}\_k\\\\ in \\\mathbb{R}^{n}\\ is known as the span of these vectors and is written as \\\mathrm{span} \\\vec{u}\_1, \cdots , \vec{u}\_k\\\\.

Consider the following example.

##### Example \\\PageIndex{1}\\: Span of Vectors

Describe the span of the vectors \\\vec{u}=\left$$ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right$$^T\\ and \\\vec{v}=\left$$ \begin{array}{rrr} 3 & 2 & 0 \end{array} \right$$^T \in \mathbb{R}^{3}\\.

###### Solution

You can see that any linear combination of the vectors \\\vec{u}\\ and \\\vec{v}\\ yields a vector of the form \\\left$$ \begin{array}{rrr} x & y & 0 \end{array} \right$$^T\\ in the \\XY\\-plane.

Moreover every vector in the \\XY\\-plane is in fact such a linear combination of the vectors \\\vec{u}\\ and \\\vec{v}\\. That’s because \\\left$$ \begin{array}{r} x \\ y \\ 0 \end{array} \right$$ = (-2x+3y) \left$$ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right$$ + (x-y)\left$$ \begin{array}{r} 3 \\ 2 \\ 0 \end{array} \right$$\nonumber \\

Thus \\\mathrm{span}\\\vec{u},\vec{v}\\\\ is precisely the \\XY\\-plane.

You can convince yourself that no single vector can span the \\XY\\-plane. In fact, take a moment to consider what is meant by the span of a single vector.

However you can make the set larger if you wish. For example consider the larger set of vectors \\\\ \vec{u}, \vec{v}, \vec{w}\\\\ where \\\vec{w}=\left$$ \begin{array}{rrr} 4 & 5 & 0 \end{array} \right$$^T\\. Since the first two vectors already span the entire \\XY\\-plane, the span is once again precisely the \\XY\\-plane and nothing has been gained. Of course if you add a new vector such as \\\vec{w}=\left$$ \begin{array}{rrr} 0 & 0 & 1 \end{array} \right$$^T\\ then it does span a different space. What is the span of \\\vec{u}, \vec{v}, \vec{w}\\ in this case?

The distinction between the sets \\\\ \vec{u}, \vec{v}\\\\ and \\\\ \vec{u}, \vec{v}, \vec{w}\\\\ will be made using the concept of linear independence.

Consider the vectors \\\vec{u}, \vec{v}\\, and \\\vec{w}\\ discussed above. In the next example, we will show how to formally demonstrate that \\\vec{w}\\ is in the span of \\\vec{u}\\ and \\\vec{v}\\.

##### Example \\\PageIndex{2}\\: Vector in a Span

Let \\\vec{u}=\left$$ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right$$^T\\ and \\\vec{v}=\left$$ \begin{array}{rrr} 3 & 2 & 0 \end{array} \right$$^T \in \mathbb{R}^{3}\\. Show that \\\vec{w} = \left$$ \begin{array}{rrr} 4 & 5 & 0 \end{array} \right$$^{T}\\ is in \\\mathrm{span} \left\\ \vec{u}, \vec{v} \right\\\\.

###### Solution

For a vector to be in \\\mathrm{span} \left\\ \vec{u}, \vec{v} \right\\\\, it must be a linear combination of these vectors. If \\\vec{w} \in \mathrm{span} \left\\ \vec{u}, \vec{v} \right\\\\, we must be able to find scalars \\a,b\\ such that\\\vec{w} = a \vec{u} +b \vec{v}\nonumber \\

We proceed as follows. \\\left$$ \begin{array}{r} 4 \\ 5 \\ 0 \end{array} \right$$ = a \left$$ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right$$ + b \left$$ \begin{array}{r} 3 \\ 2 \\ 0 \end{array} \right$$\nonumber \\ This is equivalent to the following system of equations \\\begin{aligned} a + 3b &= 4 \\ a + 2b &= 5\end{aligned}\\

We solving this system the usual way, constructing the augmented matrix and row reducing to find the reduced row-echelon form. \\\left$$ \begin{array}{rr\|r} 1 & 3 & 4 \\ 1 & 2 & 5 \end{array} \right$$ \rightarrow \cdots \rightarrow \left$$ \begin{array}{rr\|r} 1 & 0 & 7 \\ 0 & 1 & -1 \end{array} \right$$\nonumber \\ The solution is \\a=7, b=-1\\. This means that \\\vec{w} = 7 \vec{u} - \vec{v}\nonumber \\ Therefore we can say that \\\vec{w}\\ is in \\\mathrm{span} \left\\ \vec{u}, \vec{v} \right\\\\.

Linearly Independent Set of Vectors

We now turn our attention to the following question: what linear combinations of a given set of vectors \\\\ \vec{u}\_1, \cdots ,\vec{u}\_k\\\\ in \\\mathbb{R}^{n}\\ yields the zero vector? Clearly \\0\vec{u}\_1 + 0\vec{u}\_2+ \cdots + 0 \vec{u}\_k = \vec{0}\\, but is it possible to have \\\sum\_{i=1}^{k}a\_{i}\vec{u}\_{i}=\vec{0}\\ without all coefficients being zero?

You can create examples where this easily happens. For example if \\\vec{u}\_1=\vec{u}\_2\\, then \\1\vec{u}\_1 - \vec{u}\_2+ 0 \vec{u}\_3 + \cdots + 0 \vec{u}\_k = \vec{0}\\, no matter the vectors \\\\ \vec{u}\_3, \cdots ,\vec{u}\_k\\\\. 0But sometimes it can be more subtle.

##### Example \\\PageIndex{3}\\: Linearly Dependent Set of Vectors

Consider the vectors \\\vec{u}\_1=\left$$ \begin{array}{rrr} 0 & 1 & -2 \end{array} \right$$^T, \vec{u}\_2=\left$$ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right$$^T, \vec{u}\_3=\left$$ \begin{array}{rrr} -2 & 3 & 2 \end{array} \right$$^T, \mbox{ and } \vec{u}\_4=\left$$ \begin{array}{rrr} 1 & -2 & 0 \end{array} \right$$^T\nonumber \\ in \\\mathbb{R}^{3}\\.

Then verify that \\1\vec{u}\_1 +0 \vec{u}\_2+ - \vec{u}\_3 -2 \vec{u}\_4 = \vec{0}\nonumber \\

You can see that the linear combination does yield the zero vector but has some non-zero coefficients. Thus we define a set of vectors to be *linearly dependent* if this happens.

##### Definition \\\PageIndex{3}\\: Linearly Dependent Set of Vectors

A set of non-zero vectors \\\\ \vec{u}\_1, \cdots ,\vec{u}\_k\\\\ in \\\mathbb{R}^{n}\\ is said to be linearly dependent if a linear combination of these vectors without all coefficients being zero does yield the zero vector.

Note that if \\\sum\_{i=1}^{k}a\_{i}\vec{u}\_{i}=\vec{0}\\ and some coefficient is non-zero, say \\a_1 \neq 0\\, then \\\vec{u}\_1 = \frac{-1}{a_1} \sum\_{i=2}^{k}a\_{i}\vec{u}\_{i}\nonumber \\ and thus \\\vec{u}\_1\\ is in the span of the other vectors. And the converse clearly works as well, so we get that a set of vectors is linearly dependent precisely when one of its vector is in the span of the other vectors of that set.

In particular, you can show that the vector \\\vec{u}\_1\\ in the above example is in the span of the vectors \\\\ \vec{u}\_2, \vec{u}\_3, \vec{u}\_4 \\\\.

If a set of vectors is NOT linearly dependent, then it must be that any linear combination of these vectors which yields the zero vector must use all zero coefficients. This is a very important notion, and we give it its own name of *linear independence*.

##### Definition \\\PageIndex{4}\\: Linearly Independent Set of Vectors

A set of non-zero vectors \\\\ \vec{u}\_1, \cdots ,\vec{u}\_k\\\\ in \\\mathbb{R}^{n}\\ is said to be linearly independent if whenever \\\sum\_{i=1}^{k}a\_{i}\vec{u}\_{i}=\vec{0}\nonumber \\ it follows that each \\a\_{i}=0\\.

Note also that we require all vectors to be non-zero to form a linearly independent set.

To view this in a more familiar setting, form the \\n \times k\\ matrix \\A\\ having these vectors as columns. Then all we are saying is that the set \\\\ \vec{u}\_1, \cdots ,\vec{u}\_k\\\\ is linearly independent precisely when \\AX=0\\ has only the trivial solution.

Here is an example.

##### Example \\\PageIndex{4}\\: Linearly Independent Vectors

Consider the vectors \\\vec{u}=\left$$ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right$$^T\\, \\\vec{v}=\left$$ \begin{array}{rrr} 1 & 0 & 1 \end{array} \right$$^T\\, and \\\vec{w}=\left$$ \begin{array}{rrr} 0 & 1 & 1 \end{array} \right$$^T\\ in \\\mathbb{R}^{3}\\. Verify whether the set \\\\\vec{u}, \vec{v}, \vec{w}\\\\ is linearly independent.

###### Solution

So suppose that we have a linear combinations \\a\vec{u} + b \vec{v} + c\vec{w} = \vec{0}\\. Then you can see that this can only happen with \\a=b=c=0\\.

As mentioned above, you can equivalently form the \\3 \times 3\\ matrix \\A = \left$$ \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ \end{array} \right$$\\, and show that \\AX=0\\ has only the trivial solution.

Thus this means the set \\\left\\ \vec{u}, \vec{v}, \vec{w} \right\\\\ is linearly independent.

In terms of spanning, a set of vectors is linearly independent if it does not contain unnecessary vectors, that is not vector is in the span of the others.

Thus we put all this together in the following important theorem.

##### Theorem \\\PageIndex{1}\\: Linear Independence as a Linear Combination

Let \\\left\\\vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ be a collection of vectors in \\\mathbb{R}^{n}\\. Then the following are equivalent:

1. It is linearly independent, that is whenever \\\sum\_{i=1}^{k}a\_{i}\vec{u}\_{i}=\vec{0}\nonumber \\ it follows that each coefficient \\a\_{i}=0\\.

2. No vector is in the span of the others.

3. The system of linear equations \\AX=0\\ has only the trivial solution, where \\A\\ is the \\n \times k\\ matrix having these vectors as columns.

The last sentence of this theorem is useful as it allows us to use the reduced row-echelon form of a matrix to determine if a set of vectors is linearly independent. Let the vectors be columns of a matrix \\A\\. Find the reduced row-echelon form of \\A\\. If each column has a leading one, then it follows that the vectors are linearly independent.

Sometimes we refer to the condition regarding sums as follows: The set of vectors, \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ is linearly independent if and only if there is no nontrivial linear combination which equals the zero vector. A nontrivial linear combination is one in which not all the scalars equal zero. Similarly, a trivial linear combination is one in which all scalars equal zero.

Here is a detailed example in \\\mathbb{R}^{4}\\.

##### Example \\\PageIndex{5}\\: Linear Independence

Determine whether the set of vectors given by \\\left\\ \left$$ \begin{array}{r} 1 \\ 2 \\ 3 \\ 0 \end{array} \right$$, \\ \left$$ \begin{array}{r} 2 \\ 1 \\ 0 \\ 1 \end{array} \right$$ , \\ \left$$ \begin{array}{r} 0 \\ 1 \\ 1 \\ 2 \end{array} \right$$ , \\ \left$$ \begin{array}{r} 3 \\ 2 \\ 2 \\ 0 \end{array} \right$$ \right\\\nonumber \\ is linearly independent. If it is linearly dependent, express one of the vectors as a linear combination of the others.

###### Solution

In this case the matrix of the corresponding homogeneous system of linear equations is \\\left$$ \begin{array}{rrrr\|r} 1 & 2 & 0 & 3 & 0\\ 2 & 1 & 1 & 2 & 0 \\ 3 & 0 & 1 & 2 & 0 \\ 0 & 1 & 2 & 0 & 0 \end{array} \right$$\nonumber \\

The reduced row-echelon form is \\\left$$ \begin{array}{rrrr\|r} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array} \right$$\nonumber \\

and so every column is a pivot column and the corresponding system \\AX=0\\ only has the trivial solution. Therefore, these vectors are linearly independent and there is no way to obtain one of the vectors as a linear combination of the others.

Consider another example.

##### Example \\\PageIndex{6}\\: Linear Independence

Determine whether the set of vectors given by \\\left\\ \left$$ \begin{array}{r} 1 \\ 2 \\ 3 \\ 0 \end{array} \right$$, \\ \left$$ \begin{array}{r} 2 \\ 1 \\ 0 \\ 1 \end{array} \right$$, \\ \left$$ \begin{array}{r} 0 \\ 1 \\ 1 \\ 2 \end{array} \right$$, \\ \left$$ \begin{array}{r} 3 \\ 2 \\ 2 \\ -1 \end{array} \right$$ \right\\\nonumber \\ is linearly independent. If it is linearly dependent, express one of the vectors as a linear combination of the others.

###### Solution

Form the \\4 \times 4\\ matrix \\A\\ having these vectors as columns: \\A= \left$$ \begin{array}{rrrr} 1 & 2 & 0 & 3 \\ 2 & 1 & 1 & 2 \\ 3 & 0 & 1 & 2 \\ 0 & 1 & 2 & -1 \end{array} \right$$\nonumber \\ Then by Theorem $\PageIndex{1}$, the given set of vectors is linearly independent exactly if the system \\AX=0\\ has only the trivial solution.

The augmented matrix for this system and corresponding reduced row-echelon form are given by \\\left$$ \begin{array}{rrrr\|r} 1 & 2 & 0 & 3 & 0 \\ 2 & 1 & 1 & 2 & 0 \\ 3 & 0 & 1 & 2 & 0 \\ 0 & 1 & 2 & -1 & 0 \end{array} \right$$ \rightarrow \cdots \rightarrow \left$$ \begin{array}{rrrr\|r} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right$$\nonumber \\ Not all the columns of the coefficient matrix are pivot columns and so the vectors are not linearly independent. In this case, we say the vectors are linearly dependent.

It follows that there are infinitely many solutions to \\AX=0\\, one of which is \\\left$$ \begin{array}{r} 1 \\ 1 \\ -1 \\ -1 \end{array} \right$$\nonumber \\ Therefore we can write \\1\left$$ \begin{array}{r} 1 \\ 2 \\ 3 \\ 0 \end{array} \right$$ +1\left$$ \begin{array}{r} 2 \\ 1 \\ 0 \\ 1 \end{array} \right$$ -1 \left$$ \begin{array}{r} 0 \\ 1 \\ 1 \\ 2 \end{array} \right$$ -1 \left$$ \begin{array}{r} 3 \\ 2 \\ 2 \\ -1 \end{array} \right$$ = \left$$ \begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \end{array} \right$$\nonumber \\

This can be rearranged as follows \\1\left$$ \begin{array}{r} 1 \\ 2 \\ 3 \\ 0 \end{array} \right$$ +1\left$$ \begin{array}{r} 2 \\ 1 \\ 0 \\ 1 \end{array} \right$$ -1 \left$$ \begin{array}{r} 0 \\ 1 \\ 1 \\ 2 \end{array} \right$$ =\left$$ \begin{array}{r} 3 \\ 2 \\ 2 \\ -1 \end{array} \right$$\nonumber \\ This gives the last vector as a linear combination of the first three vectors.

Notice that we could rearrange this equation to write any of the four vectors as a linear combination of the other three.

When given a linearly independent set of vectors, we can determine if related sets are linearly independent.

##### Example \\\PageIndex{7}\\: Related Sets of Vectors

Let \\\\ \vec{u},\vec{v},\vec{w}\\\\ be an independent set of \\\mathbb{R}^n\\. Is \\\\\vec{u}+\vec{v}, 2\vec{u}+\vec{w}, \vec{v}-5\vec{w}\\\\ linearly independent?

###### Solution

Suppose \\a(\vec{u}+\vec{v}) + b(2\vec{u}+\vec{w}) + c(\vec{v}-5\vec{w})=\vec{0}\_n\\ for some \\a,b,c\in\mathbb{R}\\. Then \$a+2b)\vec{u} + (a+c)\vec{v} + (b-5c)\vec{w}=\vec{0}\_n.\nonumber \\

Since \\\\\vec{u},\vec{v},\vec{w}\\\\ is independent, \\\begin{aligned} a + 2b & = 0 \\ a + c & = 0 \\ b - 5c & = 0 \end{aligned}\\

This system of three equations in three variables has the unique solution \\a=b=c=0\\. Therefore, \\\\\vec{u}+\vec{v}, 2\vec{u}+\vec{w}, \vec{v}-5\vec{w}\\\\ is independent.

The following corollary follows from the fact that if the augmented matrix of a homogeneous system of linear equations has more columns than rows, the system has infinitely many solutions.

##### Corollary \\\PageIndex{1}\\: Linear Dependence in \\\mathbb{R}''\\

Let \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ be a set of vectors in \\\mathbb{R}^{n}\\. If \\k\>n\\, then the set is linearly dependent (i.e. NOT linearly independent).

Proof

Form the \\n \times k\\ matrix \\A\\ having the vectors \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ as its columns and suppose \\k \> n\\. Then \\A\\ has rank \\r \leq n \

##### Example \\\PageIndex{8}\\: Linear Dependence

Consider the vectors \\\left\\ \left$$ \begin{array}{r} 1 \\ 4 \end{array} \right$$, \left$$ \begin{array}{r} 2 \\ 3 \end{array} \right$$, \left$$ \begin{array}{r} 3 \\ 2 \end{array} \right$$ \right\\\nonumber \\ Are these vectors linearly independent?

###### Solution

This set contains three vectors in \\\mathbb{R}^2\\. By Corollary $\PageIndex{1}$ these vectors are linearly dependent. In fact, we can write \$-1) \left$$ \begin{array}{r} 1 \\ 4 \end{array} \right$$ + (2) \left$$ \begin{array}{r} 2 \\ 3 \end{array} \right$$ = \left$$ \begin{array}{r} 3 \\ 2 \end{array} \right$$\nonumber \\ showing that this set is linearly dependent.

The third vector in the previous example is in the span of the first two vectors. We could find a way to write this vector as a linear combination of the other two vectors. It turns out that the linear combination which we found is the only one, provided that the set is linearly independent.

##### Theorem \\\PageIndex{2}\\: Unique Linear Combination

Let \\U \subseteq\mathbb{R}^n\\ be an independent set. Then any vector \\\vec{x}\in\mathrm{span}(U)\\ can be written uniquely as a linear combination of vectors of \\U\\.

Proof

To prove this theorem, we will show that two linear combinations of vectors in \\U\\ that equal \\\vec{x}\\ must be the same. Let \\U =\\ \vec{u}\_1, \vec{u}\_2, \ldots, \vec{u}\_k\\\\. Suppose that there is a vector \\\vec{x}\in \mathrm{span}(U)\\ such that \\\begin{aligned} \vec{x} & = s_1\vec{u}\_1 + s_2\vec{u}\_2 + \cdots + s_k\vec{u}\_k, \mbox{ for some } s_1, s_2, \ldots, s_k\in\mathbb{R}, \mbox{ and} \\ \vec{x} & = t_1\vec{u}\_1 + t_2\vec{u}\_2 + \cdots + t_k\vec{u}\_k, \mbox{ for some } t_1, t_2, \ldots, t_k\in\mathbb{R}.\end{aligned}\\ Then \\\vec{0}\_n=\vec{x}-\vec{x} = (s_1-t_1)\vec{u}\_1 + (s_2-t_2)\vec{u}\_2 + \cdots + (s_k-t_k)\vec{u}\_k\\.

Since \\U\\ is independent, the only linear combination that vanishes is the trivial one, so \\s_i-t_i=0\\ for all \\i\\, \\1\leq i\leq k\\.

Therefore, \\s_i=t_i\\ for all \\i\\, \\1\leq i\leq k\\, and the representation is unique.Let \\U \subseteq\mathbb{R}^n\\ be an independent set. Then any vector \\\vec{x}\in\mathrm{span}(U)\\ can be written uniquely as a linear combination of vectors of \\U\\.

Suppose that \\\vec{u},\vec{v}\\ and \\\vec{w}\\ are nonzero vectors in \\\mathbb{R}^3\\, and that \\\\ \vec{v},\vec{w}\\\\ is independent. Consider the set \\\\ \vec{u},\vec{v},\vec{w}\\\\. When can we know that this set is independent? It turns out that this follows exactly when \\\vec{u}\not\in\mathrm{span}\\\vec{v},\vec{w}\\\\.

##### Example \\\PageIndex{9}\\

Suppose that \\\vec{u},\vec{v}\\ and \\\vec{w}\\ are nonzero vectors in \\\mathbb{R}^3\\, and that \\\\ \vec{v},\vec{w}\\\\ is independent. Prove that \\\\ \vec{u},\vec{v},\vec{w}\\\\ is independent if and only if \\\vec{u}\not\in\mathrm{span}\\\vec{v},\vec{w}\\\\.

###### Solution

If \\\vec{u}\in\mathrm{span}\\\vec{v},\vec{w}\\\\, then there exist \\a,b\in\mathbb{R}\\ so that \\\vec{u}=a\vec{v} + b\vec{w}\\. This implies that \\\vec{u}-a\vec{v} - b\vec{w}=\vec{0}\_3\\, so \\\vec{u}-a\vec{v} - b\vec{w}\\ is a nontrivial linear combination of \\\\ \vec{u},\vec{v},\vec{w}\\\\ that vanishes, and thus \\\\ \vec{u},\vec{v},\vec{w}\\\\ is dependent.

Now suppose that \\\vec{u}\not\in\mathrm{span}\\\vec{v},\vec{w}\\\\, and suppose that there exist \\a,b,c\in\mathbb{R}\\ such that \\a\vec{u}+b\vec{v}+c\vec{w}=\vec{0}\_3\\. If \\a\neq 0\\, then \\\vec{u}=-\frac{b}{a}\vec{v}-\frac{c}{a}\vec{w}\\, and \\\vec{u}\in\mathrm{span}\\\vec{v},\vec{w}\\\\, a contradiction. Therefore, \\a=0\\, implying that \\b\vec{v}+c\vec{w}=\vec{0}\_3\\. Since \\\\ \vec{v},\vec{w}\\\\ is independent, \\b=c=0\\, and thus \\a=b=c=0\\, i.e., the only linear combination of \\\vec{u},\vec{v}\\ and \\\vec{w}\\ that vanishes is the trivial one.

Therefore, \\\\ \vec{u},\vec{v},\vec{w}\\\\ is independent.

Consider the following useful theorem.

##### Theorem \\\PageIndex{3}\\: Invertible Matrices

Let \\A\\ be an invertible \\n \times n\\ matrix. Then the columns of \\A\\ are independent and span \\\mathbb{R}^n\\. Similarly, the rows of \\A\\ are independent and span the set of all \\1 \times n\\ vectors.

This theorem also allows us to determine if a matrix is invertible. If an \\n \times n\\ matrix \\A\\ has columns which are independent, or span \\\mathbb{R}^n\\, then it follows that \\A\\ is invertible. If it has rows that are independent, or span the set of all \\1 \times n\\ vectors, then \\A\\ is invertible.

A Short Application to Chemistry

The following section applies the concepts of spanning and linear independence to the subject of chemistry.

When working with chemical reactions, there are sometimes a large number of reactions and some are in a sense redundant. Suppose you have the following chemical reactions. \\\begin{array}{c} CO+\frac{1}{2}O\_{2}\rightarrow CO\_{2} \\ H\_{2}+\frac{1}{2}O\_{2}\rightarrow H\_{2}O \\ CH\_{4}+\frac{3}{2}O\_{2}\rightarrow CO+2H\_{2}O \\ CH\_{4}+2O\_{2}\rightarrow CO\_{2}+2H\_{2}O \end{array}\nonumber \\ There are four chemical reactions here but they are not independent reactions. There is some redundancy. What are the independent reactions? Is there a way to consider a shorter list of reactions? To analyze this situation, we can write the reactions in a matrix as follows \\\left$$ \begin{array}{cccccc} CO & O\_{2} & CO\_{2} & H\_{2} & H\_{2}O & CH\_{4} \\ 1 & 1/2 & -1 & 0 & 0 & 0 \\ 0 & 1/2 & 0 & 1 & -1 & 0 \\ -1 & 3/2 & 0 & 0 & -2 & 1 \\ 0 & 2 & -1 & 0 & -2 & 1 \end{array} \right$$\nonumber \\

Each row contains the coefficients of the respective elements in each reaction. For example, the top row of numbers comes from \\CO+\frac{1}{2}O\_{2}-CO\_{2}=0\\ which represents the first of the chemical reactions.

We can write these coefficients in the following matrix \\\left$$ \begin{array}{rrrrrr} 1 & 1/2 & -1 & 0 & 0 & 0 \\ 0 & 1/2 & 0 & 1 & -1 & 0 \\ -1 & 3/2 & 0 & 0 & -2 & 1 \\ 0 & 2 & -1 & 0 & -2 & 1 \end{array} \right$$\nonumber \\ Rather than listing all of the reactions as above, it would be more efficient to only list those which are independent by throwing out that which is redundant. We can use the concepts of the previous section to accomplish this.

First, take the reduced row-echelon form of the above matrix. \\\left$$ \begin{array}{rrrrrr} 1 & 0 & 0 & 3 & -1 & -1 \\ 0 & 1 & 0 & 2 & -2 & 0 \\ 0 & 0 & 1 & 4 & -2 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right$$\nonumber \\ The top three rows represent “independent" reactions which come from the original four reactions. One can obtain each of the original four rows of the matrix given above by taking a suitable linear combination of rows of this reduced row-echelon matrix.

With the redundant reaction removed, we can consider the simplified reactions as the following equations \\\begin{array}{c} CO+3H\_{2}-1H\_{2}O-1CH\_{4}=0 \\ O\_{2}+2H\_{2}-2H\_{2}O=0 \\ CO\_{2}+4H\_{2}-2H\_{2}O-1CH\_{4}=0 \end{array}\nonumber \\ In terms of the original notation, these are the reactions \\\begin{array}{c} CO+3H\_{2}\rightarrow H\_{2}O+CH\_{4} \\ O\_{2}+2H\_{2}\rightarrow 2H\_{2}O \\ CO\_{2}+4H\_{2}\rightarrow 2H\_{2}O+CH\_{4} \end{array}\nonumber \\

These three reactions provide an equivalent system to the original four equations. The idea is that, in terms of what happens chemically, you obtain the same information with the shorter list of reactions. Such a simplification is especially useful when dealing with very large lists of reactions which may result from experimental evidence.

Subspaces and Basis

The goal of this section is to develop an understanding of a subspace of \\\mathbb{R}^n\\. Before a precise definition is considered, we first examine the subspace test given below.

##### Theorem \\\PageIndex{4}\\: Subspace Test

A subset \\V\\ of \\\mathbb{R}^n\\ is a subspace of \\\mathbb{R}^n\\ if

1. the zero vector of \\\mathbb{R}^n\\, \\\vec{0}\_n\\, is in \\V\\;

2. \\V\\ is closed under addition, i.e., for all \\\vec{u},\vec{w}\in V\\, \\\vec{u}+\vec{w}\in V\\;

3. \\V\\ is closed under scalar multiplication, i.e., for all \\\vec{u}\in V\\ and \\k\in\mathbb{R}\\, \\k\vec{u}\in V\\.

This test allows us to determine if a given set is a subspace of \\\mathbb{R}^n\\. Notice that the subset \\V = \left\\ \vec{0} \right\\\\ is a subspace of \\\mathbb{R}^n\\ (called the zero subspace ), as is \\\mathbb{R}^n\\ itself. A subspace which is not the zero subspace of \\\mathbb{R}^n\\ is referred to as a proper subspace.

A subspace is simply a set of vectors with the property that linear combinations of these vectors remain in the set. Geometrically in \\\mathbb{R}^{3}\\, it turns out that a subspace can be represented by either the origin as a single point, lines and planes which contain the origin, or the entire space \\\mathbb{R}^{3}\\.

Consider the following example of a line in \\\mathbb{R}^3\\.

##### Example \\\PageIndex{10}\\: Subspace of \\\mathbb{R}^3\\

In \\\mathbb{R}^3\\, the line \\L\\ through the origin that is parallel to the vector \\{\vec{d}}= \left$$ \begin{array}{r} -5 \\ 1 \\ -4 \end{array}\right$$\\ has (vector) equation \\\left$$ \begin{array}{r} x \\ y \\ z \end{array}\right$$ =t\left$$ \begin{array}{r} -5 \\ 1 \\ -4 \end{array}\right$$, t\in\mathbb{R}\\, so \\L=\left\\ t{\vec{d}} ~\|~ t\in\mathbb{R}\right\\.\nonumber \\ Then \\L\\ is a subspace of \\\mathbb{R}^3\\.

###### Solution

Using the subspace test given above we can verify that \\L\\ is a subspace of \\\mathbb{R}^3\\.

• First: \\\vec{0}\_3\in L\\ since \\0\vec{d}=\vec{0}\_3\\.
• Suppose \\\vec{u},\vec{v}\in L\\. Then by definition, \\\vec{u}=s\vec{d}\\ and \\\vec{v}=t\vec{d}\\, for some \\s,t\in\mathbb{R}\\. Thus \\\vec{u}+\vec{v} = s\vec{d}+t\vec{d} = (s+t)\vec{d}.\nonumber \\ Since \\s+t\in\mathbb{R}\\, \\\vec{u}+\vec{v}\in L\\; i.e., \\L\\ is closed under addition.
• Suppose \\\vec{u}\in L\\ and \\k\in\mathbb{R}\\ (\\k\\ is a scalar). Then \\\vec{u}=t\vec{d}\\, for some \\t\in\mathbb{R}\\, so \\k\vec{u}=k(t\vec{d})=(kt)\vec{d}.\nonumber \\ Since \\kt\in\mathbb{R}\\, \\k\vec{u}\in L\\; i.e., \\L\\ is closed under scalar multiplication.

Since \\L\\ satisfies all conditions of the subspace test, it follows that \\L\\ is a subspace.

Note that there is nothing special about the vector \\\vec{d}\\ used in this example; the same proof works for any nonzero vector \\\vec{d}\in\mathbb{R}^3\\, so any line through the origin is a subspace of \\\mathbb{R}^3\\.

We are now prepared to examine the precise definition of a subspace as follows.

##### Definition \\\PageIndex{5}\\: Subspace

Let \\V\\ be a nonempty collection of vectors in \\\mathbb{R}^{n}.\\ Then \\V\\ is called a subspace if whenever \\a\\ and \\b\\ are scalars and \\\vec{u}\\ and \\\vec{v}\\ are vectors in \\V,\\ the linear combination \\a \vec{u}+ b \vec{v}\\ is also in \\V\\.

More generally this means that a subspace contains the span of any finite collection vectors in that subspace. It turns out that in \\\mathbb{R}^{n}\\, a subspace is exactly the span of finitely many of its vectors.

##### Theorem \\\PageIndex{5}\\: Subspaces are Spans

Let \\V\\ be a nonempty collection of vectors in \\\mathbb{R}^{n}.\\ Then \\V\\ is a subspace of \\\mathbb{R}^{n}\\ if and only if there exist vectors \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ in \\V\\ such that \\V= \mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\nonumber \\ Furthermore, let \\W\\ be another subspace of \\\mathbb{R}^n\\ and suppose \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\ \in W\\. Then it follows that \\V\\ is a subset of \\W\\.

Note that since \\W\\ is arbitrary, the statement that \\V \subseteq W\\ means that any other subspace of \\\mathbb{R}^n\\ that contains these vectors will also contain \\V\\.

Proof

We first show that if \\V\\ is a subspace, then it can be written as \\V= \mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\. Pick a vector \\\vec{u}\_{1}\\ in \\V\\. If \\V= \mathrm{span}\left\\ \vec{u}\_{1}\right\\ ,\\ then you have found your list of vectors and are done. If \\V\neq \mathrm{span}\left\\ \vec{u}\_{1}\right\\ ,\\ then there exists \\\vec{u}\_{2}\\ a vector of \\V\\ which is not in \\\mathrm{span}\left\\ \vec{u}\_{1}\right\\ .\\ Consider \\\mathrm{span}\left\\ \vec{u}\_{1},\vec{u}\_{2}\right\\.\\ If \\V=\mathrm{span}\left\\ \vec{u}\_{1},\vec{u}\_{2}\right\\\\, we are done. Otherwise, pick \\\vec{u}\_{3}\\ not in \\\mathrm{span}\left\\ \vec{u}\_{1},\vec{u}\_{2}\right\\ .\\ Continue this way. Note that since \\V\\ is a subspace, these spans are each contained in \\V\\. The process must stop with \\\vec{u}\_{k}\\ for some \\k\leq n\\ by Corollary $\PageIndex{1}$, and thus \\V=\mathrm{span}\left\\ \vec{u}\_{1},\cdots , \vec{u}\_{k}\right\\\\.

Now suppose \\V=\mathrm{span}\left\\ \vec{u}\_{1},\cdots , \vec{u}\_{k}\right\\\\, we must show this is a subspace. So let \\\sum\_{i=1}^{k}c\_{i}\vec{u}\_{i}\\ and \\\sum\_{i=1}^{k}d\_{i}\vec{u}\_{i}\\ be two vectors in \\V\\, and let \\a\\ and \\b\\ be two scalars. Then \\a \sum\_{i=1}^{k}c\_{i}\vec{u}\_{i}+ b \sum\_{i=1}^{k}d\_{i}\vec{u}\_{i}= \sum\_{i=1}^{k}\left( a c\_{i}+b d\_{i}\right) \vec{u}\_{i}\nonumber \\ which is one of the vectors in \\\mathrm{span}\left\\ \vec{u}\_{1},\cdots , \vec{u}\_{k}\right\\\\ and is therefore contained in \\V\\. This shows that \\\mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ has the properties of a subspace.

To prove that \\V \subseteq W\\, we prove that if \\\vec{u}\_i\in V\\, then \\\vec{u}\_i \in W\\.

Suppose \\\vec{u}\in V\\. Then \\\vec{u}=a_1\vec{u}\_1 + a_2\vec{u}\_2 + \cdots + a_k\vec{u}\_k\\ for some \\a_i\in\mathbb{R}\\, \\1\leq i\leq k\\. Since \\W\\ contain each \\\vec{u}\_i\\ and \\W\\ is a vector space, it follows that \\a_1\vec{u}\_1 + a_2\vec{u}\_2 + \cdots + a_k\vec{u}\_k \in W\\.

Since the vectors \\\vec{u}\_i\\ we constructed in the proof above are not in the span of the previous vectors (by definition), they must be linearly independent and thus we obtain the following corollary.

##### Corollary \\\PageIndex{2}\\: Subspaces are Spans of Independent Vectors

If \\V\\ is a subspace of \\\mathbb{R}^{n},\\ then there exist linearly independent vectors \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ in \\V\\ such that \\V=\mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\.

In summary, subspaces of \\\mathbb{R}^{n}\\ consist of spans of finite, linearly independent collections of vectors of \\\mathbb{R}^{n}\\. Such a collection of vectors is called a basis.

##### Definition \\\PageIndex{6}\\: Basis of a Subspace

Let \\V\\ be a subspace of \\\mathbb{R}^{n}\\. Then \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ is a basis for \\V\\ if the following two conditions hold.

1. \\\mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\ =V\\

2. \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ is linearly independent

Note the plural of basis is bases.

The following is a simple but very useful example of a basis, called the standard basis.

##### Definition \\\PageIndex{7}\\: Standard Basis of \\\mathbb{R}^n\\

Let \\\vec{e}\_i\\ be the vector in \\\mathbb{R}^n\\ which has a \\1\\ in the \\i^{th}\\ entry and zeros elsewhere, that is the \\i^{th}\\ column of the identity matrix. Then the collection \\\left\\\vec{e}\_1, \vec{e}\_2, \cdots, \vec{e}\_n \right\\\\ is a basis for \\\mathbb{R}^n\\ and is called the standard basis of \\\mathbb{R}^n\\.

The main theorem about bases is not only they exist, but that they must be of the same size. To show this, we will need the the following fundamental result, called the Exchange Theorem.

##### Theorem \\\PageIndex{6}\\: Exchange Theorem

Suppose \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{r}\right\\\\ is a linearly independent set of vectors in \\\mathbb{R}^n\\, and each \\\vec{u}\_{k}\\ is contained in \\\mathrm{span}\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{s}\right\\\\ Then \\s\geq r.\\

In words, spanning sets have at least as many vectors as linearly independent sets.

Proof

Since each \\\vec{u}\_j\\ is in \\\mathrm{span}\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{s}\right\\\\, there exist scalars \\a\_{ij}\\ such that \\\vec{u}\_{j}=\sum\_{i=1}^{s}a\_{ij}\vec{v}\_{i}\nonumber \\ Suppose for a contradiction that \\s\

We are now ready to show that any two bases are of the same size.

##### Theorem \\\PageIndex{7}\\: Bases of \\\mathbb{R}^{n}\\ are of the Same Size

Let \\V\\ be a subspace of \\\mathbb{R}^{n}\\ with two bases \\B_1\\ and \\B_2\\. Suppose \\B_1\\ contains \\s\\ vectors and \\B_2\\ contains \\r\\ vectors. Then \\s=r.\\

Proof

This follows right away from Theorem 9.4.4. Indeed observe that \\B_1 = \left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{s}\right\\\\ is a spanning set for \\V\\ while \\B_2 = \left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{r}\right\\\\ is linearly independent, so \\s \geq r.\\ Similarly \\B_2 = \left\\ \vec{v}\_{1},\cdots ,\vec{v} \_{r}\right\\\\ is a spanning set for \\V\\ while \\B_1 = \left\\ \vec{u}\_{1},\cdots , \vec{u}\_{s}\right\\\\ is linearly independent, so \\r\geq s\\.

The following definition can now be stated.

##### Definition \\\PageIndex{8}\\: Dimension of a Subspace

Let \\V\\ be a subspace of \\\mathbb{R}^{n}\\. Then the dimension of \\V\\, written \\\mathrm{dim}(V)\\ is defined to be the number of vectors in a basis.

The next result follows.

##### Corollary \\\PageIndex{3}\\: Dimension of \\\mathbb{R}^n\\

The dimension of \\\mathbb{R}^{n}\\ is \\n.\\

Proof

You only need to exhibit a basis for \\\mathbb{R}^{n}\\ which has \\n\\ vectors. Such a basis is the standard basis \\\left\\ \vec{e}\_{1},\cdots , \vec{e}\_{n}\right\\\\.

Consider the following example.

##### Example \\\PageIndex{11}\\: Basis of Subspace

Let \\V=\left\\ \left$$\begin{array}{c} a\\ b\\ c\\ d\end{array}\right$$\in\mathbb{R}^4 ~:~ a-b=d-c \right\\.\nonumber \\ Show that \\V\\ is a subspace of \\\mathbb{R}^4\\, find a basis of \\V\\, and find \\\dim(V)\\.

###### Solution

The condition \\a-b=d-c\\ is equivalent to the condition \\a=b-c+d\\, so we may write

\\V =\left\\ \left$$\begin{array}{c} b-c+d\\ b\\ c\\ d\end{array}\right$$ ~:~b,c,d \in\mathbb{R} \right\\ = \left\\ b\left$$\begin{array}{c} 1\\ 1\\ 0\\ 0\end{array}\right$$ +c\left$$\begin{array}{c} -1\\ 0\\ 1\\ 0\end{array}\right$$ +d\left$$\begin{array}{c} 1\\ 0\\ 0\\ 1\end{array}\right$$ ~:~ b,c,d\in\mathbb{R} \right\\\nonumber \\

This shows that \\V\\ is a subspace of \\\mathbb{R}^4\\, since \\V=\mathrm{span}\\ \vec{u}\_1, \vec{u}\_2, \vec{u}\_3 \\\\ where

\\\vec{u}\_1 = \left$$\begin{array}{r} 1 \\ 1 \\ 0 \\ 0 \end{array}\right$$, \vec{u}\_2 = \left$$\begin{array}{r} -1 \\ 0 \\ 1 \\ 0 \end{array}\right$$, \vec{u}\_3 = \left$$\begin{array}{r} 1 \\ 0 \\ 0 \\ 1 \end{array}\right$$\nonumber \\

Furthermore,

\\\left\\ \left$$\begin{array}{c} 1\\ 1\\ 0\\ 0\end{array}\right$$, \left$$\begin{array}{c} -1\\ 0\\ 1\\ 0\end{array}\right$$, \left$$\begin{array}{c} 1\\ 0\\ 0\\ 1\end{array}\right$$ \right\\\nonumber \\ is linearly independent, as can be seen by taking the reduced row-echelon form of the matrix whose columns are \\\vec{u}\_1, \vec{u}\_2\\ and \\\vec{u}\_3\\.

\\\left$$\begin{array}{rrr} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right$$ \rightarrow \left$$\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right$$\nonumber \\

Since every column of the reduced row-echelon form matrix has a leading one, the columns are linearly independent.

Therefore \\\\ \vec{u}\_1, \vec{u}\_2, \vec{u}\_3 \\\\ is linearly independent and spans \\V\\, so is a basis of \\V\\. Hence \\V\\ has dimension three.

We continue by stating further properties of a set of vectors in \\\mathbb{R}^{n}\\.

##### Corollary \\\PageIndex{4}\\: Linearly Independent and Spanning Sets in \\\mathbb{R}^{n}\\

The following properties hold in \\\mathbb{R}^{n}\\:

• Suppose \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\\\\ is linearly independent. Then \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\\\\ is a basis for \\\mathbb{R}^{n}\\.
• Suppose \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{m}\right\\\\ spans \\\mathbb{R}^{n}.\\ Then \\m\geq n.\\
• If \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\\\\ spans \\\mathbb{R}^{n},\\ then \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\\\\ is linearly independent.

Proof

Assume first that \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\\\\ is linearly independent, and we need to show that this set spans \\\mathbb{R}^{n}\\. To do so, let \\\vec{v}\\ be a vector of \\\mathbb{R}^{n}\\, and we need to write \\\vec{v}\\ as a linear combination of \\\vec{u}\_i\\’s. Consider the matrix \\A\\ having the vectors \\\vec{u}\_i\\ as columns: \\A = \left$$ \begin{array}{rrr} \vec{u}\_{1} & \cdots & \vec{u}\_{n} \end{array} \right$$\nonumber \\

By linear independence of the \\\vec{u}\_i\\’s, the reduced row-echelon form of \\A\\ is the identity matrix. Therefore the system \\A\vec{x}= \vec{v}\\ has a (unique) solution, so \\\vec{v}\\ is a linear combination of the \\\vec{u}\_i\\’s.

To establish the second claim, suppose that \\m\

Finally consider the third claim. If \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\\\\ is not linearly independent, then replace this list with \\\left\\ \vec{u}\_{i\_{1}},\cdots ,\vec{u}\_{i\_{k}}\right\\\\ where these are the pivot columns of the matrix \\\left$$ \begin{array}{ccc} \vec{u}\_{1} & \cdots & \vec{u}\_{n} \end{array} \right$$\nonumber \\ Then \\\left\\ \vec{u}\_{i\_{1}},\cdots ,\vec{u}\_{i\_{k}}\right\\\\ spans \\\mathbb{R}^{n}\\ and is linearly independent, so it is a basis having less than \\n\\ vectors again contrary to Corollary $\PageIndex{3}$.

The next theorem follows from the above claim.

##### Theorem \\\PageIndex{8}\\: Existence of Basis

Let \\V\\ be a subspace of \\\mathbb{R}^n\\. Then there exists a basis of \\V\\ with \\\dim(V)\leq n\\.

Consider Corollary $\PageIndex{4}$ together with Theorem $\PageIndex{8}$. Let \\\dim(V) = r\\. Suppose there exists an independent set of vectors in \\V\\. If this set contains \\r\\ vectors, then it is a basis for \\V\\. If it contains less than \\r\\ vectors, then vectors can be added to the set to create a basis of \\V\\. Similarly, any spanning set of \\V\\ which contains more than \\r\\ vectors can have vectors removed to create a basis of \\V\\.

We illustrate this concept in the next example.

##### Example \\\PageIndex{12}\\: Extending an Independent Set

Consider the set \\U\\ given by \\U=\left\\ \left.\left$$\begin{array}{c} a\\ b\\ c\\ d\end{array}\right$$ \in\mathbb{R}^4 ~\right\|~ a-b=d-c \right\\\nonumber \\ Then \\U\\ is a subspace of \\\mathbb{R}^4\\ and \\\dim(U)=3\\.

Then \\S=\left\\ \left$$\begin{array}{c} 1\\ 1\\ 1\\ 1\end{array}\right$$, \left$$\begin{array}{c} 2\\ 3\\ 3\\ 2\end{array}\right$$ \right\\,\nonumber \\ is an independent subset of \\U\\. Therefore \\S\\ can be extended to a basis of \\U\\.

###### Solution

To extend \\S\\ to a basis of \\U\\, find a vector in \\U\\ that is not in \\\mathrm{span}(S)\\. \\\left$$\begin{array}{rrr} 1 & 2 & ? \\ 1 & 3 & ? \\ 1 & 3 & ? \\ 1 & 2 & ? \end{array}\right$$\nonumber \\

\\\left$$\begin{array}{rrr} 1 & 2 & 1 \\ 1 & 3 & 0 \\ 1 & 3 & -1 \\ 1 & 2 & 0 \end{array}\right$$ \rightarrow \left$$\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right$$\nonumber \\

Therefore, \\S\\ can be extended to the following basis of \\U\\: \\\left\\ \left$$\begin{array}{r} 1\\ 1\\ 1\\ 1\end{array}\right$$, \left$$\begin{array}{r} 2\\ 3\\ 3\\ 2\end{array}\right$$, \left$$\begin{array}{r} 1\\ 0\\ -1\\ 0\end{array}\right$$ \right\\,\nonumber \\

Next we consider the case of removing vectors from a spanning set to result in a basis.

##### Theorem \\\PageIndex{9}\\: Finding a Basis from a Span

Let \\W\\ be a subspace. Also suppose that \\W=span\left\\ \vec{w} \_{1},\cdots ,\vec{w}\_{m}\right\\\\. Then there exists a subset of \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{m}\right\\\\ which is a basis for \\W\\.

Proof

Let \\S\\ denote the set of positive integers such that for \\k\in S,\\ there exists a subset of \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{m}\right\\\\ consisting of exactly \\k\\ vectors which is a spanning set for \\W\\. Thus \\m\in S\\. Pick the smallest positive integer in \\S\\. Call it \\k\\. Then there exists \\\left\\ \vec{u}\_{1},\cdots , \vec{u}\_{k}\right\\ \subseteq \left\\ \vec{w}\_{1},\cdots ,\vec{w} \_{m}\right\\\\ such that \\\text{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u} \_{k}\right\\ =W.\\ If \\\sum\_{i=1}^{k}c\_{i}\vec{w}\_{i}=\vec{0}\nonumber \\ and not all of the \\c\_{i}=0,\\ then you could pick \\c\_{j}\neq 0\\, divide by it and solve for \\\vec{u}\_{j}\\ in terms of the others, \\\vec{w}\_{j}=\sum\_{i\neq j}\left( -\frac{c\_{i}}{c\_{j}}\right) \vec{w}\_{i}\nonumber \\ Then you could delete \\\vec{w}\_{j}\\ from the list and have the same span. Any linear combination involving \\\vec{w}\_{j}\\ would equal one in which \\\vec{w}\_{j}\\ is replaced with the above sum, showing that it could have been obtained as a linear combination of \\\vec{w}\_{i}\\ for \\i\neq j\\. Thus \\k-1\in S\\ contrary to the choice of \\k\\. Hence each \\c\_{i}=0\\ and so \\\left\\ \vec{u}\_{1},\cdots ,\vec{u} \_{k}\right\\\\ is a basis for \\W\\ consisting of vectors of \\\left\\ \vec{w} \_{1},\cdots ,\vec{w}\_{m}\right\\\\.

The following example illustrates how to carry out this shrinking process which will obtain a subset of a span of vectors which is linearly independent.

##### Example \\\PageIndex{13}\\: Subset of a Span

Let \\W\\ be the subspace \\span\left\\ \left$$ \begin{array}{r} 1 \\ 2 \\ -1 \\ 1 \end{array} \right$$ ,\left$$ \begin{array}{r} 1 \\ 3 \\ -1 \\ 1 \end{array} \right$$ ,\left$$ \begin{array}{r} 8 \\ 19 \\ -8 \\ 8 \end{array} \right$$ ,\left$$ \begin{array}{r} -6 \\ -15 \\ 6 \\ -6 \end{array} \right$$ ,\left$$ \begin{array}{r} 1 \\ 3 \\ 0 \\ 1 \end{array} \right$$ ,\left$$ \begin{array}{r} 1 \\ 5 \\ 0 \\ 1 \end{array} \right$$ \right\\\nonumber \\ Find a basis for \\W\\ which consists of a subset of the given vectors.

###### Solution

You can use the reduced row-echelon form to accomplish this reduction. Form the matrix which has the given vectors as columns. \\\left$$ \begin{array}{rrrrrr} 1 & 1 & 8 & -6 & 1 & 1 \\ 2 & 3 & 19 & -15 & 3 & 5 \\ -1 & -1 & -8 & 6 & 0 & 0 \\ 1 & 1 & 8 & -6 & 1 & 1 \end{array} \right$$\nonumber \\ Then take the reduced row-echelon form

\\\left$$ \begin{array}{rrrrrr} 1 & 0 & 5 & -3 & 0 & -2 \\ 0 & 1 & 3 & -3 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right$$\nonumber \\ It follows that a basis for \\W\\ is

\\\left\\ \left$$ \begin{array}{r} 1 \\ 2 \\ -1 \\ 1 \end{array} \right$$ ,\left$$ \begin{array}{r} 1 \\ 3 \\ -1 \\ 1 \end{array} \right$$ ,\left$$ \begin{array}{c} 1 \\ 3 \\ 0 \\ 1 \end{array} \right$$ \right\\\nonumber \\ Since the first, second, and fifth columns are obviously a basis for the column space of the , the same is true for the matrix having the given vectors as columns.

Consider the following theorems regarding a subspace contained in another subspace.

##### Theorem \\\PageIndex{10}\\: Subset of a Subspace

Let \\V\\ and \\W\\ be subspaces of \\\mathbb{R}^n\\, and suppose that \\W\subseteq V\\. Then \\\dim(W) \leq \dim(V)\\ with equality when \\W=V\\.

##### Theorem \\\PageIndex{11}\\: Extending a Basis

Let \\W\\ be any non-zero subspace \\\mathbb{R}^{n}\\ and let \\W\subseteq V\\ where \\V\\ is also a subspace of \\\mathbb{R}^{n}\\. Then every basis of \\W\\ can be extended to a basis for \\V\\.

The proof is left as an exercise but proceeds as follows. Begin with a basis for \\W,\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{s}\right\\\\ and add in vectors from \\V\\ until you obtain a basis for \\V\\. Not that the process will stop because the dimension of \\V\\ is no more than \\n\\.

Consider the following example.

##### Example \\\PageIndex{14}\\: Extending a Basis

Let \\V=\mathbb{R}^{4}\\ and let \\W=\mathrm{span}\left\\ \left$$ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right$$ ,\left$$ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array} \right$$ \right\\\nonumber \\ Extend this basis of \\W\\ to a basis of \\\mathbb{R}^{n}\\.

###### Solution

An easy way to do this is to take the reduced row-echelon form of the matrix

\\\left$$ \begin{array}{cccccc} 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \end{array} \right$$ \label{basiseq1}\\

Note how the given vectors were placed as the first two columns and then the matrix was extended in such a way that it is clear that the span of the columns of this matrix yield all of \\\mathbb{R}^{4}\\. Now determine the pivot columns. The reduced row-echelon form is

\\\left$$ \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 1 & -1 \end{array} \right$$ \label{basiseq2}\\

Therefore the pivot columns are \\\left$$ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right$$ ,\left$$ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array} \right$$ ,\left$$ \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right$$ ,\left$$ \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \right$$\nonumber \\

and now this is an extension of the given basis for \\W\\ to a basis for \\\mathbb{R}^{4}\\.

Why does this work? The columns of \\\eqref{basiseq1}\\ obviously span \\\mathbb{R }^{4}\\. In fact the span of the first four is the same as the span of all six.

Consider another example.

##### Example \\\PageIndex{15}\\: Extending a Basis

Let \\W\\ be the span of \\\left$$ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right$$\\ in \\\mathbb{R}^{4}\\. Let \\V\\ consist of the span of the vectors \\\left$$ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right$$ ,\left$$ \begin{array}{c} 0 \\ 1 \\ 1 \\ 1 \end{array} \right$$ ,\left$$ \begin{array}{r} 7 \\ -6 \\ 1 \\ -6 \end{array} \right$$ ,\left$$ \begin{array}{r} -5 \\ 7 \\ 2 \\ 7 \end{array} \right$$ ,\left$$ \begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right$$\nonumber \\ Find a basis for \\V\\ which extends the basis for \\W\\.

###### Solution

Note that the above vectors are not linearly independent, but their span, denoted as \\V\\ is a subspace which does include the subspace \\W\\.

Using the process outlined in the previous example, form the following matrix

\\\left$$ \begin{array}{rrrrr} 1 & 0 & 7 & -5 & 0 \\ 0 & 1 & -6 & 7 & 0 \\ 1 & 1 & 1 & 2 & 0 \\ 0 & 1 & -6 & 7 & 1 \end{array} \right$$\nonumber \\

Next find its reduced row-echelon form \\\left$$ \begin{array}{rrrrr} 1 & 0 & 7 & -5 & 0 \\ 0 & 1 & -6 & 7 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right$$\nonumber \\

It follows that a basis for \\V\\ consists of the first two vectors and the last. \\\left\\ \left$$ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right$$ ,\left$$ \begin{array}{c} 0 \\ 1 \\ 1 \\ 1 \end{array} \right$$ ,\left$$ \begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right$$ \right\\\nonumber \\ Thus \\V\\ is of dimension 3 and it has a basis which extends the basis for \\W\\.

Row Space, Column Space, and Null Space of a Matrix

We begin this section with a new definition.

##### Definition \\\PageIndex{9}\\: Row and Column Space

Let \\A\\ be an \\m\times n\\ matrix. The column space of \\A\\, written \\\mathrm{col}(A)\\, is the span of the columns. The row space of \\A\\, written \\\mathrm{row}(A)\\, is the span of the rows.

Using the reduced row-echelon form, we can obtain an efficient description of the row and column space of a matrix. Consider the following lemma.

##### Lemma \\\PageIndex{1}\\: Effect of Row Operations on Row Space

Let \\A\\ and \\B\\ be \\m\times n\\ matrices such that \\A\\ can be carried to \\B\\ by elementary row \\\left$$ \mbox{column} \right$$\\ operations. Then \\\mathrm{row}(A)=\mathrm{row}(B)\\ \\\left$$\mathrm{col}(A)=\mathrm{col}(B) \right$$\\.

Proof

We will prove that the above is true for row operations, which can be easily applied to column operations.

Let \\\vec{r}\_1, \vec{r}\_2, \ldots, \vec{r}\_m\\ denote the rows of \\A\\.

• If \\B\\ is obtained from \\A\\ by a interchanging two rows of \\A\\, then \\A\\ and \\B\\ have exactly the same rows, so \\\mathrm{row}(B)=\mathrm{row}(A)\\.
• Suppose \\p\neq 0\\, and suppose that for some \\j\\, \\1\leq j\leq m\\, \\B\\ is obtained from \\A\\ by multiplying row \\j\\ by \\p\\. Then \\\mathrm{row}(B)=\mathrm{span}\\ \vec{r}\_1, \ldots, p\vec{r}\_{j}, \ldots, \vec{r}\_m\\.\nonumber \\ Since \\\\ \vec{r}\_1, \ldots, p\vec{r}\_{j}, \ldots, \vec{r}\_m\\ \subseteq\mathrm{row}(A),\nonumber \\ it follows that \\\mathrm{row}(B)\subseteq\mathrm{row}(A)\\. Conversely, since \\\\ \vec{r}\_1, \ldots, \vec{r}\_m\\\subseteq\mathrm{row}(B),\nonumber \\ it follows that \\\mathrm{row}(A)\subseteq\mathrm{row}(B)\\. Therefore, \\\mathrm{row}(B)=\mathrm{row}(A)\\.

Suppose \\p\neq 0\\, and suppose that for some \\i\\ and \\j\\, \\1\leq i,j\leq m\\, \\B\\ is obtained from \\A\\ by adding \\p\\ time row \\j\\ to row \\i\\. Without loss of generality, we may assume \\i\

Then \\\mathrm{row}(B)=\mathrm{span}\\ \vec{r}\_1, \ldots, \vec{r}\_{i-1}, \vec{r}\_i+p\vec{r}\_j, \ldots,\vec{r}\_j,\ldots, \vec{r}\_m\\.\nonumber \\

Since \\\\ \vec{r}\_1, \ldots, \vec{r}\_{i-1}, \vec{r}\_i+p\vec{r}\_{j}, \ldots, \vec{r}\_m\\ \subseteq\mathrm{row}(A),\nonumber \\ it follows that \\\mathrm{row}(B)\subseteq\mathrm{row}(A)\\.

Conversely, since \\\\ \vec{r}\_1, \ldots, \vec{r}\_m\\\subseteq\mathrm{row}(B),\nonumber \\ it follows that \\\mathrm{row}(A)\subseteq\mathrm{row}(B)\\. Therefore, \\\mathrm{row}(B)=\mathrm{row}(A)\\.

Consider the following lemma.

##### Lemma \\\PageIndex{2}\\: Row Space of a reduced row-echelon form Matrix

Let \\A\\ be an \\m \times n\\ matrix and let \\R\\ be its reduced row-echelon form. Then the nonzero rows of \\R\\ form a basis of \\\mathrm{row}(R)\\, and consequently of \\\mathrm{row}(A)\\.

This lemma suggests that we can examine the reduced row-echelon form of a matrix in order to obtain the row space. Consider now the column space. The column space can be obtained by simply saying that it equals the span of all the columns. However, you can often get the column space as the span of fewer columns than this. A variation of the previous lemma provides a solution. Suppose \\A\\ is row reduced to its reduced row-echelon form \\R\\. Identify the pivot columns of \\R\\ (columns which have leading ones), and take the corresponding columns of \\A\\. It turns out that this forms a basis of \\\mathrm{col}(A)\\.

Before proceeding to an example of this concept, we revisit the definition of rank.

##### Definition \\\PageIndex{10}\\: Rank of a Matrix

Previously, we defined \\\mathrm{rank}(A)\\ to be the number of leading entries in the row-echelon form of \\A\\. Using an understanding of dimension and row space, we can now define rank as follows: \\\mbox{rank}(A) = \dim(\mathrm{row}(A))\nonumber \\

Consider the following example.

##### Example \\\PageIndex{16}\\: Rank, Column and Row Space

Find the rank of the following matrix and describe the column and row spaces. \\A = \left$$ \begin{array}{rrrrr} 1 & 2 & 1 & 3 & 2 \\ 1 & 3 & 6 & 0 & 2 \\ 3 & 7 & 8 & 6 & 6 \end{array} \right$$\nonumber \\

###### Solution

The reduced row-echelon form of \\A\\ is \\\left$$ \begin{array}{rrrrr} 1 & 0 & -9 & 9 & 2 \\ 0 & 1 & 5 & -3 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right$$\nonumber \\ Therefore, the rank is \\2\\.

Notice that the first two columns of \\R\\ are pivot columns. By the discussion following Lemma $\PageIndex{2}$, we find the corresponding columns of \\A\\, in this case the first two columns. Therefore a basis for \\\mathrm{col}(A)\\ is given by \\\left\\\left$$ \begin{array}{r} 1 \\ 1 \\ 3 \end{array} \right$$ , \left$$ \begin{array}{r} 2 \\ 3 \\ 7 \end{array} \right$$ \right\\\nonumber \\

For example, consider the third column of the original matrix. It can be written as a linear combination of the first two columns of the original matrix as follows. \\\left$$ \begin{array}{r} 1 \\ 6 \\ 8 \end{array} \right$$ =-9\left$$ \begin{array}{r} 1 \\ 1 \\ 3 \end{array} \right$$ +5\left$$ \begin{array}{r} 2 \\ 3 \\ 7 \end{array} \right$$\nonumber \\

What about an efficient description of the row space? By Lemma $\PageIndex{2}$ we know that the nonzero rows of \\R\\ create a basis of \\\mathrm{row}(A)\\. For the above matrix, the row space equals \\\mathrm{row}(A) = \mathrm{span} \left\\ \left$$ \begin{array}{rrrrr} 1 & 0 & -9 & 9 & 2 \end{array} \right$$, \left$$ \begin{array}{rrrrr} 0 & 1 & 5 & -3 & 0 \end{array} \right$$ \right\\\nonumber \\

Notice that the column space of \\A\\ is given as the span of columns of the original matrix, while the row space of \\A\\ is the span of rows of the reduced row-echelon form of \\A\\.

Consider another example.

##### Example \\\PageIndex{17}\\: Rank, Column and Row Space

Find the rank of the following matrix and describe the column and row spaces. \\\left$$ \begin{array}{rrrrrr} 1 & 2 & 1 & 3 & 2 \\ 1 & 3 & 6 & 0 & 2 \\ 1 & 2 & 1 & 3 & 2 \\ 1 & 3 & 2 & 4 & 0 \end{array} \right$$\nonumber \\

###### Solution

The reduced row-echelon form is \\\left$$ \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & \frac{13}{2} \\ 0 & 1 & 0 & 2 & -\frac{5}{2} \\ 0 & 0 & 1 & -1 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{array} \right$$\nonumber \\ and so the rank is \\3\\. The row space is given by \\\mathrm{row}(A) = \mathrm{span} \left\\ \left$$ \begin{array}{ccccc} 1 & 0 & 0 & 0 & \frac{13}{2} \end{array} \right$$, \left$$ \begin{array}{rrrrr} 0 & 1 & 0 & 2 & -\frac{5}{2} \end{array} \right$$ , \left$$ \begin{array}{rrrrr} 0 & 0 & 1 & -1 & \frac{1}{2} \end{array} \right$$ \right\\\nonumber \\

Notice that the first three columns of the reduced row-echelon form are pivot columns. The column space is the span of the first three columns in the original matrix, \\\mathrm{col}(A) = \mathrm{span} \left\\ \left$$ \begin{array}{r} 1 \\ 1 \\ 1 \\ 1 \end{array} \right$$, \\ \left$$ \begin{array}{r} 2 \\ 3 \\ 2 \\ 3 \end{array} \right$$ , \\ \left$$ \begin{array}{r} 1 \\ 6 \\ 1 \\ 2 \end{array} \right$$ \right\\\nonumber \\

Consider the solution given above for Example $\PageIndex{17}$, where the rank of \\A\\ equals \\3\\. Notice that the row space and the column space each had dimension equal to \\3\\. It turns out that this is not a coincidence, and this essential result is referred to as the Rank Theorem and is given now. Recall that we defined \\\mathrm{rank}(A) = \mathrm{dim}(\mathrm{row}(A))\\.

##### Theorem \\\PageIndex{12}\\: Rank Theorem

Let \\A\\ be an \\m \times n\\ matrix. Then \\\mathrm{dim}(\mathrm{col} (A))\\, the dimension of the column space, is equal to the dimension of the row space, \\\mathrm{dim}(\mathrm{row}(A))\\.

The following statements all follow from the Rank Theorem.

##### Corollary \\\PageIndex{5}\\: Results of the Rank Theorem

Let \\A\\ be a matrix. Then the following are true:

1. \\\mathrm{rank}(A) = \mathrm{rank}(A^T)\\.

2. For \\A\\ of size \\m \times n\\, \\\mathrm{rank}(A) \leq m\\ and \\\mathrm{rank}(A) \leq n\\.

3. For \\A\\ of size \\n \times n\\, \\A\\ is invertible if and only if \\\mathrm{rank}(A) = n\\.

4. For invertible matrices \\B\\ and \\C\\ of appropriate size, \\\mathrm{rank}(A) = \mathrm{rank}(BA) = \mathrm{rank}(AC)\\.

Consider the following example.

##### Example \\\PageIndex{18}\\: Rank of the Transpose

Let \\A = \left$$ \begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array} \right$$\nonumber \\ Find \\\mathrm{rank}(A)\\ and \\\mathrm{rank}(A^T)\\.

###### Solution

To find \\\mathrm{rank}(A)\\ we first row reduce to find the reduced row-echelon form. \\A = \left$$ \begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array} \right$$ \rightarrow \cdots \rightarrow \left$$ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right$$\nonumber \\

Therefore the rank of \\A\\ is \\2\\. Now consider \\A^T\\ given by \\A^T = \left$$ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right$$\nonumber \\ Again we row reduce to find the reduced row-echelon form.

\\\left$$ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right$$ \rightarrow \cdots \rightarrow \left$$ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right$$\nonumber \\

You can see that \\\mathrm{rank}(A^T) = 2\\, the same as \\\mathrm{rank}(A)\\.

We now define what is meant by the null space of a general \\m\times n\\ matrix.

##### Definition \\\PageIndex{11}\\: Null Space, or Kernel, of \\A\\

The null space of a matrix \\A\\, also referred to as the kernel of \\A\\, is defined as follows. \\\mathrm{null} \left( A\right) =\left\\ \vec{x} :A \vec{x} =\vec{0}\right\\\nonumber \\

It can also be referred to using the notation \\\ker \left( A\right)\\. Similarly, we can discuss the image of \\A\\, denoted by \\\mathrm{im}\left( A\right)\\. The image of \\A\\ consists of the vectors of \\\mathbb{R}^{m}\\ which “get hit” by \\A\\. The formal definition is as follows.

##### Definition \\\PageIndex{12}\\: Image of \\A\\

The image of \\A\\, written \\\mathrm{im}\left( A\right)\\ is given by \\\mathrm{im}\left( A \right) = \left\\ A\vec{x} : \vec{x} \in \mathbb{R}^n \right\\\nonumber \\

Consider \\A\\ as a mapping from \\\mathbb{R}^{n}\\ to \\\mathbb{R}^{m}\\ whose action is given by multiplication. The following diagram displays this scenario. \\\overset{\mathrm{null} \left( A\right) }{\mathbb{R}^{n}}\\ \overset{A}{\rightarrow }\\ \overset{ \mathrm{im}\left( A\right) }{\mathbb{R}^{m}}\nonumber \\ As indicated, \\\mathrm{im}\left( A\right)\\ is a subset of \\\mathbb{R}^{m}\\ while \\\mathrm{null} \left( A\right)\\ is a subset of \\\mathbb{R}^{n}\\.

It turns out that the null space and image of \\A\\ are both subspaces. Consider the following example.

##### Example \\\PageIndex{19}\\: Null Space

Let \\A\\ be an \\m\times n\\ matrix. Then the null space of \\A\\, \\\mathrm{null}(A)\\ is a subspace of \\\mathbb{R}^n\\.

###### Solution

• Since \\A\vec{0}\_n=\vec{0}\_m\\, \\\vec{0}\_n\in\mathrm{null}(A)\\.
• Let \\\vec{x},\vec{y}\in\mathrm{null}(A)\\. Then \\A\vec{x}=\vec{0}\_m\\ and \\A\vec{y}=\vec{0}\_m\\, so \\A(\vec{x}+\vec{y})=A\vec{x}+A\vec{y} = \vec{0}\_m+\vec{0}\_m=\vec{0}\_m,\nonumber \\ and thus \\\vec{x}+\vec{y}\in\mathrm{null}(A)\\.
• Let \\\vec{x}\in\mathrm{null}(A)\\ and \\k\in\mathbb{R}\\. Then \\A\vec{x}=\vec{0}\_m\\, so \\A(k\vec{x}) = k(A\vec{x})=k\vec{0}\_m=\vec{0}\_m,\nonumber \\ and thus \\k\vec{x}\in\mathrm{null}(A)\\.

Therefore by the subspace test, \\\mathrm{null}(A)\\ is a subspace of \\\mathbb{R}^n\\.

The proof that \\\mathrm{im}(A)\\ is a subspace of \\\mathbb{R}^m\\ is similar and is left as an exercise to the reader.

We now wish to find a way to describe \\\mathrm{null}(A)\\ for a matrix \\A\\. However, finding \\\mathrm{null} \left( A\right)\\ is not new! There is just some new terminology being used, as \\\mathrm{null} \left( A\right)\\ is simply the solution to the system \\A\vec{x}=\vec{0}\\.

##### Theorem \\\PageIndex{13}\\: Basis of null(A)

Let \\A\\ be an \\m \times n\\ matrix such that \\\mathrm{rank}(A) = r\\. Then the system \\A\vec{x}=\vec{0}\_m\\ has \\n-r\\ basic solutions, providing a basis of \\\mathrm{null}(A)\\ with \\\dim(\mathrm{null}(A))=n-r\\.

Consider the following example.

##### Example \\\PageIndex{20}\\: Null Space of \\A\\

Let \\A=\left$$ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & -1 & 1 \\ 2 & 3 & 3 \end{array} \right$$\nonumber \\ Find \\\mathrm{null} \left( A\right)\\ and \\\mathrm{im}\left( A\right)\\.

###### Solution

In order to find \\\mathrm{null} \left( A\right)\\, we simply need to solve the equation \\A\vec{x}=\vec{0}\\. This is the usual procedure of writing the augmented matrix, finding the reduced row-echelon form and then the solution. The augmented matrix and corresponding reduced row-echelon form are \\\left$$ \begin{array}{rrr\|r} 1 & 2 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 2 & 3 & 3 & 0 \end{array} \right$$ \rightarrow \cdots \rightarrow \left$$ \begin{array}{rrr\|r} 1 & 0 & 3 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right$$\nonumber \\

The third column is not a pivot column, and therefore the solution will contain a parameter. The solution to the system \\A\vec{x}=\vec{0}\\ is given by \\\left$$ \begin{array}{r} -3t \\ t \\ t \end{array} \right$$ :t\in \mathbb{R}\nonumber \\ which can be written as \\t \left$$ \begin{array}{r} -3 \\ 1 \\ 1 \end{array} \right$$ :t\in \mathbb{R}\nonumber \\

Therefore, the null space of \\A\\ is all multiples of this vector, which we can write as \\\mathrm{null} (A) = \mathrm{span} \left\\ \left$$ \begin{array}{r} -3 \\ 1 \\ 1 \end{array} \right$$ \right\\\nonumber \\

Finally \\\mathrm{im}\left( A\right)\\ is just \\\left\\ A\vec{x} : \vec{x} \in \mathbb{R}^n \right\\\\ and hence consists of the span of all columns of \\A\\, that is \\\mathrm{im}\left( A\right) = \mathrm{col} (A)\\.

Notice from the above calculation that that the first two columns of the reduced row-echelon form are pivot columns. Thus the column space is the span of the first two columns in the original matrix, and we get \\\mathrm{im}\left( A\right) = \mathrm{col}(A) = \mathrm{span} \left\\ \left$$ \begin{array}{r} 1 \\ 0 \\ 2 \end{array} \right$$, \\ \left$$ \begin{array}{r} 2 \\ -1 \\ 3 \end{array} \right$$ \right\\\nonumber \\

Here is a larger example, but the method is entirely similar.

##### Example \\\PageIndex{21}\\: Null Space of \\A\\

Let \\A=\left$$ \begin{array}{rrrrr} 1 & 2 & 1 & 0 & 1 \\ 2 & -1 & 1 & 3 & 0 \\ 3 & 1 & 2 & 3 & 1 \\ 4 & -2 & 2 & 6 & 0 \end{array} \right$$\nonumber \\ Find the null space of \\A\\.

###### Solution

To find the null space, we need to solve the equation \\AX=0\\. The augmented matrix and corresponding reduced row-echelon form are given by

\\\left$$ \begin{array}{rrrrr\|r} 1 & 2 & 1 & 0 & 1 & 0 \\ 2 & -1 & 1 & 3 & 0 & 0 \\ 3 & 1 & 2 & 3 & 1 & 0 \\ 4 & -2 & 2 & 6 & 0 & 0 \end{array} \right$$ \rightarrow \cdots \rightarrow \left$$ \begin{array}{rrrrr\|r} 1 & 0 & \frac{3}{5} & \frac{6}{5} & \frac{1}{5} & 0 \\ 0 & 1 & \frac{1}{5} & -\frac{3}{5} & \frac{2}{5} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right$$\nonumber \\ It follows that the first two columns are pivot columns, and the next three correspond to parameters. Therefore, \\\mathrm{null} \left( A\right)\\ is given by \\\left$$ \begin{array}{c} \left( -\frac{3}{5}\right) s +\left( -\frac{6}{5}\right) t+\left( \frac{1}{5}\right) r \\ \left( -\frac{1}{5}\right) s +\left( \frac{3}{5}\right) t +\left( - \frac{2}{5}\right) r \\ s \\ t \\ r \end{array} \right$$ :s ,t ,r\in \mathbb{R}\text{.}\nonumber \\ We write this in the form \\s \left$$ \begin{array}{r} -\frac{3}{5} \\ -\frac{1}{5} \\ 1 \\ 0 \\ 0 \end{array} \right$$ + t \left$$ \begin{array}{r} -\frac{6}{5} \\ \frac{3}{5} \\ 0 \\ 1 \\ 0 \end{array} \right$$ + r \left$$ \begin{array}{r} \frac{1}{5} \\ -\frac{2}{5} \\ 0 \\ 0 \\ 1 \end{array} \right$$ :s , t , r\in \mathbb{R}\text{.}\nonumber \\ In other words, the null space of this matrix equals the span of the three vectors above. Thus \\\mathrm{null} \left( A\right) =\mathrm{span}\left\\ \left$$ \begin{array}{r} -\frac{3}{5} \\ -\frac{1}{5} \\ 1 \\ 0 \\ 0 \end{array} \right$$ ,\left$$ \begin{array}{r} -\frac{6}{5} \\ \frac{3}{5} \\ 0 \\ 1 \\ 0 \end{array} \right$$ ,\left$$ \begin{array}{r} \frac{1}{5} \\ -\frac{2}{5} \\ 0 \\ 0 \\ 1 \end{array} \right$$ \right\\\nonumber \\

Notice also that the three vectors above are linearly independent and so the dimension of \\\mathrm{null} \left( A\right)\\ is 3. The following is true in general, the number of parameters in the solution of \\AX=0\\ equals the dimension of the null space. Recall also that the number of leading ones in the reduced row-echelon form equals the number of pivot columns, which is the rank of the matrix, which is the same as the dimension of either the column or row space.

Before we proceed to an important theorem, we first define what is meant by the nullity of a matrix.

##### Definition \\\PageIndex{13}\\: Nullity

The dimension of the null space of a matrix is called the nullity, denoted \\\dim( \mathrm{null}\left(A\right))\\.

From our observation above we can now state an important theorem.

##### Theorem \\\PageIndex{14}\\: Rank and Nullity

Let \\A\\ be an \\m\times n\\ matrix. Then \\\mathrm{rank}\left( A\right) + \dim( \mathrm{null}\left(A\right)) =n\\.

##### Example \\\PageIndex{22}\\: Rank and Nullity

Let \\A=\left$$ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & -1 & 1 \\ 2 & 3 & 3 \end{array} \right$$\nonumber \\

Find \\\mathrm{rank}\left( A\right)\\ and \\\dim( \mathrm{null}\left(A\right))\\.

###### Solution

In the above Example $\PageIndex{20}$ we determined that the reduced row-echelon form of \\A\\ is given by \\\left$$ \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array} \right$$\nonumber \\

Therefore the rank of \\A\\ is \\2\\. We also determined that the null space of \\A\\ is given by \\\mathrm{null} (A) = \mathrm{span} \left\\ \left$$ \begin{array}{r} -3 \\ 1 \\ 1 \end{array} \right$$ \right\\\nonumber \\

Therefore the nullity of \\A\\ is \\1\\. It follows from Theorem $\PageIndex{14}$ that \\\mathrm{rank}\left( A\right) + \dim( \mathrm{null}\left(A\right)) = 2 + 1 = 3\\, which is the number of columns of \\A\\.

We conclude this section with two similar, and important, theorems.

##### Theorem \\\PageIndex{15}\\

Let \\A\\ be an \\m\times n\\ matrix. The following are equivalent.

1. \\\mathrm{rank}(A)=n\\.

2. \\\mathrm{row}(A)=\mathbb{R}^n\\, i.e., the rows of \\A\\ span \\\mathbb{R}^n\\.

3. The columns of \\A\\ are independent in \\\mathbb{R}^m\\.

4. The \\n\times n\\ matrix \\A^TA\\ is invertible.

5. There exists an \\n\times m\\ matrix \\C\\ so that \\CA=I_n\\.

6. If \\A\vec{x}=\vec{0}\_m\\ for some \\\vec{x}\in\mathbb{R}^n\\, then \\\vec{x}=\vec{0}\_n\\.

##### Theorem \\\PageIndex{16}\\

Let \\A\\ be an \\m\times n\\ matrix. The following are equivalent.

1. \\\mathrm{rank}(A)=m\\.

2. \\\mathrm{col}(A)=\mathbb{R}^m\\, i.e., the columns of \\A\\ span \\\mathbb{R}^m\\.

3. The rows of \\A\\ are independent in \\\mathbb{R}^n\\.

4. The \\m\times m\\ matrix \\AA^T\\ is invertible.

5. There exists an \\n\times m\\ matrix \\C\\ so that \\AC=I_m\\.

6. The system \\A\vec{x}=\vec{b}\\ is consistent for every \\\vec{b}\in\mathbb{R}^m\\.

---

4_11_3A_Orthogonality

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.11%3A_Orthogonality

Skip to main content

##### Outcomes

1. Determine if a given set is orthogonal or orthonormal.

2. Determine if a given matrix is orthogonal.

3. Given a linearly independent set, use the Gram-Schmidt Process to find corresponding orthogonal and orthonormal sets.

4. Find the orthogonal projection of a vector onto a subspace.

5. Find the least squares approximation for a collection of points.

In this section, we examine what it means for vectors (and sets of vectors) to be orthogonal and orthonormal. First, it is necessary to review some important concepts. You may recall the definitions for the span of a set of vectors and a linear independent set of vectors. We include the definitions and examples here for convenience.

##### Definition \\\PageIndex{1}\\: Span of a Set of Vectors and Subspace

The collection of all linear combinations of a set of vectors \\\\ \vec{u}\_1, \cdots ,\vec{u}\_k\\\\ in \\\mathbb{R}^{n}\\ is known as the span of these vectors and is written as \\\mathrm{span} \\\vec{u}\_1, \cdots , \vec{u}\_k\\\\.

We call a collection of the form \\\mathrm{span} \\\vec{u}\_1, \cdots , \vec{u}\_k\\\\ a subspace of \\\mathbb{R}^{n}\\.

Consider the following example.

##### Example \\\PageIndex{1}\\: Spanning Vectors

Describe the span of the vectors \\\vec{u}=\left$$ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right$$^T\\ and \\\vec{v}=\left$$ \begin{array}{rrr} 3 & 2 & 0 \end{array} \right$$^T \in \mathbb{R}^{3}\\.

###### Solution

You can see that any linear combination of the vectors \\\vec{u}\\ and \\\vec{v}\\ yields a vector \\\left$$ \begin{array}{rrr} x & y & 0 \end{array} \right$$^T\\ in the \\XY\\-plane.

Moreover every vector in the \\XY\\-plane is in fact such a linear combination of the vectors \\\vec{u}\\ and \\\vec{v}\\. That’s because \\\left$$ \begin{array}{r} x \\ y \\ 0 \end{array} \right$$ = (-2x+3y) \left$$ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right$$ + (x-y)\left$$ \begin{array}{r} 3 \\ 2 \\ 0 \end{array} \right$$\nonumber \\

Thus span\\\\\vec{u},\vec{v}\\\\ is precisely the \\XY\\-plane.

The span of a set of a vectors in \\\mathbb{R}^n\\ is what we call a subspace of \\\mathbb{R}^n\\. A subspace \\W\\ is characterized by the feature that any linear combination of vectors of \\W\\ is again a vector contained in \\W\\.

Another important property of sets of vectors is called linear independence.

##### Definition \\\PageIndex{2}\\: Linear Independence

A set of non-zero vectors \\\\ \vec{u}\_1, \cdots ,\vec{u}\_k\\\\ in \\\mathbb{R}^{n}\\ is said to be linearly independent if no vector in that set is in the span of the other vectors of that set.

Here is an example.

##### Example \\\PageIndex{2}\\: Linearly Independent Vectors

Consider vectors \\\vec{u}=\left$$ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right$$^T\\, \\\vec{v}=\left$$ \begin{array}{rrr} 3 & 2 & 0 \end{array} \right$$^T\\, and \\\vec{w}=\left$$ \begin{array}{rrr} 4 & 5 & 0 \end{array} \right$$^T \in \mathbb{R}^{3}\\. Verify whether the set \\\\\vec{u}, \vec{v}, \vec{w}\\\\ is linearly independent.

###### Solution

We already verified in Example $\PageIndex{1}$ that \\\mathrm{span} \\\vec{u}, \vec{v} \\\\ is the \\XY\\-plane. Since \\\vec{w}\\ is clearly also in the \\XY\\-plane, then the set \\\\\vec{u}, \vec{v}, \vec{w}\\\\ is not linearly independent.

In terms of spanning, a set of vectors is linearly independent if it does not contain unnecessary vectors. In the previous example you can see that the vector \\\vec{w}\\ does not help to span any new vector not already in the span of the other two vectors. However you can verify that the set \\\\\vec{u}, \vec{v}\\\\ is linearly independent, since you will not get the \\XY\\-plane as the span of a single vector.

We can also determine if a set of vectors is linearly independent by examining linear combinations. A set of vectors is linearly independent if and only if whenever a linear combination of these vectors equals zero, it follows that all the coefficients equal zero. It is a good exercise to verify this equivalence, and this latter condition is often used as the (equivalent) definition of linear independence.

If a subspace is spanned by a linearly independent set of vectors, then we say that it is a basis for the subspace.

##### Definition \\\PageIndex{3}\\: Basis

Let \\V\\ be a subspace of \\\mathbb{R}^{n}\\. Then \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ is a basis for \\V\\ if the following two conditions hold.

1. \\\mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\ =V\\

2. \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ is linearly independent

Thus the set of vectors \\\\\vec{u}, \vec{v}\\\\ from Example $\PageIndex{2}$ is a basis for \\XY\\-plane in \\\mathbb{R}^{3}\\ since it is both linearly independent and spans the \\XY\\-plane.

Recall from the properties of the dot product of vectors that two vectors \\\vec{u}\\ and \\\vec{v}\\ are orthogonal if \\\vec{u} \cdot \vec{v} = 0\\. Suppose a vector is orthogonal to a spanning set of \\\mathbb{R}^n\\. What can be said about such a vector? This is the discussion in the following example.

##### Example \\\PageIndex{3}\\: Orthogonal Vector to a Spanning Set

Let \\\\\vec{x}\_1, \vec{x}\_2, \ldots, \vec{x}\_k\\\in\mathbb{R}^n\\ and suppose \\\mathbb{R}^n=\mathrm{span}\\\vec{x}\_1, \vec{x}\_2, \ldots, \vec{x}\_k\\\\. Furthermore, suppose that there exists a vector \\\vec{u}\in\mathbb{R}^n\\ for which \\\vec{u}\cdot \vec{x}\_j=0\\ for all \\j\\, \\1\leq j\leq k\\. What type of vector is \\\vec{u}\\?

###### Solution

Write \\\vec{u}=t_1\vec{x}\_1 + t_2\vec{x}\_2 +\cdots +t_k\vec{x}\_k\\ for some \\t_1, t_2, \ldots, t_k\in\mathbb{R}\\ (this is possible because \\\vec{x}\_1, \vec{x}\_2, \ldots, \vec{x}\_k\\ span \\\mathbb{R}^n\$.

Then

\\\begin{aligned} \\ \vec{u} \\ ^2 & = \vec{u}\cdot\vec{u} \\ & = \vec{u}\cdot(t_1\vec{x}\_1 + t_2\vec{x}\_2 +\cdots +t_k\vec{x}\_k) \\ & = \vec{u}\cdot (t_1\vec{x}\_1) + \vec{u}\cdot (t_2\vec{x}\_2) + \cdots + \vec{u}\cdot (t_k\vec{x}\_k) \\ & = t_1(\vec{u}\cdot \vec{x}\_1) + t_2(\vec{u}\cdot \vec{x}\_2) + \cdots + t_k(\vec{u}\cdot \vec{x}\_k) \\ & = t_1(0) + t_2(0) + \cdots + t_k(0) = 0.\end{aligned}\\

Since \\ \\ \vec{u} \\ ^2 =0\\, \\ \\ \vec{u} \\ =0\\. We know that \\ \\ \vec{u} \\ =0\\ if and only if \\\vec{u}=\vec{0}\_n\\. Therefore, \\\vec{u}=\vec{0}\_n\\. In conclusion, the only vector orthogonal to every vector of a spanning set of \\\mathbb{R}^n\\ is the zero vector.

We can now discuss what is meant by an orthogonal set of vectors.

##### Definition \\\PageIndex{4}\\: Orthogonal Set of Vectors

Let \\\\ \vec{u}\_1, \vec{u}\_2, \cdots, \vec{u}\_m \\\\ be a set of vectors in \\\mathbb{R}^n\\. Then this set is called an orthogonal set if the following conditions hold:

1. \\\vec{u}\_i \cdot \vec{u}\_j = 0\\ for all \\i \neq j\\

2. \\\vec{u}\_i \neq \vec{0}\\ for all \\i\\

If we have an orthogonal set of vectors and normalize each vector so they have length 1, the resulting set is called an orthonormal set of vectors. They can be described as follows.

##### Definition \\\PageIndex: {5}\\ Orthonormal Set of Vectors

A set of vectors, \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{m}\right\\\\ is said to be an orthonormal set if \\\vec{w}\_i \cdot \vec{w}\_j = \delta \_{ij} = \left\\ \begin{array}{c} 1\text{ if }i=j \\ 0\text{ if }i\neq j \end{array} \right.\nonumber \\

Note that all orthonormal sets are orthogonal, but the reverse is not necessarily true since the vectors may not be normalized. In order to normalize the vectors, we simply need divide each one by its length.

##### Definition \\\PageIndex{6}\\: Normalizing an Orthogonal Set

Normalizing an orthogonal set is the process of turning an orthogonal (but not orthonormal) set into an orthonormal set. If \\\\ \vec{u}\_1, \vec{u}\_2, \ldots, \vec{u}\_k\\\\ is an orthogonal subset of \\\mathbb{R}^n\\, then \\\left\\ \frac{1}{ \\ \vec{u}\_1 \\ }\vec{u}\_1, \frac{1}{ \\ \vec{u}\_2 \\ }\vec{u}\_2, \ldots, \frac{1}{ \\ \vec{u}\_k \\ }\vec{u}\_k \right\\\nonumber \\ is an orthonormal set.

We illustrate this concept in the following example.

##### Example \\\PageIndex{4}\\: Orthonormal Set

Consider the set of vectors given by \\\left\\ \vec{u}\_1, \vec{u}\_2 \right\\ = \left\\ \left$$ \begin{array}{c} 1 \\ 1 \end{array} \right$$, \left$$ \begin{array}{r} -1 \\ 1 \end{array} \right$$ \right\\\nonumber \\ Show that it is an orthogonal set of vectors but not an orthonormal one. Find the corresponding orthonormal set.

###### Solution

One easily verifies that \\\vec{u}\_1 \cdot \vec{u}\_2 = 0\\ and \\\left\\ \vec{u}\_1, \vec{u}\_2 \right\\\\ is an orthogonal set of vectors. On the other hand one can compute that \\ \\ \vec{u}\_1 \\ = \\ \vec{u}\_2 \\ = \sqrt{2} \neq 1\\ and thus it is not an orthonormal set.

Thus to find a corresponding orthonormal set, we simply need to normalize each vector. We will write \\\\ \vec{w}\_1, \vec{w}\_2 \\\\ for the corresponding orthonormal set. Then, \\\begin{aligned} \vec{w}\_1 &= \frac{1}{ \\ \vec{u}\_1 \\ } \vec{u}\_1\\ &= \frac{1}{\sqrt{2}} \left$$ \begin{array}{c} 1 \\ 1 \end{array} \right$$ \\ &= \left$$ \begin{array}{c} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{array} \right$$\end{aligned}\\

Similarly, \\\begin{aligned} \vec{w}\_2 &= \frac{1}{ \\ \vec{u}\_2 \\ } \vec{u}\_2\\ &= \frac{1}{\sqrt{2}} \left$$ \begin{array}{r} -1 \\ 1 \end{array} \right$$ \\ &= \left$$ \begin{array}{r} -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{array} \right$$\end{aligned}\\

Therefore the corresponding orthonormal set is \\\left\\ \vec{w}\_1, \vec{w}\_2 \right\\ = \left\\ \left$$ \begin{array}{c} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{array} \right$$, \left$$ \begin{array}{r} -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{array} \right$$ \right\\\nonumber \\

You can verify that this set is orthogonal.

Consider an orthogonal set of vectors in \\\mathbb{R}^n\\, written \\\\ \vec{w}\_1, \cdots, \vec{w}\_k \\\\ with \\k \leq n\\. The span of these vectors is a subspace \\W\\ of \\\mathbb{R}^n\\. If we could show that this orthogonal set is also linearly independent, we would have a basis of \\W\\. We will show this in the next theorem.

##### Theorem \\\PageIndex{1}\\: Orthogonal Basis of a Subspace

Let \\\\ \vec{w}\_1, \vec{w}\_2, \cdots, \vec{w}\_k \\\\ be an orthonormal set of vectors in \\\mathbb{R}^n\\. Then this set is linearly independent and forms a basis for the subspace \\W = \mathrm{span} \\ \vec{w}\_1, \vec{w}\_2, \cdots, \vec{w}\_k \\\\.

Proof

To show it is a linearly independent set, suppose a linear combination of these vectors equals \\\vec{0}\\, such as: \\a_1 \vec{w}\_1 + a_2 \vec{w}\_2 + \cdots + a_k \vec{w}\_k = \vec{0}, a_i \in \mathbb{R}\nonumber \\ We need to show that all \\a_i = 0\\. To do so, take the dot product of each side of the above equation with the vector \\\vec{w}\_i\\ and obtain the following.

\\\begin{aligned} \vec{w}\_i \cdot (a_1 \vec{w}\_1 + a_2 \vec{w}\_2 + \cdots + a_k \vec{w}\_k ) &= \vec{w}\_i \cdot \vec{0}\\ a_1 (\vec{w}\_i \cdot \vec{w}\_1) + a_2 (\vec{w}\_i \cdot \vec{w}\_2) + \cdots + a_k (\vec{w}\_i \cdot \vec{w}\_k) &= 0 \end{aligned}\\

Now since the set is orthogonal, \\\vec{w}\_i \cdot \vec{w}\_m = 0\\ for all \\m \neq i\\, so we have: \\a_1 (0) + \cdots + a_i(\vec{w}\_i \cdot \vec{w}\_i) + \cdots + a_k (0) = 0\nonumber \\ \\a_i \\ \vec{w}\_i \\ ^2 = 0\nonumber \\

Since the set is orthogonal, we know that \\ \\ \vec{w}\_i \\ ^2 \neq 0\\. It follows that \\a_i =0\\. Since the \\a_i\\ was chosen arbitrarily, the set \\\\ \vec{w}\_1, \vec{w}\_2, \cdots, \vec{w}\_k \\\\ is linearly independent.

Finally since \\W = \mbox{span} \\ \vec{w}\_1, \vec{w}\_2, \cdots, \vec{w}\_k \\\\, the set of vectors also spans \\W\\ and therefore forms a basis of \\W\\.

If an orthogonal set is a basis for a subspace, we call this an orthogonal basis. Similarly, if an orthonormal set is a basis, we call this an orthonormal basis.

We conclude this section with a discussion of Fourier expansions. Given any orthogonal basis \\B\\ of \\\mathbb{R}^n\\ and an arbitrary vector \\\vec{x} \in \mathbb{R}^n\\, how do we express \\\vec{x}\\ as a linear combination of vectors in \\B\\? The solution is Fourier expansion.

##### Theorem \\\PageIndex{2}\\: Fourier Expansion

Let \\V\\ be a subspace of \\\mathbb{R}^n\\ and suppose \\\\ \vec{u}\_1, \vec{u}\_2, \ldots, \vec{u}\_m \\\\ is an orthogonal basis of \\V\\. Then for any \\\vec{x}\in V\\,

\\\vec{x} = \left(\frac{\vec{x}\cdot \vec{u}\_1}{ \\ \vec{u}\_1 \\ ^2}\right) \vec{u}\_1 + \left(\frac{\vec{x}\cdot \vec{u}\_2}{ \\ \vec{u}\_2 \\ ^2}\right) \vec{u}\_2 + \cdots + \left(\frac{\vec{x}\cdot \vec{u}\_m}{ \\ \vec{u}\_m \\ ^2}\right) \vec{u}\_m\nonumber \\

This expression is called the Fourier expansion of \\\vec{x}\\, and \\\frac{\vec{x}\cdot \vec{u}\_j}{ \\ \vec{u}\_j \\ ^2},\nonumber \\ \\j=1,2,\ldots,m\\ are the Fourier coefficients.

Consider the following example.

##### Example \\\PageIndex{5}\\: Fourier Expansion

Let \\\vec{u}\_1= \left$$\begin{array}{r} 1 \\ -1 \\ 2 \end{array}\right$$, \vec{u}\_2= \left$$\begin{array}{r} 0 \\ 2 \\ 1 \end{array}\right$$\\, and \\\vec{u}\_3 =\left$$\begin{array}{r} 5 \\ 1 \\ -2 \end{array}\right$$\\, and let \\\vec{x} =\left$$\begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right$$\\.

Then \\B=\\ \vec{u}\_1, \vec{u}\_2, \vec{u}\_3\\\\ is an orthogonal basis of \\\mathbb{R}^3\\.

Compute the Fourier expansion of \\\vec{x}\\, thus writing \\\vec{x}\\ as a linear combination of the vectors of \\B\\.

###### Solution

Since \\B\\ is a basis (verify!) there is a unique way to express \\\vec{x}\\ as a linear combination of the vectors of \\B\\. Moreover since \\B\\ is an orthogonal basis (verify!), then this can be done by computing the Fourier expansion of \\\vec{x}\\.

That is:

\\\vec{x} = \left(\frac{\vec{x}\cdot \vec{u}\_1}{ \\ \vec{u}\_1 \\ ^2}\right) \vec{u}\_1 + \left(\frac{\vec{x}\cdot \vec{u}\_2}{ \\ \vec{u}\_2 \\ ^2}\right) \vec{u}\_2 + \left(\frac{\vec{x}\cdot \vec{u}\_3}{ \\ \vec{u}\_3 \\ ^2}\right) \vec{u}\_3. \nonumber\\

We readily compute:

\\\frac{\vec{x}\cdot\vec{u}\_1}{ \\ \vec{u}\_1 \\ ^2} = \frac{2}{6}, \\ \frac{\vec{x}\cdot\vec{u}\_2}{ \\ \vec{u}\_2 \\ ^2} = \frac{3}{5}, \mbox{ and } \frac{\vec{x}\cdot\vec{u}\_3}{ \\ \vec{u}\_3 \\ ^2} = \frac{4}{30}. \nonumber\\

Therefore, \\\left$$\begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right$$ = \frac{1}{3}\left$$\begin{array}{r} 1 \\ -1 \\ 2 \end{array}\right$$ +\frac{3}{5}\left$$\begin{array}{r} 0 \\ 2 \\ 1 \end{array}\right$$ +\frac{2}{15}\left$$\begin{array}{r} 5 \\ 1 \\ -2 \end{array}\right$$. \nonumber\\

Orthogonal Matrices

Recall that the process to find the inverse of a matrix was often cumbersome. In contrast, it was very easy to take the transpose of a matrix. Luckily for some special matrices, the transpose equals the inverse. When an \\n \times n\\ matrix has all real entries and its transpose equals its inverse, the matrix is called an orthogonal matrix.

The precise definition is as follows.

##### Definition \\\PageIndex{7}\\: Orthogonal Matrices

A real \\n\times n\\ matrix \\U\\ is called an orthogonal matrix if

\\UU^{T}=U^{T}U=I.\nonumber \\

Note since \\U\\ is assumed to be a square matrix, it suffices to verify only one of these equalities \\UU^{T}=I\\ or \\U^{T}U=I\\ holds to guarantee that \\U^T\\ is the inverse of \\U\\.

Consider the following example.

##### Example \\\PageIndex{6}\\: Orthogonal Matrix

Orthogonal Matrix Show the matrix \\U=\left$$ \begin{array}{rr} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{array} \right$$\nonumber \\ is orthogonal.

###### Solution

All we need to do is verify (one of the equations from) the requirements of Definition $\PageIndex{7}$.

\\UU^{T}=\left$$ \begin{array}{rr} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{array} \right$$ \left$$ \begin{array}{rr} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{array} \right$$ = \left$$ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right$$\nonumber \\

Since \\UU^{T} = I\\, this matrix is orthogonal.

Here is another example.

##### Example \\\PageIndex{7}\\: Orthogonal Matrix

Orthogonal Matrix Let \\U=\left$$ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right$$ .\\ Is \\U\\ orthogonal?

###### Solution

Again the answer is yes and this can be verified simply by showing that \\U^{T}U=I\\:

\\\begin{aligned} U^{T}U&=\left$$ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right$$ ^{T}\left$$ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right$$ \\ &=\left$$ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right$$ \left$$ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right$$ \\ &=\left$$ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right$$\end{aligned}\\

When we say that \\U\\ is orthogonal, we are saying that \\UU^T=I\\, meaning that \\\sum\_{j}u\_{ij}u\_{jk}^{T}=\sum\_{j}u\_{ij}u\_{kj}=\delta \_{ik}\nonumber \\ where \\\delta \_{ij}\\ is the Kronecker symbol defined by \\\delta \_{ij}=\left\\ \begin{array}{c} 1 \text{ if }i=j \\ 0\text{ if }i\neq j \end{array} \right.\nonumber \\

In words, the product of the \\i^{th}\\ row of \\U\\ with the \\k^{th}\\ row gives \\1\\ if \\i=k\\ and \\0\\ if \\i\neq k.\\ The same is true of the columns because \\U^{T}U=I\\ also. Therefore, \\\sum\_{j}u\_{ij}^{T}u\_{jk}=\sum\_{j}u\_{ji}u\_{jk}=\delta \_{ik}\nonumber \\ which says that the product of one column with another column gives \\1\\ if the two columns are the same and \\0\\ if the two columns are different.

More succinctly, this states that if \\\vec{u}\_{1},\cdots ,\vec{u}\_{n}\\ are the columns of \\U,\\ an orthogonal matrix, then \\\vec{u}\_{i}\cdot \vec{u}\_{j}=\delta \_{ij} = \left\\ \begin{array}{c} 1\text{ if }i=j \\ 0\text{ if }i\neq j \end{array} \right.\nonumber \\

We will say that the columns form an orthonormal set of vectors, and similarly for the rows. Thus a matrix is orthogonal if its rows (or columns) form an orthonormal set of vectors. Notice that the convention is to call such a matrix orthogonal rather than orthonormal (although this may make more sense!).

##### Proposition \\\PageIndex{1}\\: Orthonormal Basis

The rows of an \\n \times n\\ orthogonal matrix form an orthonormal basis of \\\mathbb{R}^n\\. Further, any orthonormal basis of \\\mathbb{R}^n\\ can be used to construct an \\n \times n\\ orthogonal matrix.

Proof

Recall from Theorem $\PageIndex{1}$ that an orthonormal set is linearly independent and forms a basis for its span. Since the rows of an \\n \times n\\ orthogonal matrix form an orthonormal set, they must be linearly independent. Now we have \\n\\ linearly independent vectors, and it follows that their span equals \\\mathbb{R}^n\\. Therefore these vectors form an orthonormal basis for \\\mathbb{R}^n\\.

Suppose now that we have an orthonormal basis for \\\mathbb{R}^n\\. Since the basis will contain \\n\\ vectors, these can be used to construct an \\n \times n\\ matrix, with each vector becoming a row. Therefore the matrix is composed of orthonormal rows, which by our above discussion, means that the matrix is orthogonal. Note we could also have construct a matrix with each vector becoming a column instead, and this would again be an orthogonal matrix. In fact this is simply the transpose of the previous matrix.

Consider the following proposition.

##### Proposition \\\PageIndex{2}\\: Determinant of Orthogonal Matrices

Det Suppose \\U\\ is an orthogonal matrix. Then \\\det \left( U\right) = \pm 1.\\

Proof

This result follows from the properties of determinants. Recall that for any matrix \\A\\, \\\det(A)^T = \det(A)\\. Now if \\U\\ is orthogonal, then: \$\det \left( U\right)) ^{2}=\det \left( U^{T}\right) \det \left( U\right) =\det \left( U^{T}U\right) =\det \left( I\right) =1\nonumber \\

Therefore \$\det (U))^2 = 1\\ and it follows that \\\det \left( U\right) = \pm 1\\.

Orthogonal matrices are divided into two classes, proper and improper. The proper orthogonal matrices are those whose determinant equals 1 and the improper ones are those whose determinant equals \\-1\\. The reason for the distinction is that the improper orthogonal matrices are sometimes considered to have no physical significance. These matrices cause a change in orientation which would correspond to material passing through itself in a non physical manner. Thus in considering which coordinate systems must be considered in certain applications, you only need to consider those which are related by a proper orthogonal transformation. Geometrically, the linear transformations determined by the proper orthogonal matrices correspond to the composition of rotations.

We conclude this section with two useful properties of orthogonal matrices.

##### Example \\\PageIndex{8}\\: Product and Inverse of Orthogonal Matrices

Suppose \\A\\ and \\B\\ are orthogonal matrices. Then \\AB\\ and \\A^{-1}\\ both exist and are orthogonal.

###### Solution

First we examine the product \\AB\\. \$AB)(B^TA^T)=A(BB^T)A^T =AA^T=I\nonumber \\ Since \\AB\\ is square, \\B^TA^T=(AB)^T\\ is the inverse of \\AB\\, so \\AB\\ is invertible, and \$AB)^{-1}=(AB)^T\\ Therefore, \\AB\\ is orthogonal.

Next we show that \\A^{-1}=A^T\\ is also orthogonal. \$A^{-1})^{-1} = A = (A^T)^{T} =(A^{-1})^{T}\nonumber \\ Therefore \\A^{-1}\\ is also orthogonal.

Gram-Schmidt Process

The Gram-Schmidt process is an algorithm to transform a set of vectors into an orthonormal set spanning the same subspace, that is generating the same collection of linear combinations (see Definition 9.2.2).

The goal of the Gram-Schmidt process is to take a linearly independent set of vectors and transform it into an orthonormal set with the same span. The first objective is to construct an orthogonal set of vectors with the same span, since from there an orthonormal set can be obtained by simply dividing each vector by its length.

##### Algorithm \\\PageIndex{1}\\: Gram-Schmidt Process

Let \\\\ \vec{u}\_1,\cdots ,\vec{u}\_n \\\\ be a set of linearly independent vectors in \\\mathbb{R}^{n}\\.

I: Construct a new set of vectors \\\\ \vec{v}\_1,\cdots ,\vec{v}\_n \\\\ as follows: \\\begin{array}{ll} \vec{v}\_1 & = \vec{u}\_1 \\ \vec{v}\_{2} & = \vec{u}\_{2} - \left( \dfrac{ \vec{u}\_2 \cdot \vec{v}\_1}{ \\ \vec{v}\_1 \\ ^2} \right) \vec{v}\_1\\ \vec{v}\_{3} & = \vec{u}\_{3} - \left( \dfrac{\vec{u}\_3 \cdot \vec{v}\_1}{ \\ \vec{v}\_1 \\ ^2} \right) \vec{v}\_1 - \left( \dfrac{\vec{u}\_3 \cdot \vec{v}\_2}{ \\ \vec{v}\_2 \\ ^2} \right) \vec{v}\_2\\ \vdots \\ \vec{v}\_{n} & = \vec{u}\_{n} - \left( \dfrac{\vec{u}\_n \cdot \vec{v}\_1}{ \\ \vec{v}\_1 \\ ^2} \right) \vec{v}\_1 - \left( \dfrac{\vec{u}\_n \cdot \vec{v}\_2}{ \\ \vec{v}\_2 \\ ^2} \right) \vec{v}\_2 - \cdots - \left( \dfrac{\vec{u}\_{n} \cdot \vec{v}\_{n-1}}{ \\ \vec{v}\_{n-1} \\ ^2} \right) \vec{v}\_{n-1} \\ \end{array}\nonumber \\

II: Now let \\\vec{w}\_i = \dfrac{\vec{v}\_i}{ \\ \vec{v}\_i \\ }\\ for \\i=1, \cdots ,n\\.

Then

1. \\\left\\ \vec{v}\_1, \cdots, \vec{v}\_n \right\\\\ is an orthogonal set.

2. \\\left\\ \vec{w}\_1,\cdots , \vec{w}\_n \right\\\\ is an orthonormal set.

3. \\\mathrm{span}\left\\ \vec{u}\_1,\cdots ,\vec{u}\_n \right\\ = \mathrm{span} \left\\ \vec{v}\_1, \cdots, \vec{v}\_n \right\\ = \mathrm{span}\left\\ \vec{w}\_1,\cdots ,\vec{w}\_n \right\\\\.

###### Solution

The full proof of this algorithm is beyond this material, however here is an indication of the arguments.

To show that \\\left\\ \vec{v}\_1,\cdots , \vec{v}\_n \right\\\\ is an orthogonal set, let \\a_2 = \dfrac{ \vec{u}\_2 \cdot \vec{v}\_1}{ \\ \vec{v}\_1 \\ ^2}\nonumber \\ then: \\\begin{array}{ll} \vec{v}\_1 \cdot \vec{v}\_2 & = \vec{v}\_1 \cdot \left( \vec{u}\_2 - a_2 \vec{v}\_1 \right) \\ & = \vec{v}\_1 \cdot \vec{u}\_2 - a_2 (\vec{v}\_1 \cdot \vec{v}\_1 \\ & = \vec{v}\_1 \cdot \vec{u}\_2 - \dfrac{ \vec{u}\_2 \cdot \vec{v}\_1}{ \\ \vec{v}\_1 \\ ^2} \\ \vec{v}\_1 \\ ^2 \\ & = ( \vec{v}\_1 \cdot \vec{u}\_2 ) - ( \vec{u}\_2 \cdot \vec{v}\_1 ) =0\\ \end{array}\nonumber \\ Now that you have shown that \\\\ \vec{v}\_1, \vec{v}\_2\\\\ is orthogonal, use the same method as above to show that \\\\ \vec{v}\_1, \vec{v}\_2, \vec{v}\_3\\\\ is also orthogonal, and so on.

Then in a similar fashion you show that \\\mathrm{span}\left\\ \vec{u}\_1,\cdots ,\vec{u}\_n \right\\ = \mathrm{span}\left\\ \vec{v}\_1,\cdots ,\vec{v}\_n \right\\\\.

Finally defining \\\vec{w}\_i = \dfrac{\vec{v}\_i}{ \\ \vec{v}\_i \\ }\\ for \\i=1, \cdots ,n\\ does not affect orthogonality and yields vectors of length 1, hence an orthonormal set. You can also observe that it does not affect the span either and the proof would be complete.

Consider the following example.

##### Example \\\PageIndex{9}\\: Find Orthonormal Set with Same Span

Consider the set of vectors \\\\\vec{u}\_1, \vec{u}\_2\\\\ given as in Example $\PageIndex{1}$. That is \\\vec{u}\_1=\left$$ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right$$, \vec{u}\_2=\left$$ \begin{array}{r} 3 \\ 2 \\ 0 \end{array} \right$$ \in \mathbb{R}^{3}\nonumber \\

Use the Gram-Schmidt algorithm to find an orthonormal set of vectors \\\\\vec{w}\_1, \vec{w}\_2\\\\ having the same span.

###### Solution

We already remarked that the set of vectors in \\\\\vec{u}\_1, \vec{u}\_2\\\\ is linearly independent, so we can proceed with the Gram-Schmidt algorithm: \\\begin{aligned} \vec{v}\_1 &= \vec{u}\_1 = \left$$ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right$$ \\ \vec{v}\_{2} &= \vec{u}\_{2} - \left( \dfrac{\vec{u}\_2 \cdot \vec{v}\_1}{ \\ \vec{v}\_1 \\ ^2} \right) \vec{v}\_1\\ &= \left$$ \begin{array}{r} 3 \\ 2 \\ 0 \end{array} \right$$ - \frac{5}{2} \left$$ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right$$ \\ &= \left$$ \begin{array}{r} \frac{1}{2} \\ - \frac{1}{2} \\ 0 \end{array} \right$$ \end{aligned}\\

Now to normalize simply let \\\begin{aligned} \vec{w}\_1 &= \frac{\vec{v}\_1}{ \\ \vec{v}\_1 \\ } = \left$$ \begin{array}{r} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \right$$ \\ \vec{w}\_2 &= \frac{\vec{v}\_2}{ \\ \vec{v}\_2 \\ } = \left$$ \begin{array}{r} \frac{1}{\sqrt{2}} \\ - \frac{1}{\sqrt{2}} \\ 0 \end{array} \right$$\end{aligned}\\

You can verify that \\\\\vec{w}\_1, \vec{w}\_2\\\\ is an orthonormal set of vectors having the same span as \\\\\vec{u}\_1, \vec{u}\_2\\\\, namely the \\XY\\-plane.

In this example, we began with a linearly independent set and found an orthonormal set of vectors which had the same span. It turns out that if we start with a basis of a subspace and apply the Gram-Schmidt algorithm, the result will be an orthogonal basis of the same subspace. We examine this in the following example.

##### Example \\\PageIndex{10}\\: Find a Corresponding Orthogonal Basis

Let \\\vec{x}\_1=\left$$\begin{array}{c} 1\\ 0\\ 1\\ 0 \end{array}\right$$, \vec{x}\_2=\left$$\begin{array}{c} 1\\ 0\\ 1\\ 1 \end{array}\right$$, \mbox{ and } \vec{x}\_3=\left$$\begin{array}{c} 1\\ 1\\ 0\\ 0 \end{array}\right$$,\nonumber \\ and let \\U=\mathrm{span}\\\vec{x}\_1, \vec{x}\_2,\vec{x}\_3\\\\. Use the Gram-Schmidt Process to construct an orthogonal basis \\B\\ of \\U\\.

###### Solution

First \\\vec{f}\_1=\vec{x}\_1\\.

Next, \\\vec{f}\_2=\left$$\begin{array}{c} 1\\ 0\\ 1\\ 1 \end{array}\right$$ -\frac{2}{2}\left$$\begin{array}{c} 1\\ 0\\ 1\\ 0 \end{array}\right$$ =\left$$\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right$$.\nonumber \\

Finally, \\\vec{f}\_3=\left$$\begin{array}{c} 1\\ 1\\ 0\\ 0 \end{array}\right$$ -\frac{1}{2}\left$$\begin{array}{c} 1\\ 0\\ 1\\ 0 \end{array}\right$$ -\frac{0}{1}\left$$\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right$$ =\left$$\begin{array}{c} 1/2\\ 1\\ -1/2\\ 0 \end{array}\right$$.\nonumber \\

Therefore, \\\left\\ \left$$\begin{array}{c} 1\\ 0\\ 1\\ 0 \end{array}\right$$, \left$$\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right$$, \left$$\begin{array}{c} 1/2\\ 1\\ -1/2\\ 0 \end{array}\right$$ \right\\\nonumber \\ is an orthogonal basis of \\U\\. However, it is sometimes more convenient to deal with vectors having integer entries, in which case we take \\B=\left\\ \left$$\begin{array}{c} 1\\ 0\\ 1\\ 0 \end{array}\right$$, \left$$\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right$$, \left$$\begin{array}{r} 1\\ 2\\ -1\\ 0 \end{array}\right$$ \right\\.\nonumber \\

Orthogonal Projections

An important use of the Gram-Schmidt Process is in orthogonal projections, the focus of this section.

You may recall that a subspace of \\\mathbb{R}^n\\ is a set of vectors which contains the zero vector, and is closed under addition and scalar multiplication. Let’s call such a subspace \\W\\. In particular, a plane in \\\mathbb{R}^n\\ which contains the origin, \\\left(0,0, \cdots, 0 \right)\\, is a subspace of \\\mathbb{R}^n\\.

Suppose a point \\Y\\ in \\\mathbb{R}^n\\ is not contained in \\W\\, then what point \\Z\\ in \\W\\ is closest to \\Y\\? Using the Gram-Schmidt Process, we can find such a point. Let \\\vec{y}, \vec{z}\\ represent the position vectors of the points \\Y\\ and \\Z\\ respectively, with \\\vec{y}-\vec{z}\\ representing the vector connecting the two points \\Y\\ and \\Z\\. It will follow that if \\Z\\ is the point on \\W\\ closest to \\Y\\, then \\\vec{y} - \vec{z}\\ will be perpendicular to \\W\\ (can you see why?); in other words, \\\vec{y} - \vec{z}\\ is orthogonal to \\W\\ (and to every vector contained in \\W\$ as in the following diagram.

Figure $\PageIndex{1}$

The vector \\\vec{z}\\ is called the orthogonal projection of \\\vec{y}\\ on \\W\\. The definition is given as follows.

##### Definition \\\PageIndex{8}\\: Orthogonal Projection

Let \\W\\ be a subspace of \\\mathbb{R}^n\\, and \\Y\\ be any point in \\\mathbb{R}^n\\. Then the orthogonal projection of \\Y\\ onto \\W\\ is given by \\\vec{z} = \mathrm{proj}\_{W}\left( \vec{y}\right) = \left( \frac{\vec{y} \cdot \vec{w}\_1}{ \\ \vec{w}\_1 \\ ^2}\right) \vec{w}\_1 + \left( \frac{\vec{y} \cdot \vec{w}\_2}{ \\ \vec{w}\_2 \\ ^2}\right) \vec{w}\_2 + \cdots + \left( \frac{\vec{y} \cdot \vec{w}\_m}{ \\ \vec{w}\_m \\ ^2}\right) \vec{w}\_m\nonumber \\ where \\\\\vec{w}\_1, \vec{w}\_2, \cdots, \vec{w}\_m \\\\ is any orthogonal basis of \\W\\.

Therefore, in order to find the orthogonal projection, we must first find an orthogonal basis for the subspace. Note that one could use an orthonormal basis, but it is not necessary in this case since as you can see above the normalization of each vector is included in the formula for the projection.

Before we explore this further through an example, we show that the orthogonal projection does indeed yield a point \\Z\\ (the point whose position vector is the vector \\\vec{z}\\ above) which is the point of \\W\\ closest to \\Y\\.

##### Theorem \\\PageIndex{3}\\: Approximation Theorem

Let \\W\\ be a subspace of \\\mathbb{R}^n\\ and \\Y\\ any point in \\\mathbb{R}^n\\. Let \\Z\\ be the point whose position vector is the orthogonal projection of \\Y\\ onto \\W\\.

Then, \\Z\\ is the point in \\W\\ closest to \\Y\\.

Proof

First \\Z\\ is certainly a point in \\W\\ since it is in the span of a basis of \\W\\.

To show that \\Z\\ is the point in \\W\\ closest to \\Y\\, we wish to show that \\\|\vec{y}-\vec{z}\_1\| \> \|\vec{y}-\vec{z}\|\\ for all \\\vec{z}\_1 \neq \vec{z} \in W\\. We begin by writing \\\vec{y}-\vec{z}\_1 = (\vec{y} - \vec{z}) + (\vec{z} - \vec{z}\_1)\\. Now, the vector \\\vec{y} - \vec{z}\\ is orthogonal to \\W\\, and \\\vec{z} - \vec{z}\_1\\ is contained in \\W\\. Therefore these vectors are orthogonal to each other. By the Pythagorean Theorem, we have that \\ \\ \vec{y} - \vec{z}\_1 \\ ^2 = \\ \vec{y} - \vec{z} \\ ^2 + \\ \vec{z} -\vec{z}\_1 \\ ^2 \> \\ \vec{y} - \vec{z} \\ ^2\nonumber \\ This follows because \\\vec{z} \neq \vec{z}\_1\\ so \\ \\ \vec{z} -\vec{z}\_1 \\ ^2 \> 0.\\

Hence, \\ \\ \vec{y} - \vec{z}\_1 \\ ^2 \> \\ \vec{y} - \vec{z} \\ ^2\\. Taking the square root of each side, we obtain the desired result.

Consider the following example.

##### Example \\\PageIndex{11}\\: Orthogonal Projection

Let \\W\\ be the plane through the origin given by the equation \\x - 2y + z = 0\\. Find the point in \\W\\ closest to the point \\Y = (1,0,3)\\.

###### Solution

We must first find an orthogonal basis for \\W\\. Notice that \\W\\ is characterized by all points \$a,b,c)\\ where \\c = 2b-a\\. In other words, \\W = \left$$ \begin{array}{c} a \\ b \\ 2b - a \end{array} \right$$ = a \left$$ \begin{array}{c} 1 \\ 0 \\ -1 \end{array} \right$$ + b \left$$ \begin{array}{c} 0 \\ 1 \\ 2 \end{array} \right$$, \\ a,b \in \mathbb{R}\nonumber \\

We can thus write \\W\\ as \\\begin{aligned} W &= \mbox{span} \left\\ \vec{u}\_1, \vec{u}\_2 \right\\ \\ &= \mbox{span} \left\\ \left$$ \begin{array}{r} 1 \\ 0 \\ -1 \end{array} \right$$, \left$$ \begin{array}{c} 0 \\ 1 \\ 2 \end{array} \right$$ \right\\\end{aligned}\\

Notice that this span is a basis of \\W\\ as it is linearly independent. We will use the Gram-Schmidt Process to convert this to an orthogonal basis, \\\left\\\vec{w}\_1, \vec{w}\_2 \right\\\\. In this case, as we remarked it is only necessary to find an orthogonal basis, and it is not required that it be orthonormal.

\\\vec{w}\_1 = \vec{u}\_1 = \left$$ \begin{array}{r} 1 \\ 0 \\ -1 \end{array} \right$$\nonumber \\ \\\begin{aligned} \vec{w}\_2 &= \vec{u}\_2 - \left( \frac{ \vec{u}\_2 \cdot \vec{w}\_1}{ \\ \vec{w}\_1 \\ ^2} \right) \vec{w}\_1\\ &= \left$$ \begin{array}{c} 0 \\ 1 \\ 2 \end{array} \right$$ - \left( \frac{-2}{2}\right) \left$$ \begin{array}{r} 1 \\ 0 \\ -1 \end{array} \right$$ \\ &= \left$$ \begin{array}{c} 0 \\ 1 \\ 2 \end{array} \right$$ + \left$$ \begin{array}{r} 1 \\ 0 \\ -1 \end{array} \right$$ \\ &= \left$$ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right$$\end{aligned}\\

Therefore an orthogonal basis of \\W\\ is \\\left\\ \vec{w}\_1, \vec{w}\_2 \right\\ = \left\\ \left$$ \begin{array}{r} 1 \\ 0 \\ -1 \end{array} \right$$, \left$$ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right$$ \right\\\nonumber \\

We can now use this basis to find the orthogonal projection of the point \\Y=(1,0,3)\\ on the subspace \\W\\. We will write the position vector \\\vec{y}\\ of \\Y\\ as \\\vec{y} = \left$$ \begin{array}{c} 1 \\ 0 \\ 3 \end{array} \right$$\\. Using Definition $\PageIndex{8}$, we compute the projection as follows: \\\begin{aligned} \vec{z} &= \mathrm{proj}\_{W}\left( \vec{y}\right)\\ &= \left( \frac{\vec{y} \cdot \vec{w}\_1}{ \\ \vec{w}\_1 \\ ^2}\right) \vec{w}\_1 + \left( \frac{\vec{y} \cdot \vec{w}\_2}{ \\ \vec{w}\_2 \\ ^2}\right) \vec{w}\_2 \\ &= \left( \frac{-2}{2} \right) \left$$ \begin{array}{r} 1 \\ 0 \\ -1 \end{array} \right$$ + \left( \frac{4}{3} \right) \left$$ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right$$ \\ &= \left$$ \begin{array}{c} \frac{1}{3} \\ \frac{4}{3} \\ \frac{7}{3} \end{array} \right$$\end{aligned}\\

Therefore the point \\Z\\ on \\W\\ closest to the point \$1,0,3)\\ is \\\left( \frac{1}{3}, \frac{4}{3}, \frac{7}{3} \right)\\.

Recall that the vector \\\vec{y} - \vec{z}\\ is perpendicular (orthogonal) to all the vectors contained in the plane \\W\\. Using a basis for \\W\\, we can in fact find all such vectors which are perpendicular to \\W\\. We call this set of vectors the orthogonal complement of \\W\\ and denote it \\W^{\perp}\\.

##### Definition \\\PageIndex{9}\\: Orthogonal Complement

Let \\W\\ be a subspace of \\\mathbb{R}^n\\. Then the orthogonal complement of \\W\\, written \\W^{\perp}\\, is the set of all vectors \\\vec{x}\\ such that \\\vec{x} \cdot \vec{z} = 0\\ for all vectors \\\vec{z}\\ in \\W\\. \\W^{\perp} = \\ \vec{x} \in \mathbb{R}^n \\ \mbox{such that} \\ \vec{x} \cdot \vec{z} = 0 \\ \mbox{for all} \\ \vec{z} \in W \\\nonumber \\

The orthogonal complement is defined as the set of all vectors which are orthogonal to all vectors in the original subspace. It turns out that it is sufficient that the vectors in the orthogonal complement be orthogonal to a spanning set of the original space.

##### Proposition \\\PageIndex{3}\\: Orthogonal to Spanning Set

Let \\W\\ be a subspace of \\\mathbb{R}^n\\ such that \\W = \mathrm{span} \left\\ \vec{w}\_1, \vec{w}\_2, \cdots, \vec{w}\_m \right\\\\. Then \\W^{\perp}\\ is the set of all vectors which are orthogonal to each \\\vec{w}\_i\\ in the spanning set.

The following proposition demonstrates that the orthogonal complement of a subspace is itself a subspace.

##### Proposition \\\PageIndex{4}\\: The Orthogonal Complement

Let \\W\\ be a subspace of \\\mathbb{R}^n\\. Then the orthogonal complement \\W^{\perp}\\ is also a subspace of \\\mathbb{R}^n\\.

Consider the following proposition.

##### Proposition \\\PageIndex{5}\\: Orthogonal Complement of \\\mathbb{R}^n\\

The complement of \\\mathbb{R}^n\\ is the set containing the zero vector: \$\mathbb{R}^n)^{\perp} = \left\\ \vec{0} \right\\\nonumber \\ Similarly, \\\left\\ \vec{0} \right\\^{\perp} = (\mathbb{R}^n). \nonumber \\

Proof

Here, \\\vec{0}\\ is the zero vector of \\\mathbb{R}^n\\. Since \\\vec{x}\cdot\vec{0}=0\\ for all \\\vec{x}\in\mathbb{R}^n\\, \\\mathbb{R}^n\subseteq\\ \vec{0}\\^{\perp}\\. Since \\\\ \vec{0}\\^{\perp}\subseteq\mathbb{R}^n\\, the equality follows, i.e., \\\\ \vec{0}\\^{\perp}=\mathbb{R}^n\\.

Again, since \\\vec{x}\cdot\vec{0}=0\\ for all \\\vec{x}\in\mathbb{R}^n\\, \\\vec{0}\in (\mathbb{R}^n)^{\perp}\\, so \\\\ \vec{0}\\\subseteq(\mathbb{R}^n)^{\perp}\\. Suppose \\\vec{x}\in\mathbb{R}^n\\, \\\vec{x}\neq\vec{0}\\. Since \\\vec{x}\cdot\vec{x}=\|\|\vec{x}\|\|^2\\ and \\\vec{x}\neq\vec{0}\\, \\\vec{x}\cdot\vec{x}\neq 0\\, so \\\vec{x}\not\in(\mathbb{R}^n)^{\perp}\\. Therefore \$\mathbb{R}^n)^{\perp}\subseteq \\\vec{0}\\\\, and thus \$\mathbb{R}^n)^{\perp}=\\\vec{0}\\\\.

In the next example, we will look at how to find \\W^{\perp}\\.

##### Example \\\PageIndex{12}\\: Orthogonal Complement

Let \\W\\ be the plane through the origin given by the equation \\x - 2y + z = 0\\. Find a basis for the orthogonal complement of \\W\\.

###### Solution

From Example $\PageIndex{11}$ we know that we can write \\W\\ as \\W = \mbox{span} \left\\ \vec{u}\_1, \vec{u}\_2 \right\\ = \mbox{span} \left\\ \left$$ \begin{array}{r} 1 \\ 0 \\ -1 \end{array} \right$$, \left$$ \begin{array}{c} 0 \\ 1 \\ 2 \end{array} \right$$ \right\\\nonumber \\

In order to find \\W^{\perp}\\, we need to find all \\\vec{x}\\ which are orthogonal to every vector in this span.

Let \\\vec{x} = \left$$ \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right$$\\. In order to satisfy \\\vec{x} \cdot \vec{u}\_1 = 0\\, the following equation must hold. \\x_1 - x_3 = 0\nonumber \\

In order to satisfy \\\vec{x} \cdot \vec{u}\_2 = 0\\, the following equation must hold. \\x_2 + 2x_3 = 0\nonumber \\

Both of these equations must be satisfied, so we have the following system of equations. \\\begin{array}{c} x_1 - x_3 = 0 \\ x_2 + 2x_3 = 0 \end{array}\nonumber \\

To solve, set up the augmented matrix.

\\\left$$ \begin{array}{rrr\|r} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \end{array} \right$$\nonumber \\

Using Gaussian Elimination, we find that \\W^{\perp} = \mbox{span} \left\\ \left$$ \begin{array}{r} 1 \\ -2 \\ 1 \end{array} \right$$ \right\\\\, and hence \\\left\\ \left$$ \begin{array}{r} 1 \\ -2 \\ 1 \end{array} \right$$ \right\\\\ is a basis for \\W^{\perp}\\.

The following results summarize the important properties of the orthogonal projection.

##### Theorem \\\PageIndex{4}\\: Orthogonal Projection

Let \\W\\ be a subspace of \\\mathbb{R}^n\\, \\Y\\ be any point in \\\mathbb{R}^n\\, and let \\Z\\ be the point in \\W\\ closest to \\Y\\. Then,

1. The position vector \\\vec{z}\\ of the point \\Z\\ is given by \\\vec{z} = \mathrm{proj}\_{W}\left( \vec{y}\right)\\

2. \\\vec{z} \in W\\ and \\\vec{y} - \vec{z} \in W^{\perp}\\

3. \\\| Y - Z \| \< \| Y - Z_1 \|\\ for all \\Z_1 \neq Z \in W\\

Consider the following example of this concept.

##### Example \\\PageIndex{13}\\: Find a Vector Closest to a Given Vector

Let \\\vec{x}\_1=\left$$\begin{array}{c} 1\\ 0\\ 1\\ 0 \end{array}\right$$, \vec{x}\_2=\left$$\begin{array}{c} 1\\ 0\\ 1\\ 1 \end{array}\right$$, \vec{x}\_3=\left$$\begin{array}{c} 1\\ 1\\ 0\\ 0 \end{array}\right$$, \mbox{ and } \vec{v}=\left$$\begin{array}{c} 4\\ 3\\ -2\\ 5 \end{array}\right$$.\nonumber \\ We want to find the vector in \\W =\mathrm{span}\\\vec{x}\_1, \vec{x}\_2,\vec{x}\_3\\\\ closest to \\\vec{y}\\.

###### Solution

We will first use the Gram-Schmidt Process to construct the orthogonal basis, \\B\\, of \\W\\: \\B=\left\\ \left$$\begin{array}{c} 1\\ 0\\ 1\\ 0 \end{array}\right$$, \left$$\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right$$, \left$$\begin{array}{r} 1\\ 2\\ -1\\ 0 \end{array}\right$$ \right\\.\nonumber \\

By Theorem $\PageIndex{4}$, \\\mathrm{proj}\_U(\vec{v}) = \frac{2}{2} \left$$\begin{array}{c} 1\\ 0\\ 1\\ 0 \end{array}\right$$ + \frac{5}{1}\left$$\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right$$ + \frac{12}{6}\left$$\begin{array}{r} 1\\ 2\\ -1\\ 0 \end{array}\right$$ = \left$$\begin{array}{r} 3\\ 4\\ -1\\ 5 \end{array}\right$$\nonumber \\ is the vector in \\U\\ closest to \\\vec{y}\\.

Consider the next example.

##### Example \\\PageIndex{14}\\: Vector Written as a Sum of Two Vectors

Let \\W\\ be a subspace given by \\W = \mbox{span} \left\\ \left$$ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \\ \end{array} \right$$, \left$$ \begin{array}{c} 0 \\ 1 \\ 0 \\ 2 \\ \end{array} \right$$ \right\\\\, and \\Y = (1,2,3,4)\\.

Find the point \\Z\\ in \\W\\ closest to \\Y\\, and moreover write \\\vec{y}\\ as the sum of a vector in \\W\\ and a vector in \\W^{\perp}\\.

###### Solution

From Theorem $\PageIndex{3}$ the point \\Z\\ in \\W\\ closest to \\Y\\ is given by \\\vec{z} = \mathrm{proj}\_{W}\left( \vec{y}\right)\\.

Notice that since the above vectors already give an orthogonal basis for \\W\\, we have:

\\\begin{aligned} \vec{z} &= \mathrm{proj}\_{W}\left( \vec{y}\right)\\ &= \left( \frac{\vec{y} \cdot \vec{w}\_1}{ \\ \vec{w}\_1 \\ ^2}\right) \vec{w}\_1 + \left( \frac{\vec{y} \cdot \vec{w}\_2}{ \\ \vec{w}\_2 \\ ^2}\right) \vec{w}\_2 \\ &= \left( \frac{4}{2} \right) \left$$ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right$$ + \left( \frac{10}{5} \right) \left$$ \begin{array}{c} 0 \\ 1 \\ 0 \\ 2 \end{array} \right$$ \\ &= \left$$ \begin{array}{c} 2 \\ 2 \\ 2 \\ 4 \end{array} \right$$\end{aligned}\\

Therefore the point in \\W\\ closest to \\Y\\ is \\Z = (2,2,2,4)\\.

Now, we need to write \\\vec{y}\\ as the sum of a vector in \\W\\ and a vector in \\W^{\perp}\\. This can easily be done as follows: \\\vec{y} = \vec{z} + (\vec{y} - \vec{z})\nonumber \\ since \\\vec{z}\\ is in \\W\\ and as we have seen \\\vec{y} - \vec{z}\\ is in \\W^{\perp}\\.

The vector \\\vec{y} - \vec{z}\\ is given by \\\vec{y} - \vec{z} = \left$$ \begin{array}{c} 1 \\ 2 \\ 3 \\ 4 \end{array} \right$$ - \left$$ \begin{array}{c} 2 \\ 2 \\ 2 \\ 4 \end{array} \right$$ = \left$$ \begin{array}{r} -1 \\ 0 \\ 1 \\ 0 \end{array} \right$$\nonumber \\ Therefore, we can write \\\vec{y}\\ as \\\left$$ \begin{array}{c} 1 \\ 2 \\ 3 \\ 4 \end{array} \right$$ = \left$$ \begin{array}{c} 2 \\ 2 \\ 2 \\ 4 \end{array} \right$$ + \left$$ \begin{array}{r} -1 \\ 0 \\ 1 \\ 0 \end{array} \right$$\nonumber \\

##### Example \\\PageIndex{15}\\: Point in a Plane Closest to a Given Point

Find the point \\Z\\ in the plane \\3x+y-2z=0\\ that is closest to the point \\Y=(1,1,1)\\.

###### Solution

The solution will proceed as follows.

1. Find a basis \\X\\ of the subspace \\W\\ of \\\mathbb{R}^3\\ defined by the equation \\3x+y-2z=0\\.

2. Orthogonalize the basis \\X\\ to get an orthogonal basis \\B\\ of \\W\\.

3. Find the projection on \\W\\ of the position vector of the point \\Y\\.

We now begin the solution.

1. \\3x+y-2z=0\\ is a system of one equation in three variables. Putting the augmented matrix in reduced row-echelon form: \\\left$$\begin{array}{rrr\|r} 3 & 1 & -2 & 0 \end{array}\right$$ \rightarrow \left$$\begin{array}{rrr\|r} 1 & \frac{1}{3} & -\frac{2}{3} & 0 \end{array}\right$$\nonumber \\ gives general solution \\x=\frac{1}{3}s+\frac{2}{3}t\\, \\y=s\\, \\z=t\\ for any \\s,t\in\mathbb{R}\\. Then \\W=\mathrm{span} \left\\ \left$$\begin{array}{r} -\frac{1}{3} \\ 1 \\ 0 \end{array}\right$$, \left$$\begin{array}{r} \frac{2}{3} \\ 0 \\ 1 \end{array}\right$$\right\\\nonumber \\ Let \\X=\left\\ \left$$\begin{array}{r} -1 \\ 3 \\ 0 \end{array}\right$$, \left$$\begin{array}{r} 2 \\ 0 \\ 3 \end{array}\right$$\right\\\\. Then \\X\\ is linearly independent and \\\mathrm{span}(X)=W\\, so \\X\\ is a basis of \\W\\.

2. Use the Gram-Schmidt Process to get an orthogonal basis of \\W\\: \\\vec{f}\_1=\left$$\begin{array}{r} -1\\3\\0\end{array}\right$$\mbox{ and }\vec{f}\_2=\left$$\begin{array}{r}2\\0\\3\end{array}\right$$-\frac{-2}{10}\left$$\begin{array}{r}-1\\3\\0\end{array}\right$$=\frac{1}{5}\left$$\begin{array}{r}9\\3\\15\end{array}\right$$.\nonumber\\ Therefore \\B=\left\\\left$$\begin{array}{r}-1\\3\\0\end{array}\right$$, \left$$\begin{array}{r}3\\1\\5 \end{array}\right$$\right\\\\ is an orthogonal basis of \\W\\.

3. To find the point \\Z\\ on \\W\\ closest to \\Y=(1,1,1)\\, compute \\\begin{aligned} \mathrm{proj}\_{W}\left$$\begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right$$ & = \frac{2}{10} \left$$\begin{array}{r} -1 \\ 3 \\ 0 \end{array}\right$$ + \frac{9}{35}\left$$\begin{array}{r} 3 \\ 1 \\ 5 \end{array}\right$$\\ & = \frac{1}{7}\left$$\begin{array}{r} 4 \\ 6 \\ 9 \end{array}\right$$.\end{aligned}\\ Therefore, \\Z=\left( \frac{4}{7}, \frac{6}{7}, \frac{9}{7}\right)\\.

Least Squares Approximation

It should not be surprising to hear that many problems do not have a perfect solution, and in these cases the objective is always to try to do the best possible. For example what does one do if there are no solutions to a system of linear equations \\A\vec{x}=\vec{b}\\? It turns out that what we do is find \\\vec{x}\\ such that \\A\vec{x}\\ is as close to \\\vec{b}\\ as possible. A very important technique that follows from orthogonal projections is that of the least square approximation, and allows us to do exactly that.

We begin with a lemma.

Recall that we can form the image of an \\m \times n\\ matrix \\A\\ by \\\mathrm{im}\left( A\right) = = \left\\ A\vec{x} : \vec{x} \in \mathbb{R}^n \right\\\\. Rephrasing Theorem $\PageIndex{4}$ using the subspace \\W=\mathrm{im}\left( A\right)\\ gives the equivalence of an orthogonality condition with a minimization condition. The following picture illustrates this orthogonality condition and geometric meaning of this theorem.

Figure $\PageIndex{2}$

##### Theorem \\\PageIndex{5}\\: Existence of Minimizers

Let \\\vec{y}\in \mathbb{R}^{m}\\ and let \\A\\ be an \\m\times n\\ matrix.

Choose \\\vec{z}\in W= \mathrm{im}\left( A\right)\\ given by \\\vec{z} = \mathrm{proj}\_{W}\left( \vec{y}\right)\\, and let \\\vec{x} \in \mathbb{R}^{n}\\ such that \\\vec{z}=A\vec{x}\\.

Then

1. \\\vec{y} - A\vec{x} \in W^{\perp}\\

2. \\ \\ \vec{y} - A\vec{x} \\ \< \\ \vec{y} - \vec{u} \\ \\ for all \\\vec{u} \neq \vec{z} \in W\\

We note a simple but useful observation.

##### Lemma \\\PageIndex{1}\\: Transpose and Dot Product

Let \\A\\ be an \\m\times n\\ matrix. Then \\A\vec{x} \cdot \vec{y} = \vec{x}\cdot A^T\vec{y}\nonumber \\

Proof

This follows from the definitions: \\A\vec{x} \cdot \vec{y}=\sum\_{i,j}a\_{ij}x\_{j} y\_{i} =\sum\_{i,j}x\_{j} a\_{ji} y\_{i}= \vec{x} \cdot A^T\vec{y}\nonumber \\

The next corollary gives the technique of least squares.

##### Corollary \\\PageIndex{1}\\: Least Squares and Normal Equation

A specific value of \\\vec{x}\\ which solves the problem of Theorem $\PageIndex{5}$ is obtained by solving the equation \\A^TA\vec{x}=A^T\vec{y}\nonumber \\ Furthermore, there always exists a solution to this system of equations.

Proof

For \\\vec{x}\\ the minimizer of Theorem $\PageIndex{5}$, \\\left( \vec{y}-A\vec{x}\right) \cdot A \vec{u} =0\\ for all \\\vec{u} \in \mathbb{R}^{n}\\ and from Lemma $\PageIndex{1}$, this is the same as saying \\A^T\left( \vec{y}-A\vec{x}\right) \cdot \vec{u}=0\nonumber \\ for all \\u \in \mathbb{R}^{n}.\\ This implies \\A^T\vec{y}-A^TA\vec{x}=\vec{0}.\nonumber \\ Therefore, there is a solution to the equation of this corollary, and it solves the minimization problem of Theorem $\PageIndex{5}$.

Note that \\\vec{x}\\ might not be unique but \\A\vec{x}\\, the closest point of \\A\left(\mathbb{R}^{n}\right)\\ to \\\vec{y}\\ is unique as was shown in the above argument.

Consider the following example.

##### Example \\\PageIndex{16}\\: Least Squares Solution to a System

Find a least squares solution to the system \\\left$$ \begin{array}{rr} 2 & 1 \\ -1 & 3 \\ 4 & 5 \end{array} \right$$ \left$$ \begin{array}{c} x \\ y \end{array} \right$$ =\left$$ \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right$$\nonumber \\

###### Solution

First, consider whether there exists a real solution. To do so, set up the augmnented matrix given by \\\left$$ \begin{array}{rr\|r} 2 & 1 & 2 \\ -1 & 3 & 1 \\ 4 & 5 & 1 \end{array} \right$$\nonumber \\ The reduced row-echelon form of this augmented matrix is \\\left$$ \begin{array}{rr\|r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right$$\nonumber \\

It follows that there is no real solution to this system. Therefore we wish to find the least squares solution. The normal equations are \\\begin{aligned} A^T A \vec{x} &= A^T \vec{y} \\ \left$$ \begin{array}{rrr} 2 & -1 & 4 \\ 1 & 3 & 5 \end{array} \right$$ \left$$ \begin{array}{rr} 2 & 1 \\ -1 & 3 \\ 4 & 5 \end{array} \right$$ \left$$ \begin{array}{c} x \\ y \end{array} \right$$ &=\left$$ \begin{array}{rrr} 2 & -1 & 4 \\ 1 & 3 & 5 \end{array} \right$$ \left$$ \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right$$\end{aligned}\\ and so we need to solve the system \\\left$$ \begin{array}{rr} 21 & 19 \\ 19 & 35 \end{array} \right$$ \left$$ \begin{array}{c} x \\ y \end{array} \right$$ =\left$$ \begin{array}{r} 7 \\ 10 \end{array} \right$$\nonumber \\ This is a familiar exercise and the solution is \\\left$$ \begin{array}{c} x \\ y \end{array} \right$$ =\left$$ \begin{array}{c} \frac{5}{34} \\ \frac{7}{34} \end{array} \right$$\nonumber \\

Consider another example.

##### Example \\\PageIndex{17}\\: Least Squares Solution to a System

Find a least squares solution to the system \\\left$$ \begin{array}{rr} 2 & 1 \\ -1 & 3 \\ 4 & 5 \end{array} \right$$ \left$$ \begin{array}{c} x \\ y \end{array} \right$$ =\left$$ \begin{array}{c} 3 \\ 2 \\ 9 \end{array} \right$$\nonumber \\

###### Solution

First, consider whether there exists a real solution. To do so, set up the augmnented matrix given by \\\left$$ \begin{array}{rr\|r} 2 & 1 & 3 \\ -1 & 3 & 2 \\ 4 & 5 & 9 \end{array} \right$$\nonumber\\ The reduced row-echelon form of this augmented matrix is \\\left$$ \begin{array}{rr\|r} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right$$\nonumber \\

It follows that the system has a solution given by \\x=y=1\\. However we can also use the normal equations and find the least squares solution. \\\left$$ \begin{array}{rrr} 2 & -1 & 4 \\ 1 & 3 & 5 \end{array} \right$$ \left$$ \begin{array}{rr} 2 & 1 \\ -1 & 3 \\ 4 & 5 \end{array} \right$$ \left$$ \begin{array}{c} x \\ y \end{array} \right$$ =\left$$ \begin{array}{rrr} 2 & -1 & 4 \\ 1 & 3 & 5 \end{array} \right$$ \left$$ \begin{array}{r} 3 \\ 2 \\ 9 \end{array} \right$$\nonumber \\ Then \\\left$$ \begin{array}{rr} 21 & 19 \\ 19 & 35 \end{array} \right$$ \left$$ \begin{array}{c} x \\ y \end{array} \right$$ =\left$$ \begin{array}{c} 40 \\ 54 \end{array} \right$$\nonumber \\

The least squares solution is \\\left$$ \begin{array}{c} x \\ y \end{array} \right$$ =\left$$ \begin{array}{c} 1 \\ 1 \end{array} \right$$\nonumber \\ which is the same as the solution found above.

An important application of Corollary $\PageIndex{1}$ is the problem of finding the least squares regression line in statistics. Suppose you are given points in the \\xy\\ plane \\\left\\ \left( x\_{1},y\_{1}\right), \left( x\_{2},y\_{2}\right), \cdots, \left( x\_{n},y\_{n}\right) \right\\\nonumber \\ and you would like to find constants \\m\\ and \\b\\ such that the line \\\vec{y}=m\vec{x}+b\\ goes through all these points. Of course this will be impossible in general. Therefore, we try to find \\m,b\\ such that the line will be as close as possible. The desired system is

\\\left$$ \begin{array}{c} y\_{1} \\ \vdots \\ y\_{n} \end{array} \right$$ =\left$$ \begin{array}{cc} x\_{1} & 1 \\ \vdots & \vdots \\ x\_{n} & 1 \end{array} \right$$ \left$$ \begin{array}{c} m \\ b \end{array} \right$$\nonumber \\

which is of the form \\\vec{y}=A\vec{x}\\. It is desired to choose \\m\\ and \\b\\ to make

\\\left \\ A\left$$ \begin{array}{c} m \\ b \end{array} \right$$ -\left$$ \begin{array}{c} y\_{1} \\ \vdots \\ y\_{n} \end{array} \right$$ \right \\ ^{2}\nonumber \\

as small as possible. According to Theorem $\PageIndex{5}$ and Corollary $\PageIndex{1}$, the best values for \\m\\ and \\b\\ occur as the solution to

\\A^{T}A\left$$ \begin{array}{c} m \\ b \end{array} \right$$ =A^{T}\left$$ \begin{array}{c} y\_{1} \\ \vdots \\ y\_{n} \end{array} \right$$ ,\\ \\\mbox{where}\\ A=\left$$ \begin{array}{cc} x\_{1} & 1 \\ \vdots & \vdots \\ x\_{n} & 1 \end{array} \right$$\nonumber \\

Thus, computing \\A^{T}A,\\

\\\left$$ \begin{array}{cc} \sum\_{i=1}^{n}x\_{i}^{2} & \sum\_{i=1}^{n}x\_{i} \\ \sum\_{i=1}^{n}x\_{i} & n \end{array} \right$$ \left$$ \begin{array}{c} m \\ b \end{array} \right$$ =\left$$ \begin{array}{c} \sum\_{i=1}^{n}x\_{i}y\_{i} \\ \sum\_{i=1}^{n}y\_{i} \end{array} \right$$\nonumber \\

Solving this system of equations for \\m\\ and \\b\\ (using Cramer’s rule for example) yields:

\\m= \frac{-\left( \sum\_{i=1}^{n}x\_{i}\right) \left( \sum\_{i=1}^{n}y\_{i}\right) +\left( \sum\_{i=1}^{n}x\_{i}y\_{i}\right) n}{\left( \sum\_{i=1}^{n}x\_{i}^{2}\right) n-\left( \sum\_{i=1}^{n}x\_{i}\right) ^{2}}\nonumber \\ and \\b=\frac{-\left( \sum\_{i=1}^{n}x\_{i}\right) \sum\_{i=1}^{n}x\_{i}y\_{i}+\left( \sum\_{i=1}^{n}y\_{i}\right) \sum\_{i=1}^{n}x\_{i}^{2}}{\left( \sum\_{i=1}^{n}x\_{i}^{2}\right) n-\left( \sum\_{i=1}^{n}x\_{i}\right) ^{2}}.\nonumber \\

Consider the following example.

##### Example \\\PageIndex{18}\\: Least Squares Regression

Find the least squares regression line \\\vec{y}=m\vec{x}+b\\ for the following set of data points: \\\left\\ (0,1), (1,2), (2,2), (3,4), (4,5) \right\\ \nonumber\\

###### Solution

In this case we have \\n=5\\ data points and we obtain: \\\begin{array}{ll} \sum\_{i=1}^{5}x\_{i} = 10 & \sum\_{i=1}^{5}y\_{i} = 14 \\ \\ \sum\_{i=1}^{5}x\_{i}y\_{i} = 38 & \sum\_{i=1}^{5}x\_{i}^{2} = 30\\ \end{array}\nonumber \\ and hence \\\begin{aligned} m &= \frac{- 10 \* 14 + 5\*38}{5\*30-10^2} = 1.00 \\ \\ b &= \frac{- 10 \* 38 + 14\*30}{5\*30-10^2} = 0.80 \\\end{aligned}\\

The least squares regression line for the set of data points is: \\\vec{y} = \vec{x}+.8\nonumber \\

One could use this line to approximate other values for the data. For example for \\x=6\\ one could use \\y(6)=6+.8=6.8\\ as an approximate value for the data.

The following diagram shows the data points and the corresponding regression line.

Figure $\PageIndex{3}$

One could clearly do a least squares fit for curves of the form \\y=ax^{2}+bx+c\\ in the same way. In this case you want to solve as well as possible for \\a,b,\\ and \\c\\ the system \\\left$$ \begin{array}{ccc} x\_{1}^{2} & x\_{1} & 1 \\ \vdots & \vdots & \vdots \\ x\_{n}^{2} & x\_{n} & 1 \end{array} \right$$ \left$$ \begin{array}{c} a \\ b \\ c \end{array} \right$$ =\left$$ \begin{array}{c} y\_{1} \\ \vdots \\ y\_{n} \end{array} \right$$\nonumber \\ and one would use the same technique as above. Many other similar problems are important, including many in higher dimensions and they are all solved the same way.

---

4_12_3A_Applications

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.12%3A_Applications

Skip to main content

Applications

##### Outcomes

1. Apply the concepts of vectors in \\\mathbb{R}^n\\ to the applications of physics and work.

Vectors and Physics

Suppose you push on something. Then, your push is made up of two components, how hard you push and the direction you push. This illustrates the concept of force.

##### Definition \\\PageIndex{1}\\: Force

Force is a vector. The magnitude of this vector is a measure of how hard it is pushing. It is measured in units such as Newtons or pounds or tons. The direction of this vector is the direction in which the push is taking place.

Vectors are used to model force and other physical vectors like velocity. As with all vectors, a vector modeling force has two essential ingredients, its magnitude and its direction.

Recall the special vectors which point along the coordinate axes. These are given by \\\vec{e}\_{i} = \left $$ 0 \cdots 0 \\ 1 \\ 0 \cdots 0 \right $$^T\nonumber \\ where the \\1\\ is in the \\i^{th}\\ slot and there are zeros in all the other spaces. The direction of \\\vec{e}\_{i}\\ is referred to as the \\i^{th}\\ direction.

Consider the following picture which illustrates the case of \\\mathbb{R}^{3}.\\ Recall that in \\\mathbb{R}^3\\, we may refer to these vectors as \\\vec{i}, \vec{j},\\ and \\\vec{k}\\.

Figure $\PageIndex{1}$

Given a vector \\\vec{u}=\left $$ u\_{1} \cdots u\_{n}\right $$^T ,\\ it follows that \\\vec{u}=u\_{1}\vec{e}\_{1}+\cdots +u\_{n}\vec{e}\_{n}= \sum\_{k=1}^{n}u\_{i}\vec{e}\_{i}\nonumber \\

What does addition of vectors mean physically? Suppose two forces are applied to some object. Each of these would be represented by a force vector and the two forces acting together would yield an overall force acting on the object which would also be a force vector known as the resultant. Suppose the two vectors are \\\vec{u}=\sum\_{k=1}^{n}u\_{i}\vec{e}\_{i}\\ and \\\vec{v}=\sum\_{k=1}^{n}v\_{i}\vec{e}\_{i}\\. Then the vector \\\vec{u}\\ involves a component in the \\i^{th}\\ direction given by \\u\_{i}\vec{e}\_{i}\\, while the component in the \\i^{th}\\ direction of \\\vec{v}\\ is \\v\_{i}\vec{e}\_{i}.\\ Then the vector \\\vec{u} + \vec{v}\\ should have a component in the \\i^{th}\\ direction equal to \\\left( u\_{i}+v\_{i}\right) \vec{e}\_{i}.\\ This is exactly what is obtained when the vectors, \\\vec{u}\\ and \\\vec{v}\\ are added. \\\begin{aligned} \vec{u}+\vec{v}& =\left $$ u\_{1}+v\_{1} \cdots u\_{n}+v\_{n}\right $$^T \\ & =\sum\_{i=1}^{n}\left( u\_{i}+v\_{i}\right) \vec{e}\_{i}\end{aligned}\\

Thus the addition of vectors according to the rules of addition in \\\mathbb{R }^{n}\\ which were presented earlier, yields the appropriate vector which duplicates the cumulative effect of all the vectors in the sum.

Consider now some examples of vector addition.

##### Example \\\PageIndex{1}\\: The Resultant of Three Forces

There are three ropes attached to a car and three people pull on these ropes. The first exerts a force of \\\vec{F}\_1 = 2\vec{i} + 3\vec{j} -2 \vec{k}\\ Newtons, the second exerts a force of \\\vec{F}\_2 = 3\vec{i}+5\vec{j}+\vec{k}\\ Newtons and the third exerts a force of \\5\vec{i}-\vec{j}+2\vec{k}\\ Newtons. Find the total force in the direction of \\\vec{i}\\.

###### Solution

To find the total force, we add the vectors as described above. This is given by \\\begin{aligned} &(2\vec{i}+3\vec{j}-2\vec{k}) + (3\vec{i}+5\vec{j}+\vec{k}) + (5\vec{i}-\vec{j}+2\vec{k})\\ &= (2 + 3 + 5) \vec{i} + (3 + 5 + -1) \vec{j} + (-2+1+2) \vec{k} \\ &= 10 \vec{i} + 7 \vec{j} + \vec{k}\end{aligned}\\ Hence, the total force is \\10\vec{i}+7\vec{j}+\vec{k}\\ Newtons. Therefore, the force in the \\\vec{i}\\ direction is \\10\\ Newtons.

Consider another example.

##### Example \\\PageIndex{2}\\: Finding a Vector from Geometric Description

An airplane flies North East at 100 miles per hour. Write this as a vector.

###### Solution

A picture of this situation follows.

Figure $\PageIndex{2}$

Therefore, we need to find the vector \\\vec{u}\\ which has length 100 and direction as shown in this diagram. We can consider the vector \\\vec{u}\\ as the hypotenuse of a right triangle having equal sides, since the direction of \\\vec{u}\\ corresponds with the \\45 ^{\circ}\\ line. The sides, corresponding to the \\\vec{i}\\ and \\\vec{j}\\ directions, should be each of length 100/\\\sqrt{2}.\\ Therefore, the vector is given by \\\vec{u} = \frac{100}{\sqrt{2}} \vec{i}+ \frac{100}{\sqrt{2 }}\vec{j} = \left $$ \begin{array}{rr} {0.05in} \frac{100}{\sqrt{2}} & {0.05in}\frac{100}{\sqrt{2}} \end{array} \right $$^T \nonumber \\

This example also motivates the concept of velocity, defined below.

##### Definition \\\PageIndex{2}\\: Speed and Velocity

The speed of an object is a measure of how fast it is going. It is measured in units of length per unit time. For example, miles per hour, kilometers per minute, feet per second. The velocity is a vector having the speed as the magnitude but also specifying the direction.

Thus the velocity vector in the above example is \\ {0.05in}\frac{100}{\sqrt{2}}\vec{i}+ {0.05in}\frac{100}{\sqrt{2}}\vec{j}\\, while the speed is \\100\\ miles per hour.

Consider the following example.

The velocity of an airplane is \\100\vec{i}+\vec{j}+\vec{k}\\ measured in kilometers per hour and at a certain instant of time its position is \\\left( 1,2,1\right) .\\

Find the position of this airplane one minute later.

###### Solution

Here imagine a Cartesian coordinate system in which the third component is altitude and the first and second components are measured on a line from West to East and a line from South to North.

Consider the vector \\\left $$ \begin{array}{rrr} 1 & 2 & 1 \end{array} \right $$^T ,\\ which is the initial position vector of the airplane. As the plane moves, the position vector changes according to the velocity vector. After one minute (considered as \\\frac{1}{60}\\ of an hour) the airplane has moved in the \\\vec{i}\\ direction a distance of \\100\times \frac{1}{60}= \frac{5}{3}\\ kilometer. In the \\\vec{j}\\ direction it has moved \\\frac{1}{60}\\ kilometer during this same time, while it moves \\\frac{1}{60}\\ kilometer in the \\\vec{k}\\ direction. Therefore, the new displacement vector for the airplane is \\\left $$ \begin{array}{rrr} 1 & 2 & 1 \end{array} \right $$^T + \left $$ \begin{array}{rrr} \frac{5}{3} & \frac{1}{60} & \frac{1}{60} \end{array} \right $$^T =\left $$ \begin{array}{rrr} \frac{8}{3} & \frac{121}{60} & \frac{121}{60} \end{array} \right $$^T\nonumber \\

Now consider an example which involves combining two velocities.

##### Example \\\PageIndex{4}\\: Sum of Two Velocities

A certain river is one half kilometer wide with a current flowing at 4 kilometers per hour from East to West. A man swims directly toward the opposite shore from the South bank of the river at a speed of 3 kilometers per hour. How far down the river does he find himself when he has swam across? How far does he end up swimming?

###### Solution

Consider the following picture which demonstrates the above scenario.

Figure $\PageIndex{3}$

First we want to know the total time of the swim across the river. The velocity in the direction across the river is \\3\\ kilometers per hour, and the river is \\\frac{1}{2}\\ kilometer wide. It follows the trip takes \\1/6\\ hour or \\10\\ minutes.

Now, we can compute how far downstream he will end up. Since the river runs at a rate of \\4\\ kilometers per hour, and the trip takes \\1/6\\ hour, the distance traveled downstream is given by \\4 \left(\frac{1}{6}\right) = \frac{2}{3}\\ kilometers.

The distance traveled by the swimmer is given by the hypotenuse of a right triangle. The two arms of the triangle are given by the distance across the river, \\\frac{1}{2}\\km, and the distance traveled downstream, \\\frac{2}{3}\\ km. Then, using the Pythagorean Theorem, we can calculate the total distance \\d\\ traveled. \\d = \sqrt{ \left(\frac{2}{3} \right)^2 + \left( \frac{1}{2} \right) ^2 } = \frac{5}{6} \mbox{km}\nonumber \\

Therefore, the swimmer travels a total distance of \\\frac{5}{6}\\ kilometers.

Work

The mathematical concept of work is an application of vectors in \\\mathbb{R}^n\\. The physical concept of work differs from the notion of work employed in ordinary conversation. For example, suppose you were to slide a 150 pound weight off a table which is three feet high and shuffle along the floor for 50 yards, keeping the height always three feet and then deposit this weight on another three foot high table. The physical concept of work would indicate that the force exerted by your arms did no work during this project. The reason for this definition is that even though your arms exerted considerable force on the weight, the direction of motion was at right angles to the force they exerted. The only part of a force which does work in the sense of physics is the component of the force in the direction of motion.

Work is defined to be the magnitude of the component of this force times the distance over which it acts, when the component of force points in the direction of motion. In the case where the force points in exactly the opposite direction of motion work is given by \\\left( -1\right)\\ times the magnitude of this component times the distance. Thus the work done by a force on an object as the object moves from one point to another is a measure of the extent to which the force contributes to the motion. This is illustrated in the following picture in the case where the given force contributes to the motion.

Figure $\PageIndex{4}$

Recall that for any vector \\\vec{u}\\ in \\\mathbb{R}^n\\, we can write \\\vec{u}\\ as a sum of two vectors, as in \\\vec{u} = \vec{u}\_{\|\|} + \vec{u}\_{\perp}\nonumber \\ For any force \\\vec{F}\\, we can write this force as the sum of a vector in the direction of the motion and a vector perpendicular to the motion. In other words, \\\vec{F} = \vec{F}\_{\|\|} + \vec{F}\_{\bot}\nonumber \\

In the above picture the force, \\\vec{F}\\ is applied to an object which moves on the straight line from \\P\\ to \\Q.\\ There are two vectors shown, \\\vec{F}\_{\|\|}\\ and \\\vec{F}\_{\bot }\\ and the picture is intended to indicate that when you add these two vectors you get \\\vec{F}\\. In other words, \\\vec{F} = \vec{F}\_{\|\|} + \vec{F}\_{\bot}\\. Notice that \\\vec{F}\_{\|\|}\\ acts in the direction of motion and \\\vec{F}\_{\bot }\\ acts perpendicular to the direction of motion. Only \\\vec{F}\_{\|\|}\\ contributes to the work done by \\\vec{F}\\ on the object as it moves from \\P\\ to \\Q\\. \\\vec{F}\_{\|\|}\\ is called the component of the force in the direction of motion. From trigonometry, you see the magnitude of \\\vec{F}\_{\|\|}\\ should equal \\ \\ \vec{F} \\ \left\| \cos \theta \right\| .\\ Thus, since \\\vec{F}\_{\|\|}\\ points in the direction of the vector from \\P\\ to \\Q,\\ the total work done should equal \\ \\ \vec{F} \\ \\ \vec{PQ} \\ \cos \theta = \\ \vec{F} \\ \\ \vec{q}-\vec{p} \\ \cos \theta\nonumber \\

Now, suppose the included angle had been obtuse. Then the work done by the force \\\vec{F}\\ on the object would have been negative because \\\vec{F}\_{\|\|}\\ would point in \\-1\\ times the direction of the motion. In this case, \\\cos \theta\\ would also be negative and so it is still the case that the work done would be given by the above formula. Thus from the geometric description of the dot product given above, the work equals \\ \\ \vec{F} \\ \\ \vec{q}-\vec{p} \\ \cos \theta =\vec{F}\bullet \left( \vec{q}-\vec{p}\right)\nonumber \\ This explains the following definition.

##### Definition \\\PageIndex{3}\\: Work Done on an Object by a Force

Let \\\vec{F}\\ be a force acting on an object which moves from the point \\P\\ to the point \\Q\\, which have position vectors given by \\\vec{p}\\ and \\\vec{q}\\ respectively. Then the work done on the object by the given force equals \\\vec{F}\bullet \left( \vec{q}-\vec{p}\right) .\\

Consider the following example.

##### Example \\\PageIndex{4}\\: Finding Work

Let \\\vec{F}= \left $$ \begin{array}{rrr} 2 & 7 & -3 \end{array} \right $$^T\\ Newtons. Find the work done by this force in moving from the point \\\left( 1,2,3\right)\\ to the point \\\left( -9,-3,4\right)\\ along the straight line segment joining these points where distances are measured in meters.

###### Solution

First, compute the vector \\\vec{q} - \vec{p}\\, given by \\\left $$ \begin{array}{rrr} -9 & -3 & 4 \end{array} \right $$^T - \left $$ \begin{array}{rrr} 1 & 2 & 3 \end{array} \right $$^T = \left $$ \begin{array}{rrr} -10 & -5 & 1 \end{array} \right $$^T\nonumber \\

According to Definition $\PageIndex{3}$ the work done is \\\begin{aligned} \left $$ \begin{array}{rrr} 2 & 7 & 3 \end{array} \right $$^T \bullet \left $$ \begin{array}{rrr} -10 & -5 & 1 \end{array} \right $$^T & =-20+\left( -35\right) +\left( -3\right) \\ & =-58\text{ Newton meters}\end{aligned}\\

Note that if the force had been given in pounds and the distance had been given in feet, the units on the work would have been foot pounds. In general, work has units equal to units of a force times units of a length. Recall that \\1\\ Newton meter is equal to \\1\\ Joule. Also notice that the work done by the force can be negative as in the above example.

---

4_E_3A_Exercises

> 来源: LibreTexts

> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/04%3A_R/4.E%3A_Exercises

Skip to main content

Exercise \\\PageIndex{1}\\

Find \\-3\left$$\begin{array}{c}5\\-1\\2\\-3\end{array}\right$$+5\left$$\begin{array}{c}-8\\2\\-3\\6\end{array}\right$$\\.

Answer

\\\left$$\begin{array}{c}-55\\13\\-21\\39\end{array}\right$$\\

Exercise \\\PageIndex{2}\\

Find \\-7\left$$\begin{array}{c}6\\0\\4\\-1\end{array}\right$$+6\left$$\begin{array}{c}-13\\-1\\1\\6\end{array}\right$$\\.

Exercise \\\PageIndex{3}\\

Decide whether \\\vec{v}=\left$$\begin{array}{c}4\\4\\-3\end{array}\right$$\nonumber\\ is a linear combination of the vectors \\\vec{u}\_{1}=\left$$\begin{array}{c}3\\1\\-1\end{array}\right$$\quad\text{and}\quad\vec{u}\_{2}=\left$$\begin{array}{c}2\\-2\\1\end{array}\right$$.\nonumber\\

Answer

\\\left$$\begin{array}{c}4\\4\\-3\end{array}\right$$=2\left$$\begin{array}{c}3\\1\\-1\end{array}\right$$-\left$$\begin{array}{c}2\\-2\\1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{4}\\

Decide whether \\\vec{v}=\left$$\begin{array}{c}4\\4\\4\end{array}\right$$\nonumber\\ is a linear combination of the vectors \\\vec{u}\_1=\left$$\begin{array}{c}3\\1\\-1\end{array}\right$$\quad\text{and}\quad\vec{u}\_2=\left$$\begin{array}{c}2\\-2\\1\end{array}\right$$.\nonumber\\

Answer

The system \\\left$$\begin{array}{c}4\\4\\4\end{array}\right$$=a_1\left$$\begin{array}{c}3\\1\\-1\end{array}\right$$+a_2\left$$\begin{array}{c}2\\-2\\1\end{array}\right$$\nonumber\\ has no solution.

Exercise \\\PageIndex{5}\\

Find the vector equation for the line through \$−7, 6, 0)\\ and \$−1, 1, 4)\\. Then, find the parametric equations for this line.

Exercise \\\PageIndex{6}\\

Find parametric equations for the line through the point \$7, 7, 1)\\ with a direction vector \\\vec{d}=\left$$\begin{array}{c}1\\6\\2\end{array}\right$$\\.

Exercise \\\PageIndex{7}\\

Parametric equations of the line are \\\begin{aligned}x&=t+2 \\ y&=6-3t \\ x&=-t=6\end{aligned}\\ Find a direction vector for the line and a point on the line.

Exercise \\\PageIndex{8}\\

Find the vector equation for the line through the two points \$−5, 5, 1),\\ (2, 2, 4)\\. Then, find the parametric equations.

Exercise \\\PageIndex{9}\\

The equation of a line in two dimensions is written as \\y = x−5\\. Find parametric equations for this line.

Exercise \\\PageIndex{10}\\

Find parametric equations for the line through \$6, 5,−2)\\ and \$5, 1, 2)\\.

Exercise \\\PageIndex{11}\\

Find the vector equation and parametric equations for the line through the point \$−7, 10,−6)\\ with a direction vector \\\vec{d}=\left$$\begin{array}{c}1\\1\\3\end{array}\right$$\\.

Exercise \\\PageIndex{12}\\

Parametric equations of the line are \\\begin{aligned}x&=2t+2 \\ y&=5-4t \\ z&=-t-3\end{aligned}\\ Find a direction vector for the line and a point on the line, and write the vector equation of the line.

Exercise \\\PageIndex{13}\\

Find the vector equation and parametric equations for the line through the two points \$4, 10, 0),\\ (1,−5,−6)\\.

Exercise \\\PageIndex{14}\\

Find the point on the line segment from \\P = (−4, 7, 5)\\ to \\Q = (2,−2,−3)\\ which is \\\frac{1}{7}\\ of the way from \\P\\ to \\Q\\.

Exercise \\\PageIndex{15}\\

Suppose a triangle in \\\mathbb{R}^n\\ has vertices at \\P_1,\\ P_2,\\ and \\P_3\\. Consider the lines which are drawn from a vertex to the mid point of the opposite side. Show these three lines intersect in a point and find the coordinates of this point.

Exercise \\\PageIndex{16}\\

Find \\\left$$\begin{array}{c}1\\2\\3\\4\end{array}\right$$\bullet\left$$\begin{array}{c}2\\0\\1\\3\end{array}\right$$\\.

Answer

\\\left$$\begin{array}{c}1\\2\\3\\4\end{array}\right$$\bullet\left$$\begin{array}{c}2\\0\\1\\3\end{array}\right$$=17\\

Exercise \\\PageIndex{17}\\

Use the formula given in Proposition 4.7.2 to verify the Cauchy Schwarz inequality and to show that equality occurs if and only if one of the vectors is a scalar multiple of the other

Answer

This formula says that \\\vec{u}\bullet\vec{v} = \|\|\vec{u}\|\|\\\|\|\vec{v}\|\|\cos\theta\\ where \\θ\\ is the included angle between the two vectors. Thus \\\|\|\vec{u}\bullet\vec{v}\|\|=\|\|\vec{u}\|\|\\\|\|\vec{v}\|\|\\\|\|\cos\theta\|\|\leq \|\|\vec{u}\|\|\\\|\|\vec{v}\|\|\nonumber\\ and equality holds if and only if \\\theta = 0\\ or \\π\\. This means that the two vectors either point in the same direction or opposite directions. Hence one is a multiple of the other.

Exercise \\\PageIndex{18}\\

For \\\vec{u}\\, \\\vec{v}\\ vectors in \\\mathbb{R}^3\\, define the product, \\\vec{u}\ast\vec{v} = u_1v_1 +2u_2v_2 +3u_3v_3\\. Show the axioms for a dot product all hold for this product. Prove \\\|\|\vec{u}\ast\vec{v}\|\|\leq (\vec{u}\ast\vec{u})^{1/2}(\vec{v}\ast\vec{v})^{1/2}\nonumber\\

Answer

This follows from the Cauchy Schwarz inequality and the proof of Theorem 4.7.1 which only used the properties of the dot product. Since this new product has the same properties the Cauchy Schwarz inequality holds for it as well.

Exercise \\\PageIndex{19}\\

Let \\\vec{a}\\, \\\vec{b}\\ be vectors. Show that \\\left(\vec{a}\bullet\vec{b}\right)=\frac{1}{4}\left(\|\|\vec{a}+\vec{b}\|\|^2-\|\|\vec{a}-\vec{b}\|\|^2\right).\\

Exercise \\\PageIndex{20}\\

Using the axioms of the dot product, prove the parallelogram identity: \\\|\|\vec{a}+\vec{b}\|\|^2+\|\|\vec{a}-\vec{b}\|\|^2=2\|\|\vec{a}\|\|^2+2\|\|\vec{b}\|\|^2\nonumber\\

Exercise \\\PageIndex{21}\\

Let \\A\\ be a real \\m\times n\\ matrix and let \\\vec{u} ∈ \mathbb{R}^n\\ and \\\vec{v} ∈ \mathbb{R}^m\\. Show \\A\vec{u}\bullet\vec{v} =\vec{u}\bullet A^T\vec{v}\\. Hint: Use the definition of matrix multiplication to do this.

Answer

\\A\vec{x}\bullet\vec{y}=\sum_k(A\vec{x})\_ky_k=\sum_k\sum_iA\_{ki}x_iy_k=\sum_i\sum_kA^T\_{ik}x_iy_k=\vec{x}\bullet A^T\vec{y}\\

Exercise \\\PageIndex{22}\\

Use the result of Problem $\PageIndex{21}$ to verify directly that \$AB)^T = B^TA^T\\ without making any reference to subscripts.

Answer

\\\begin{aligned}AB\vec{x}\bullet\vec{y}&=B\vec{x}\bullet A^T\vec{y} \\ &=\vec{x}\bullet B^TA^T\vec{y} \\ &=\vec{x}\bullet (AB)^T\vec{y}\end{aligned}\\ Since this is true for all \\\vec{x}\\, it follows that, in particular, it holds for \\\vec{x}=B^TA^T\vec{y}-(AB)^T\vec{y}\nonumber\\ and so from the axioms of the dot product, \\\left(B^TA^T\vec{y}-(AB)^T\vec{y}\right)\bullet\left(B^TA^T\vec{y}-(AB)^T\vec{y}\right)=0\nonumber\\ and so \\B^TA^T\vec{y}-(AB)^T\vec{y}=\vec{0}\\. However, this is true for all \\\vec{y}\\ and so \\B^TA^T-(AB)^T=0\\.

Exercise \\\PageIndex{23}\\

Find the angle between the vectors \\\vec{u}=\left$$\begin{array}{r}3\\-1\\-1\end{array}\right$$,\\\vec{v}=\left$$\begin{array}{c}1\\4\\2\end{array}\right$$\nonumber\\

Answer

\\\frac{\left$$\begin{array}{ccc}3&-1&-1\end{array}\right$$^T\bullet\left$$\begin{array}{ccc}1&4&2\end{array}\right$$^T}{\sqrt{9+1+1}\sqrt{1+16+4}}=\frac{-3}{\sqrt{11}\sqrt{21}}=-0.19739=\cos\theta\\ Therefore we need to solve \\-0.19739=\cos\theta\nonumber\\ Thus \\\theta=1.7695\\ radians.

Exercise \\\PageIndex{24}\\

Find the angle between the vectors \\\vec{u}=\left$$\begin{array}{r}1\\-2\\1\end{array}\right$$,\\\vec{v}=\left$$\begin{array}{r}1\\2\\-7\end{array}\right$$\nonumber\\

Answer

\\\frac{-10}{\sqrt{1+4+1}\sqrt{1+4+49}}=-0.55555=\cos\theta\\ Therefore we need to solve \\−0.55555 = \cos θ\\, which gives \\θ = 2.0313\\ radians.

Exercise \\\PageIndex{25}\\

Find \\\text{proj}\_{\vec{v}}(\vec{w})\\ where \\\vec{w}=\left$$\begin{array}{r}1\\0\\-2\end{array}\right$$\\ and \\\vec{v}=\left$$\begin{array}{c}1\\2\\3\end{array}\right$$\\.

Answer

\\\frac{\vec{u}\bullet\vec{v}}{\vec{u}\bullet\vec{u}}\vec{u}=\frac{-5}{14}\left$$\begin{array}{c}1\\2\\3\end{array}\right$$=\left$$\begin{array}{r}-\frac{5}{14}\\-\frac{5}{7}\\-\frac{15}{14}\end{array}\right$$\\

Exercise \\\PageIndex{26}\\

Find \\\text{proj}\_{\vec{v}}(\vec{w})\\ where \\\vec{w}=\left$$\begin{array}{r}1\\2\\-2\end{array}\right$$\\ and \\\vec{v}=\left$$\begin{array}{c}1\\0\\3\end{array}\right$$\\.

Answer

\\\frac{\vec{u}\bullet\vec{v}}{\vec{u}\bullet\vec{u}}\vec{u}=\frac{-5}{10}\left$$\begin{array}{c}1\\0\\3\end{array}\right$$=\left$$\begin{array}{r}-\frac{1}{2}\\0\\-\frac{3}{2}\end{array}\right$$\\

Exercise \\\PageIndex{27}\\

Find \\\text{proj}\_{\vec{v}}(\vec{w})\\ where \\\vec{w}=\left$$\begin{array}{r}1\\2\\-2\\1\end{array}\right$$\\ and \\\vec{v}=\left$$\begin{array}{c}1\\2\\3\\0\end{array}\right$$\\.

Answer

\\\frac{\vec{u}\bullet\vec{v}}{\vec{u}\bullet\vec{u}}\vec{u}=\frac{\left$$\begin{array}{cccc}1&2&-2&1\end{array}\right$$^T\bullet\left$$\begin{array}{cccc}1&2&3&0\end{array}\right$$^T}{1+4+9}\left$$\begin{array}{c}1\\2\\3\\0\end{array}\right$$=\left$$\begin{array}{r}-\frac{1}{14}\\-\frac{1}{7}\\-\frac{3}{14}\\0\end{array}\right$$\\

Exercise \\\PageIndex{28}\\

Let \\P = (1, 2, 3)\\ be a point in \\\mathbb{R}^3\\. Let \\L\\ be the line through the point \\P_0 = (1, 4, 5)\\ with direction vector \\\vec{d} =\left$$\begin{array}{r}1\\-1\\1\end{array}\right$$\\. Find the shortest distance from \\P\\ to \\L\\, and find the point \\Q\\ on \\L\\ that is closest to \\P\\.

Exercise \\\PageIndex{29}\\

Let \\P = (0, 2, 1)\\ be a point in \\\mathbb{R}^3\\. Let \\L\\ be the line through the point \\P_0 = (1, 1, 1)\\ with direction vector \\\vec{d} =\left$$\begin{array}{c}3\\0\\1\end{array}\right$$\\. Find the shortest distance from \\P\\ to \\L\\, and find the point \\Q\\ on \\L\\ that is closest to \\P\\.

Exercise \\\PageIndex{30}\\

Does it make sense to speak of \\\text{proj}\_{\vec{0}} (\vec{w})\\?

Answer

No, it does not. The \\0\\ vector has no direction. The formula for \\\text{proj}\_{\vec{0}} (\vec{w})\\ doesn’t make sense either.

Exercise \\\PageIndex{31}\\

Prove the Cauchy Schwarz inequality in \\\mathbb{R}^n\\ as follows. For \\\vec{u}\\,\\\vec{v}\\ vectors, consider \$\vec{w}-\text{proj}\_{\vec{v}}\vec{w})\bullet (\vec{w}-\text{proj}\_{\vec{v}}\vec{w})\geq 0\nonumber\\ Simplify using the axioms of the dot product and then put in the formula for the projection. Notice that this expression equals \\0\\ and you get equality in the Cauchy Schwarz inequality if and only if \\\vec{w} = \text{proj}\_{\vec{v}}\vec{w}\\. What is the geometric meaning of \\\vec{w}= \text{proj}\_{\vec{v}}\vec{w}\\?

Answer

\\\left(\vec{u}-\frac{\vec{u}\bullet\vec{v}}{\|\|\vec{v}\|\|^2}\vec{v}\right)\bullet\left(\vec{u}-\frac{\vec{u}\bullet\vec{v}}{\|\|\vec{v}\|\|^2}\vec{v}\right)=\|\|\vec{u}\|\|^2-2(\vec{u}\bullet\vec{v})^2\frac{1}{\|\|\vec{v}\|\|^2}+(\vec{u}\bullet\vec{v})^2\frac{1}{\|\|\vec{v}\|\|^2}\geq 0\nonumber\\ And so \\\|\|\vec{u}\|\|^2\|\|\vec{v}\|\|^2\geq (\vec{u}\bullet\vec{v})^2\nonumber\\ You get equality exactly when \\\vec{u}=\text{proj}\_{\vec{v}}\vec{u}=\frac{\vec{u}\bullet\vec{v}}{\|\|\vec{v}\|\|^2}\vec{v}\\ in other words, when \\\vec{u}\\ is a multiple of \\\vec{v}\\.

Exercise \\\PageIndex{32}\\

Let \\\vec{v},\\\vec{w},\\\vec{u}\\ be vectors. Show that \$\vec{w}+\vec{u})\_{\perp}=\vec{w}\_\perp +\vec{u}\_\perp\\ where \\\vec{w}\_\perp =\vec{w}-\text{proj}\_{\vec{v}}(\vec{w})\\.

Answer

\\\begin{aligned}\vec{w}-\text{proj}\_{\vec{v}}(\vec{w})+\vec{u}-\text{proj}\_{\vec{v}}(\vec{u})&=\vec{w}+\vec{u}-(\text{proj}\_{\vec{v}}(\vec{w})+\text{proj}\_{\vec{v}}(\vec{u})) \\ &=\vec{w}+\vec{u}-\text{proj}\_{\vec{v}}(\vec{w}+\vec{u})\end{aligned}\\ This follows because \\\begin{aligned}\text{proj}\_{\vec{v}}(\vec{w})+\text{proj}\_{\vec{v}}(\vec{u})&=\frac{\vec{u}\bullet\vec{v}}{\|\|\vec{v}\|\|^2}\vec{v}+\frac{\vec{w}\bullet\vec{v}}{\|\|\vec{v}\|\|^2}\vec{v} \\ &=\frac{(\vec{u}+\vec{w})\bullet\vec{v}}{\|\|\vec{v}\|\|^2}\vec{v} \\ &=\text{proj}\_{\vec{v}}(\vec{w}+\vec{u})\end{aligned}\\

Exercise \\\PageIndex{33}\\

Show that \$\vec{v}-\text{proj}\_{\vec{u}}(\vec{v}),\vec{u})=(\vec{v}-\text{proj}\_{\vec{u}}(\vec{v}))\bullet\vec{u}=0\nonumber\\ and conclude every vector in \\\mathbb{R}^n\\ can be written as the sum of two vectors, one which is perpendicular and one which is parallel to the given vector.

Answer

\$\vec{v}-\text{proj}\_{\vec{u}}(\vec{v}))\bullet\vec{u}=\vec{v}\bullet\vec{u}-\left(\frac{(\vec{v}\cdot\vec{u}}{\|\|\vec{u}\|\|^2}\vec{u}\right)\bullet\vec{u}=\vec{v}\bullet\vec{u}-\vec{v}\bullet\vec{u}=0\\. Therefore, \\\vec{v}=\vec{v}-\text{proj}\_{\vec{u}}(\vec{v})+\text{proj}\_{\vec{u}}(\vec{v})\\. The first is perpendicular to \\\vec{u}\\ and the second is a multiple of \\\vec{u}\\ so it is parallel to \\\vec{u}\\.

Exercise \\\PageIndex{34}\\

Show that if \\\vec{a}\times\vec{u}=\vec{0}\\ for any unit vector \\\vec{u}\\, then \\\vec{a}=\vec{0}\\.

Answer

If \\\vec{a}\neq\vec{0}\\, then the condition says that \\\|\|\vec{a}\times\vec{u}\|\|=\|\|\vec{a}\|\|\sin\theta =0\\ for all angles \\θ\\. Hence \\\vec{a}=\vec{0}\\ after all.

Exercise \\\PageIndex{35}\\

Find the area of the triangle determined by the three points \$1, 2, 3),\\ (4, 2, 0)\\ and \$−3, 2, 1)\\.

Answer

\\\left$$\begin{array}{r}3\\0\\-3\end{array}\right$$\times\left$$\begin{array}{r}-4\\0\\-2\end{array}\right$$=\left$$\begin{array}{r}0\\18\\0\end{array}\right$$\\. So the area is \\9\\.

Exercise \\\PageIndex{36}\\

Find the area of the triangle determined by the three points \$1, 0, 3),\\ (4, 1, 0)\\ and \$−3, 1, 1)\\.

Answer

\\\left$$\begin{array}{r}3\\1\\-3\end{array}\right$$\times\left$$\begin{array}{r}-4\\1\\-2\end{array}\right$$=\left$$\begin{array}{c}1\\18\\7\end{array}\right$$\\. The area is given by \\\frac{1}{2}\sqrt{1+(18)^2+49}=\frac{1}{2}\sqrt{374}\nonumber\\

Exercise \\\PageIndex{37}\\

Find the area of the triangle determined by the three points, \$1, 2, 3),\\ (2, 3, 4)\\ and \$3, 4, 5)\\. Did something interesting happen here? What does it mean geometrically?

Answer

\\\left$$\begin{array}{ccc}1&1&1\end{array}\right$$\times\left$$\begin{array}{ccc}2&2&2\end{array}\right$$=\left$$\begin{array}{ccc}0&0&0\end{array}\right$$\\. The area is \\0\\. It means the three points are on the same line.

Exercise \\\PageIndex{38}\\

Find the area of the parallelogram determined by the vectors \\\left$$\begin{array}{c}1\\2\\3\end{array}\right$$\\, \\\left$$\begin{array}{r}3\\-2\\1\end{array}\right$$\\.

Answer

\\\left$$\begin{array}{c}1\\2\\3\end{array}\right$$\times\left$$\begin{array}{r}3\\-2\\1\end{array}\right$$=\left$$\begin{array}{r}8\\8\\-8\end{array}\right$$\\. The area is \\8\sqrt{3}\\.

Exercise \\\PageIndex{39}\\

Find the area of the parallelogram determined by the vectors \\\left$$\begin{array}{c}1\\0\\3\end{array}\right$$\\, \\\left$$\begin{array}{r}4\\-2\\1\end{array}\right$$\\.

Answer

\\\left$$\begin{array}{c}1\\0\\3\end{array}\right$$\times\left$$\begin{array}{r}4\\-2\\1\end{array}\right$$=\left$$\begin{array}{r}6\\11\\-2\end{array}\right$$\\. The area is \\\sqrt{36+121+4}=\sqrt{161}\\.

Exercise \\\PageIndex{40}\\

Is \\\vec{u}\times (\vec{v}\times\vec{w})=(\vec{u}\times\vec{v})\times\vec{w}\\? What is the meaning of \\\vec{u}\times\vec{v}\times\vec{w}\\? Explain. Hint: Try \\\left(\vec{i}\times\vec{j}\right)\times\vec{k}\\.

Answer

\\\left(\vec{i}\times\vec{j}\right)\times\vec{j}=\vec{k}\times\vec{j}=i\vec{i}\\. However, \\\vec{i}\times\left(\vec{j}\times\vec{j}\right)=\vec{0}\\ and so the cross product is not associative.

Exercise \\\PageIndex{41}\\

Verify directly that the coordinate description of the cross product, \\\vec{u}\times\vec{v}\\ has the property that it is perpendicular to both \\\vec{u}\\ and \\\vec{v}\\. Then show by direct computation that this coordinate description satisfies \\\begin{aligned} \|\|\vec{u}\times\vec{v}\|\|^2&=\|\|\vec{u}\|\|^2\|\|\vec{v}\|\|^2-(\vec{u}\bullet\vec{v})^2 \\ &=\|\|\vec{u}\|\|^2\|\|\vec{v}\|\|^2(1-\cos^2(\theta ))\end{aligned}\\ where \\\theta\\ is the angle included between the two vectors. Explain why \\\|\|\vec{u}\times\vec{v}\|\|\\ has the correct magnitude.

Answer

Verify directly from the coordinate description of the cross product that the right hand rule applies to the vectors \\\vec{i},\vec{j},\vec{k}\\. Next verify that the distributive law holds for the coordinate description of the cross product. This gives another way to approach the cross product. First define it in terms of coordinates and then get the geometric properties from this. However, this approach does not yield the right hand rule property very easily. From the coordinate description, \\\vec{a}\times\vec{b}\cdot\vec{a}=\epsilon\_{ijk}a_jb_ka_i=-\epsilon\_{jik}a_kb_ka_i=-\epsilon\_{jik}b_ka_ia_j=-\vec{a}\times\vec{b}\cdot\vec{a}\nonumber\\ and so \\\vec{a}\times\vec{b}\\ is perpendicular to \\\vec{a}\\. Similarly, \\\vec{a}\times\vec{b}\\ is perpendicular to \\\vec{b}\\. Now we need that \\\|\|\vec{a}\times\vec{b}\|\|^2=\|\|\vec{a}\|\|^2\|\|\vec{b}\|\|^2(1-\cos^2\theta )=\|\|\vec{a}\|\|^2\|\|\vec{b}\|\|^2\sin^2\theta\nonumber\\ and so \\\|\|\vec{a}\times\vec{b}\|\|=\|\|\vec{a}\|\|\\\|\|\vec{b}\|\|\sin\theta\\, the area of the parallelogram determined by \\\vec{a}\\, \\\vec{b}\\. Only the right hand rule is a little problematic. However, you can see right away from the component definition that the right hand rule holds for each of the standard unit vectors. Thus \\\vec{i}\times\vec{j}=\vec{k}\\ etc. \\\left\|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\1&0&0\\0&1&0\end{array}\right\|=\vec{k}\nonumber\\

Exercise \\\PageIndex{42}\\

Suppose \\A\\ is a \\3\times 3\\ skew symmetric matrix such that \\A^T = −A\\. Show there exists a vector \\\vec{Ω}\\ such that for all \\\vec{u} ∈ \mathbb{R}^3\\ \\A\vec{u}=\vec{\Omega}\times\vec{u}\nonumber\\ Hint: Explain why since \\A\\ is skew symmetric it is of the form \\A=\left$$\begin{array}{ccc}0&-\omega_3&\omega_2 \\ \omega_3&0&-\omega_1 \\ -\omega_2&\omega_1&0\end{array}\right$$\nonumber\\ where the \\\omega_i\\ are numbers. Then consider \\\omega_1\vec{i}+\omega_2\vec{j}+\omega_3\vec{k}\\.

Exercise \\\PageIndex{43}\\

Find the volume of the parallelepiped determined by the vectors \\\left$$\begin{array}{r}1\\-7\\-5\end{array}\right$$\\, \\\left$$\begin{array}{r}1\\-2\\-6\end{array}\right$$\\, and \\\left$$\begin{array}{c}3\\2\\3\end{array}\right$$\\.

Answer

\\\left\|\begin{array}{ccc}1&-7&-5 \\ 1&-2&-6 \\ 3&2&3\end{array}\right\|=113\\

Exercise \\\PageIndex{44}\\

Suppose \\\vec{u}\\, \\\vec{v}\\, and \\\vec{w}\\ are three vectors whose components are all integers. Can you conclude the volume of the parallelepiped determined from these three vectors will always be an integer?

Answer

Yes. It will involve the sum of product of integers and so it will be an integer.

Exercise \\\PageIndex{45}\\

What does it mean geometrically if the box product of three vectors gives zero?

Answer

It means that if you place them so that they all have their tails at the same point, the three will lie in the same plane.

Exercise \\\PageIndex{46}\\

Using Problem $\PageIndex{45}$, find an equation of a plane containing the two position vectors, \\\vec{p}\\ and \\\vec{q}\\ and the point \\0\\. Hint: If \$x, y,z)\\ is a point on this plane, the volume of the parallelepiped determined by \$x, y,z)\\ and the vectors \\\vec{p}\\, \\\vec{q}\\ equals \\0\\.

Answer

\\\vec{x}\bullet\left(\vec{a}\times\vec{b}\right)=0\\

Exercise \\\PageIndex{47}\\

Using the notion of the box product yielding either plus or minus the volume of the parallelepiped determined by the given three vectors, show that \$\vec{u}\times\vec{v})\bullet\vec{w}=\vec{u}\bullet (\vec{v}\times\vec{w})\nonumber\\ In other words, the dot and the cross can be switched as long as the order of the vectors remains the same. Hint: There are two ways to do this, by the coordinate description of the dot and cross product and by geometric reasoning.

Exercise \\\PageIndex{48}\\

Simplify \$\vec{u}\times\vec{v})\bullet (\vec{v}\times\vec{w})\times (\vec{w}\times\vec{z})\\.

Answer

Here \\$$\vec{v},\vec{w},\vec{z}$$\\ denotes the box product. Consider the cross product term. From the above, \\\begin{aligned}(\vec{v}\times\vec{w})\times(\vec{w}\times\vec{z})&=$$\vec{v},\vec{w},\vec{z}$$\vec{w}-$$\vec{w},\vec{w},\vec{z}$$\vec{v} \\ &=$$\vec{v},\vec{w},\vec{z}$$\vec{w}\end{aligned}\\ Thus it reduces to \$\vec{u}\times\vec{v})\bullet $$\vec{v},\vec{w},\vec{z}$$\vec{w}=$$\vec{v},\vec{w},\vec{z}$$$$\vec{u},\vec{v},\vec{w}$$\nonumber\\

Exercise \\\PageIndex{49}\\

Simplify \\\|\|\vec{u}\times\vec{v}\|\|^2+(\vec{u}\bullet\vec{v})^2-\|\|\vec{u}\|\|^2\|\|\vec{v}\|\|^2\\.

Answer

\\\begin{aligned}\|\|\vec{u}\times\vec{v}\|\|^2&=\epsilon\_{ijk}u_jv_k\epsilon\_{irs}u_rv_s=(\delta\_{jr}\delta\_{ks}-\delta\_{kr}\delta\_{js})u_rv_su_jv_k \\ &=u_jv_ku_jv_k-u_kv_ju_jv_k=\|\|\vec{u}\|\|^2\|\|\vec{v}\|\|^2-(\vec{u}\bullet\vec{v})^2\end{aligned}\\ It follows that the expression reduces to \\0\\. You can also do the following. \\\begin{aligned}\|\|\vec{u}\times\vec{v}\|\|^2&=\|\|\vec{u}\|\|^2\|\|\vec{v}\|\|^2\sin^2\theta \\ &=\|\|\vec{u}\|\|^2\|\|\vec{v}\|\|^2(1-\cos^2\theta ) \\ &=\|\|\vec{u}\|\|^2\|\|\vec{v}\|\|^2-\|\|\vec{u}\|\|^2\|\|\vec{v}\|\|^2\cos^2\theta \\ &=\|\|\vec{u}\|\|^2\|\|\vec{v}\|\|^2-(\vec{u}\bullet\vec{v})^2\end{aligned}\\ which implies the expression equals \\0\\.

Exercise \\\PageIndex{50}\\

For \\\vec{u},\\\vec{v},\\\vec{w}\\ functions of \\t\\, prove the following product rules: \\\begin{aligned}(\vec{u}\times\vec{v})'&=\vec{u}'\times\vec{v}+\vec{u}\times\vec{v}' \\ (\vec{u}\bullet\vec{v})'&=\vec{u}'\bullet\vec{v}+\vec{u}\bullet\vec{v}'\end{aligned}\\

Answer

We will show it using the summation convention and permutation symbol \\\begin{aligned}((\vec{u}\times\vec{v})')\_i&=((\vec{u}\times\vec{v})\_i)'=(\epsilon\_{ijk}u_jv_k)' \\ &=\epsilon\_{ijk}u_j'v_k+\epsilon\_{ijk}u_kv_k'=(\vec{u}'\times\vec{v}+\vec{u}\times\vec{v}')\_i\end{aligned}\\ and so \$\vec{u}\times\vec{v})'=\vec{u}'\times\vec{v}+\vec{u}\times\vec{v}'\\.

Exercise \\\PageIndex{51}\\

Here are some vectors. \\\left$$\begin{array}{r}1\\1\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\2\\-2\end{array}\right$$,\\\left$$\begin{array}{r}2\\7\\-4\end{array}\right$$,\\\left$$\begin{array}{r}5\\7\\-10\end{array}\right$$,\\\left$$\begin{array}{r}12\\17\\-24\end{array}\right$$\nonumber\\ Describe the span of these vectors as the span of as few vectors as possible.

Exercise \\\PageIndex{52}\\

Here are some vectors. \\\left$$\begin{array}{r}1\\2\\-2\end{array}\right$$,\\\left$$\begin{array}{r}12\\29\\-24\end{array}\right$$,\\\left$$\begin{array}{r}1\\3\\-2\end{array}\right$$,\\\left$$\begin{array}{r}2\\9\\-4\end{array}\right$$,\\\left$$\begin{array}{r}5\\12\\-10\end{array}\right$$.\nonumber\\ Describe the span of these vectors as the span of as few vectors as possible.

Exercise \\\PageIndex{53}\\

Here are some vectors. \\\left$$\begin{array}{r}1\\2\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\3\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\-2\\-2\end{array}\right$$,\\\left$$\begin{array}{r}-1\\0\\2\end{array}\right$$,\\\left$$\begin{array}{r}1\\3\\-1\end{array}\right$$\nonumber\\ Describe the span of these vectors as the span of as few vectors as possible.

Exercise \\\PageIndex{54}\\

Here are some vectors. \\\left$$\begin{array}{r}1\\1\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\2\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\-3\\-2\end{array}\right$$,\\\left$$\begin{array}{r}-1\\1\\2\end{array}\right$$\nonumber\\ Now here is another vector: \\\left$$\begin{array}{r}1\\2\\-1\end{array}\right$$\nonumber\\ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Exercise \\\PageIndex{55}\\

Here are some vectors. \\\left$$\begin{array}{r}1\\1\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\2\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\-3\\-2\end{array}\right$$,\\\left$$\begin{array}{r}-1\\1\\2\end{array}\right$$\nonumber\\ Now here is another vector: \\\left$$\begin{array}{r}2\\-3\\-4\end{array}\right$$\nonumber\\ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Exercise \\\PageIndex{56}\\

Here are some vectors. \\\left$$\begin{array}{r}1\\1\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\2\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\-3\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\2\\-1\end{array}\right$$\nonumber\\ Now here is another vector: \\\left$$\begin{array}{r}1\\9\\1\end{array}\right$$\nonumber\\ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Exercise \\\PageIndex{57}\\

Here are some vectors, \\\left$$\begin{array}{r}1\\-1\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\-5\\-2\end{array}\right$$,\\\left$$\begin{array}{r}-1\\5\\2\end{array}\right$$\nonumber\\ Now here is another vector: \\\left$$\begin{array}{r}1\\1\\-1\end{array}\right$$\nonumber\\ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Exercise \\\PageIndex{58}\\

Here are some vectors. \\\left$$\begin{array}{r}1\\-1\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\-5\\-2\end{array}\right$$,\\\left$$\begin{array}{r}-1\\5\\2\end{array}\right$$\nonumber\\ Now here is another vector: \\\left$$\begin{array}{r}1\\1\\-1\end{array}\right$$\nonumber\\ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Exercise \\\PageIndex{59}\\

Here are some vectors. \\\left$$\begin{array}{r}1\\0\\-2\end{array}\right$$,\\\left$$\begin{array}{r}1\\1\\-2\end{array}\right$$,\\\left$$\begin{array}{r}2\\-2\\-3\end{array}\right$$,\\\left$$\begin{array}{r}-1\\4\\2\end{array}\right$$\nonumber\\ Now here is another vector: \\\left$$\begin{array}{r}-1\\-4\\2\end{array}\right$$\nonumber\\ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

Exercise \\\PageIndex{60}\\

Suppose \\\\\vec{x}\_1,\cdots ,\vec{x}\_k\\\\ is a set of vectors from \\\mathbb{R}^n\\. Show that \\\vec{0}\\ is in \\span\\\vec{x}\_1,\cdots ,\vec{x}\_k\\\\.

Answer

\\\sum\limits\_{i=1}^k 0\vec{x}\_k=\vec{0}\\

Exercise \\\PageIndex{61}\\

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \\\left$$\begin{array}{r}1\\3\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\4\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\4\\0\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\10\\2\\1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{62}\\

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \\\left$$\begin{array}{r}-1\\-2\\2\\3\end{array}\right$$,\\\left$$\begin{array}{r}-3\\-4\\3\\3\end{array}\right$$,\\\left$$\begin{array}{r}0\\-1\\4\\3\end{array}\right$$,\\\left$$\begin{array}{r}0\\-1\\6\\4\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{63}\\

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \\\left$$\begin{array}{r}1\\5\\-2\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\6\\-3\\1\end{array}\right$$,\\\left$$\begin{array}{r}-1\\-4\\1\\-1\end{array}\right$$,\\\left$$\begin{array}{r}1\\6\\-2\\1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{64}\\

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \\\left$$\begin{array}{r}1\\-1\\3\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\6\\34\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\7\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\8\\1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{65}\\

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. \\\left$$\begin{array}{r}1\\3\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\4\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}-3\\-10\\3\\-3\end{array}\right$$,\\\left$$\begin{array}{r}1\\4\\0\\1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{66}\\

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \\\left$$\begin{array}{r}1\\3\\-3\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\4\\-5\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\4\\-4\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\10\\-14\\1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{67}\\

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \\\left$$\begin{array}{r}1\\0\\3\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\1\\8\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\7\\34\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\1\\7\\1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{68}\\

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \\\left$$\begin{array}{r}1\\4\\-2\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\5\\-3\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\7\\-5\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\5\\-2\\1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{69}\\

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. \\\left$$\begin{array}{r}1\\2\\2\\-4\end{array}\right$$,\\\left$$\begin{array}{r}3\\4\\1\\-4\end{array}\right$$,\\\left$$\begin{array}{r}0\\-1\\0\\4\end{array}\right$$,\\\left$$\begin{array}{r}0\\-1\\-2\\5\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{70}\\

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. \\\left$$\begin{array}{r}2\\3\\1\\-3\end{array}\right$$,\\\left$$\begin{array}{r}-5\\-6\\0\\3\end{array}\right$$,\\\left$$\begin{array}{r}-1\\-2\\1\\3\end{array}\right$$,\\\left$$\begin{array}{r}-1\\-2\\0\\4\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{71}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\1\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\2\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\-2\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\2\\0\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\-1\\-1\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Exercise \\\PageIndex{72}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\2\\-2\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\3\\-3\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\3\\-2\\1\end{array}\right$$,\\\left$$\begin{array}{r}4\\3\\-1\\4\end{array}\right$$,\\\left$$\begin{array}{r}1\\3\\-2\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Exercise \\\PageIndex{73}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\1\\0\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\2\\1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\-2\\-3\\1\end{array}\right$$,\\\left$$\begin{array}{r}2\\-5\\-7\\2\end{array}\right$$,\\\left$$\begin{array}{r}1\\2\\2\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Exercise \\\PageIndex{74}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\2\\-2\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\3\\-3\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\-1\\1\\1\end{array}\right$$,\\\left$$\begin{array}{r}2\\-3\\3\\2\end{array}\right$$,\\\left$$\begin{array}{r}1\\3\\-2\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Exercise \\\PageIndex{75}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\4\\-2\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\5\\-3\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\5\\-2\\1\end{array}\right$$,\\\left$$\begin{array}{r}4\\11\\-1\\4\end{array}\right$$,\\\left$$\begin{array}{r}1\\5\\-3\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Exercise \\\PageIndex{76}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\3\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}-\frac{3}{2}\\-\frac{9}{2}\\ \frac{3}{2}\\ -\frac{3}{2}\end{array}\right$$,\\\left$$\begin{array}{r}1\\4\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}2\\-1\\-2\\2\end{array}\right$$,\\\left$$\begin{array}{r}1\\4\\0\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Exercise \\\PageIndex{77}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\3\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\4\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}2\\-1\\-2\\2\end{array}\right$$,\\\left$$\begin{array}{r}1\\4\\0\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors

Exercise \\\PageIndex{78}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\4\\-2\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\5\\-3\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\1\\1\\1\end{array}\right$$,\\\left$$\begin{array}{r}2\\1\\3\\2\end{array}\right$$,\\\left$$\begin{array}{r}1\\5\\-2\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Exercise \\\PageIndex{79}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\-1\\3\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\7\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\8\\1\end{array}\right$$,\\\left$$\begin{array}{r}4\\-9\\-6\\4\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\8\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Exercise \\\PageIndex{80}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\-1\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}-3\\3\\3\\-3\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}2\\-9\\-2\\2\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\0\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Exercise \\\PageIndex{81}\\

Here are some vectors in \\\mathbb{R}^4\\. \\\left$$\begin{array}{r}1\\b+1\\a\\1\end{array}\right$$,\\\left$$\begin{array}{r}3\\3b+3\\3a\\3\end{array}\right$$,\\\left$$\begin{array}{r}1\\b+2\\2a+1\\1\end{array}\right$$,\\\left$$\begin{array}{r}2\\2b-5\\-5a-7\\2\end{array}\right$$,\\\left$$\begin{array}{r}1\\b+2\\2a+2\\1\end{array}\right$$\nonumber\\ These vectors can’t possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Exercise \\\PageIndex{82}\\

Let \\H=span\left\\\left$$\begin{array}{r}2\\1\\1\\1\end{array}\right$$,\\\left$$\begin{array}{r}-1\\0\\-1\\-1\end{array}\right$$,\\\left$$\begin{array}{r}5\\2\\3\\3\end{array}\right$$,\\\left$$\begin{array}{r}-1\\1\\-2\\-2\end{array}\right$$\right\\\\. Find the dimension of \\H\\ and determine a basis.

Exercise \\\PageIndex{83}\\

Let \\H\\ denote \\span\left\\\left$$\begin{array}{r}0\\1\\1\\-1\end{array}\right$$,\\\left$$\begin{array}{r}-1\\-1\\-2\\2\end{array}\right$$,\\\left$$\begin{array}{r}2\\3\\5\\-5\end{array}\right$$,\\\left$$\begin{array}{r}0\\1\\2\\-2\end{array}\right$$\right\\\\. Find the dimension of \\H\\ and determine a basis.

Exercise \\\PageIndex{84}\\

Let \\H\\ denote \\span\left\\\left$$\begin{array}{r}-2\\1\\1\\-3\end{array}\right$$,\\\left$$\begin{array}{r}-9\\4\\3\\-9\end{array}\right$$,\\\left$$\begin{array}{r}-33\\15\\12\\-36\end{array}\right$$,\\\left$$\begin{array}{r}-22\\10\\8\\-24\end{array}\right$$\right\\\\. Find the dimension of \\H\\ and determine a basis.

Exercise \\\PageIndex{85}\\

Let \\H\\ denote \\span\left\\\left$$\begin{array}{r}-1\\1\\-1\\-2\end{array}\right$$,\\\left$$\begin{array}{r}-4\\3\\-2\\-4\end{array}\right$$,\\\left$$\begin{array}{r}-3\\2\\-1\\-2\end{array}\right$$,\\\left$$\begin{array}{r}-1\\1\\-2\\-4\end{array}\right$$,\\\left$$\begin{array}{r}-7\\5\\-3\\-6\end{array}\right$$\right\\\\. Find the dimension of \\H\\ and determine a basis.

Exercise \\\PageIndex{86}\\

Let \\H\\ denote \\span\left\\\left$$\begin{array}{r}2\\3\\2\\1\end{array}\right$$,\\\left$$\begin{array}{r}8\\15\\6\\3\end{array}\right$$,\\\left$$\begin{array}{r}3\\6\\2\\1\end{array}\right$$,\\\left$$\begin{array}{r}4\\6\\6\\3\end{array}\right$$,\\\left$$\begin{array}{r}8\\15\\6\\3\end{array}\right$$\right\\\\. Find the dimension of \\H\\ and determine a basis.

Exercise \\\PageIndex{87}\\

Let \\H\\ denote \\span\left\\\left$$\begin{array}{r}0\\2\\0\\-1\end{array}\right$$,\\\left$$\begin{array}{r}-1\\6\\0\\-2\end{array}\right$$,\\\left$$\begin{array}{r}-2\\16\\0\\-6\end{array}\right$$,\\\left$$\begin{array}{r}-3\\22\\0\\-8\end{array}\right$$\right\\\\. Find the dimension of \\H\\ and determine a basis.

Exercise \\\PageIndex{88}\\

Let \\H\\ denote \\span\left\\\left$$\begin{array}{r}5\\1\\1\\4\end{array}\right$$,\\\left$$\begin{array}{r}14\\3\\2\\8\end{array}\right$$,\\\left$$\begin{array}{r}38\\8\\6\\24\end{array}\right$$,\\\left$$\begin{array}{r}47\\10\\7\\28\end{array}\right$$,\\\left$$\begin{array}{r}10\\2\\3\\12\end{array}\right$$\right\\\\. Find the dimension of \\H\\ and determine a basis.

Exercise \\\PageIndex{89}\\

Let \\H\\ denote \\span\left\\\left$$\begin{array}{r}6\\1\\1\\5\end{array}\right$$,\\\left$$\begin{array}{r}17\\3\\2\\10\end{array}\right$$,\\\left$$\begin{array}{r}52\\9\\7\\35\end{array}\right$$,\\\left$$\begin{array}{r}18\\3\\4\\20\end{array}\right$$\right\\\\. Find the dimension of \\H\\ and determine a basis.

Exercise \\\PageIndex{90}\\

Let \\M=\left\\\vec{u}=\left$$\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right$$\in\mathbb{R}^4:\sin(u_1)=1\right\\\\. Is \\M\\ a subspace? Explain.

Answer

No. Let \\\vec{u}=\left$$\begin{array}{c}\frac{\pi}{2} \\ 0\\0\\0\end{array}\right$$\\. Then \\2\vec{u}\cancel{\in}M\\ although \\\vec{u}\in M\\.

Exercise \\\PageIndex{91}\\

Let \\M=\left\\\vec{u}=\left$$\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right$$\in\mathbb{R}^4:\|\|u_1\|\|\leq 4\right\\\\. Is \\M\\ a subspace? Explain.

Answer

No. \\\left$$\begin{array}{c}1\\0\\0\\0\end{array}\right$$\in M\\ but \\10\left$$\begin{array}{c}1\\0\\0\\0\end{array}\right$$\cancel{\in }M\\.

Exercise \\\PageIndex{92}\\

Let \\M=\left\\\vec{u}=\left$$\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right$$\in\mathbb{R}^4:u_1\geq 0\text{ for each }i=1,2,3,4 \right\\\\. Is \\M\\ a subspace? Explain.

Answer

This is not a subspace. \\\left$$\begin{array}{c}1\\1\\1\\1\end{array}\right$$\\ is in it. However, \$-1)\left$$\begin{array}{c}1\\1\\1\\1\end{array}\right$$\\ is not.

Exercise \\\PageIndex{93}\\

Let \\\vec{w}\\, \\\vec{w}\_1\\ be given vectors in \\\mathbb{R}^4\\ and define \\M=\left\\\vec{u}=\left$$\begin{array}{c}u_1\\u_2\\u_3\\u_4\end{array}\right$$\in\mathbb{R}^4 :\vec{w}\bullet\vec{u}=0\text{ and }\vec{w}\_1\bullet\vec{u}=0\right\\.\nonumber\\ Is \\M\\ a subspace? Explain.

Answer

This is a subspace because it is closed with respect to vector addition and scalar multiplication.

Exercise \\\PageIndex{94}\\

Let \\\vec{w}\in\mathbb{R}^4\\ and let \\M=\left\\\vec{u}=\left$$\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right$$\in\mathbb{R}^4:\vec{w}\bullet\vec{u}=0\right\\\\. Is \\M\\ a subspace? Explain.

Answer

Yes, this is a subspace because it is closed with respect to vector addition and scalar multiplication.

Exercise \\\PageIndex{95}\\

Let \\M=\left\\\vec{u}=\left$$\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right$$\in\mathbb{R}^4:u_3\geq u_1\right\\\\. Is \\M\\ a subspace? Explain.

Answer

This is not a subspace. \\\left$$\begin{array}{c}0\\0\\1\\0\end{array}\right$$\\ is in it. However \$-1)\left$$\begin{array}{c}0\\0\\1\\0\end{array}\right$$=\left$$\begin{array}{r}0\\0\\-1\\0\end{array}\right$$\\ is not.

Exercise \\\PageIndex{96}\\

Let \\M=\left\\\vec{u}=\left$$\begin{array}{c}u_1 \\ u_2\\u_3\\u_4\end{array}\right$$\in\mathbb{R}^4:u_3=u_1=0\right\\\\. Is \\M\\ a subspace? Explain.

Answer

This is a subspace. It is closed with respect to vector addition and scalar multiplication.

Exercise \\\PageIndex{97}\\

Consider the set of vectors \\S\\ given by \\S=\left\\\left$$\begin{array}{c}4u+v-5w \\ 12u+6v-6w \\ 4u+4v+4w\end{array}\right$$ :u,v,w\in\mathbb{R}\right\\.\nonumber\\ Is \\S\\ a subspace of \\\mathbb{R}^3\\? If so, explain why, give a basis for the subspace and find its dimension.

Exercise \\\PageIndex{98}\\

Consider the set of vectors \\S\\ given by \\S=\left\\\left$$\begin{array}{c}2u+6v+7w \\ -3u-9v-12w \\ 2u+6v+6w \\ u+3v+3w \end{array}\right$$ :u,v,w\in\mathbb{R}\right\\.\nonumber\\ Is \\S\\ a subspace of \\\mathbb{R}^4\\? If so, explain why, give a basis for the subspace and find its dimension.

Exercise \\\PageIndex{99}\\

Consider the set of vectors \\S\\ given by \\S=\left\\\left$$\begin{array}{c}2u+v \\ 6v-3u+3w \\ 3v-6u+3w \end{array}\right$$ :u,v,w\in\mathbb{R}\right\\.\nonumber\\ Is this set of vectors a subspace of \\\mathbb{R}^3\\? If so, explain why, give a basis for the subspace and find its dimension.

Exercise \\\PageIndex{100}\\

Consider the vectors of the form \\\left\\\left$$\begin{array}{c}2u+v+7w \\ u-2v+w \\ -6v-6w \end{array}\right$$ :u,v,w\in\mathbb{R}\right\\.\nonumber\\ Is this set of vectors a subspace of \\\mathbb{R}^3\\? If so, explain why, give a basis for the subspace and find its dimension.

Exercise \\\PageIndex{101}\\

Consider the vectors of the form \\\left\\\left$$\begin{array}{c}3u+v+11w \\ 18u+6v+66w \\ 28u+8v+100w \end{array}\right$$ :u,v,w\in\mathbb{R}\right\\.\nonumber\\ Is this set of vectors a subspace of \\\mathbb{R}^3\\? If so, explain why, give a basis for the subspace and find its dimension.

Exercise \\\PageIndex{102}\\

Consider the vectors of the form \\\left\\\left$$\begin{array}{c}3u+v \\ 2w-4u \\ 2w-2v-8u \end{array}\right$$ :u,v,w\in\mathbb{R}\right\\.\nonumber\\ Is this set of vectors a subspace of \\\mathbb{R}^3\\? If so, explain why, give a basis for the subspace and find its dimension.

Exercise \\\PageIndex{103}\\

Consider the set of vectors \\S\\ given by \\\left\\\left$$\begin{array}{c}u+v+w \\ 2u+2v+4w \\ u+v+w \\ 0 \end{array}\right$$ :u,v,w\in\mathbb{R}\right\\.\nonumber\\ Is \\S\\ is a subspace of \\\mathbb{R}^4\\? If so, explain why, give a basis for the subspace and find its dimension.

Exercise \\\PageIndex{104}\\

Consider the set of vectors \\S\\ given by \\\left\\\left$$\begin{array}{c}v \\ -3u-3w \\ 8u-4v+4w \end{array}\right$$ :u,v,w\in\mathbb{R}\right\\.\nonumber\\ Is \\S\\ is a subspace of \\\mathbb{R}^4\\? If so, explain why, give a basis for the subspace and find its dimension.

Exercise \\\PageIndex{105}\\

If you have \\5\\ vectors in \\\mathbb{R}^5\\ and the vectors are linearly independent, can it always be concluded they span \\\mathbb{R}^5\\? Explain.

Answer

Yes. If not, there would exist a vector not in the span. But then you could add in this vector and obtain a linearly independent set of vectors with more vectors than a basis.

Exercise \\\PageIndex{106}\\

If you have \\6\\ vectors in \\\mathbb{R}^5\\, is it possible they are linearly independent? Explain.

Answer

They can't be.

Exercise \\\PageIndex{107}\\

Suppose \\A\\ is an \\m\times n\\ matrix and \\\\\vec{w}\_1,\cdots ,\vec{w}\_k\\\\ is a linearly independent set of vectors in \\A(\mathbb{R}^n ) ⊆ \mathbb{R}^m\\. Now suppose \\A\vec{z}\_i = \vec{w}\_i\\. Show \\\\\vec{z}\_1 ,\cdots ,\vec{z}\_k\\\\ is also independent.

Answer

Say \\\sum\limits\_{i=1}^k c_i\vec{z}\_i=\vec{0}\\. Then apply \\A\\ to it as follows. \\\sum\limits\_{i=1}^k c_aA\vec{z}\_i=\sum\limits\_{i=1}^kc_i\vec{w}\_i=\vec{0}\nonumber\\ and so, by linear independence of the \\\vec{w}\_i\\, it follows that each \\c_i=0\\.

Exercise \\\PageIndex{108}\\

Suppose \\V,\\ W\\ are subspaces of \\\mathbb{R}^n\\. Let \\V ∩W\\ be all vectors which are in both \\V\\ and \\W\\. Show that \\V ∩W\\ is a subspace also.

Answer

If \\\vec{x},\vec{y} ∈ V ∩W\\, then for scalars \\α,β\\, the linear combination \\α\vec{x} + β\vec{y}\\ must be in both \\V\\ and \\W\\ since they are both subspaces.

Exercise \\\PageIndex{109}\\

Suppose \\V\\ and \\W\\ both have dimension equal to \\7\\ and they are subspaces of \\\mathbb{R}^{10}\\. What are the possibilities for the dimension of \\V ∩W\\? Hint: Remember that a linear independent set can be extended to form a basis.

Exercise \\\PageIndex{110}\\

Suppose \\V\\ has dimension \\p\\ and \\W\\ has dimension \\q\\ and they are each contained in a subspace, \\U\\ which has dimension equal to \\n\\ where \\n \> \text{max}(p,q)\\. What are the possibilities for the dimension of \\V ∩W\\? Hint: Remember that a linearly independent set can be extended to form a basis.

Answer

Let \\\\x_1,\cdots ,x_k\\\\ be a basis for \\V∩W\\. Then there is a basis for \\V\\ and \\W\\ which are respectively \\\\x_1, \cdots ,x_k, y\_{k+1},\cdots ,y_p\\,\\\\x_1,\cdots ,x_k, z\_{k+1},\cdots z_q\\\nonumber\\ It follows that you must have \\k+p-k+q-k\leq n\\ and so you must have \\p+q-n\leq k\nonumber\\

Exercise \\\PageIndex{111}\\

Suppose \\A\\ is an \\m\times n\\ matrix and \\B\\ is an \\n\times p\\ matrix. Show that \\\text{dim}(\text{ker}(AB))\leq\text{dim}(\text{ker}(A))+\text{dim}(\text{ker}(B)).\nonumber\\ Consider the subspace, \\B(\mathbb{R}^p )∩\text{ker}(A)\\ and suppose a basis for this subspace is \\\\\vec{w}\_1,\cdots ,\vec{w}\_k\\\\. Now suppose \\\\\vec{u}\_1,\cdots ,\vec{u}\_r\\\\ is a basis for \\\text{ker}(B)\\. Let \\\\\vec{z}\_1,\cdots ,\vec{z}\_k\\\\ be such that \\B\vec{z}\_1 =\vec{w}\_i\\ and argue that \\\text{ker}(AB)⊆ span\\\vec{u}\_1,\cdots ,\vec{u}\_r,\vec{z}\_1,\cdots ,\vec{z}\_k\\.\nonumber\\

Answer

Here is how you do this. Suppose \\AB\vec{x} =\vec{0}\\. Then \\B\vec{x} ∈ \text{ker}(A) ∩ B(\mathbb{R}^p)\\ and so \\B\vec{x} =\sum\limits\_{i=1}^k B\vec{z}\_i\\ showing that \\\vec{x}-\sum\limits\_{i=1}^k\vec{z}\_i\in\text{ker}(B)\nonumber\\ Consider \\B(\mathbb{R}^p )∩\text{ker}(A)\\ and let a basis be \\\\\vec{w}\_1,\cdots ,\vec{w}\_k\\\\. Then each \\\vec{w}\_i\\ is of the form \\B\vec{z}\_i =\vec{w}\_i\\. Therefore, \\\\\vec{z}\_1,\cdots ,\vec{z}\_k\\\\ is linearly independent and \\AB\vec{z}\_i = 0\\. Now let \\\\\vec{u}\_1,\cdots ,\vec{u}\_r\\\\ be a basis for \\\text{ker}(B)\\. If \\AB\vec{x} =\vec{0}\\, then \\B\vec{x} ∈ \text{ker}(A)∩B(\mathbb{R}^p)\\ and so \\B\vec{x} =\sum\limits\_{i=1}^k c_iB\vec{z}\_1\\ which implies \\\vec{x}-\sum\limits\_{i=1}^k c_i\vec{z}\_i\in\text{ker}(B)\nonumber\\ and so it is of the form \\\vec{x}-\sum\limits\_{i=1}^kc_i\vec{z}\_i=\sum\limits\_{j=1}^r d_j\vec{u}\_j\nonumber\\ It follows that if \\AB\vec{x} =\vec{0}\\ so that \\\vec{x} ∈ \text{ker}(AB)\\, then \\\vec{x}\in span (\vec{z}\_1,\cdots ,\vec{z}\_k,\vec{u}\_1, \cdots ,\vec{u}\_r ).\nonumber\\ Therefore, \\\begin{aligned}\text{dim}(\text{ker}(AB))&\leq k+r=\text{dim}(B(\mathbb{R}^p)∩\text{ker}(A))+\text{dim}(\text{ker}(B)) \\ &\leq\text{dim}(\text{ker}(A))+\text{dim}(\text{ker}(B))\end{aligned}\\

Exercise \\\PageIndex{112}\\

Show that if \\A\\ is an \\m\times n\\ matrix, then \\\text{ker}(A)\\ is a subspace of \\\mathbb{R}^n\\.

Answer

If \\\vec{x}\\, \\\vec{y}\in\text{ker}(A)\\ then \\A(a\vec{x}+b\vec{y})=aA\vec{x}+bA\vec{y}=a\vec{0}+b\vec{0}=\vec{0}\nonumber\\ and so \\\text{ker}(A)\\ is closed under linear combinations. Hence it is a subspace.

Exercise \\\PageIndex{113}\\

Find the rank of the following matrix. Also find a basis for the row and column spaces. \\\left$$\begin{array}{rrrrrr}1&3&0&-2&0&3 \\ 3&9&1&-7&0&8 \\ 1&3&1&-3&1&-1 \\ 1&3&-1&-1&-2&10\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{114}\\

Find the rank of the following matrix. Also find a basis for the row and column spaces. \\\left$$\begin{array}{rrrrrr}1&3&0&-2&7&3 \\ 3&9&1&-7&23&8 \\ 1&3&1&-3&9&2 \\ 1&3&-1&-1&5&4\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{115}\\

Find the rank of the following matrix. Also find a basis for the row and column spaces. \\\left$$\begin{array}{rrrrrr}1&0&3&0&7&0 \\ 3&1&10&0&23&0 \\ 1&1&4&1&7&0 \\ 1&-1&2&-2&9&1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{116}\\

Find the rank of the following matrix. Also find a basis for the row and column spaces. \\\left$$\begin{array}{rrr}1&0&3 \\ 3&1&10 \\ 1&1&4 \\ 1&-1&2\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{117}\\

Find the rank of the following matrix. Also find a basis for the row and column spaces. \\\left$$\begin{array}{rrrrr}0&0&-1&0&1 \\ 1&2&3&-2&-18 \\ 1&2&2&-1&-11 \\ -1&-2&-2&1&11\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{118}\\

Find the rank of the following matrix. Also find a basis for the row and column spaces. \\\left$$\begin{array}{rrrr}1&0&3&0 \\ 3&1&10&0 \\ -1&1&-2&1 \\ 1&-1&2&-2\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{119}\\

Find \\\text{ker}(A)\\ for the following matrices.

1. \\A=\left$$\begin{array}{rr}2&3 \\ 4&6\end{array}\right$$\\

2. \\A=\left$$\begin{array}{rrr}1&0&-1 \\ -1&1&3 \\ 3&2&1\end{array}\right$$\\

3. \\A=\left$$\begin{array}{rrr}2&4&0 \\ 3&6&-2 \\ 1&2&-2\end{array}\right$$\\

4. \\A=\left$$\begin{array}{rrrr}2&-1&3&5 \\ 2&0&1&2 \\ 6&4&-5&-6 \\ 0&2&-4&-6\end{array}\right$$\\

Exercise \\\PageIndex{120}\\

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \\\left$$\begin{array}{c}\frac{1}{6}\sqrt{2}\sqrt{3} \\ \frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3}\end{array}\right$$,\\ \left$$\begin{array}{c}\frac{1}{2}\sqrt{2} \\ 0 \\ \frac{1}{2}\sqrt{2}\end{array}\right$$,\\ \left$$\begin{array}{c}-\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3}\end{array}\right$$\nonumber\\ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Exercise \\\PageIndex{121}\\

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \\\left$$\begin{array}{r}1\\2\\-1\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\1\end{array}\right$$,\\\left$$\begin{array}{r}-1\\1\\1\end{array}\right$$\nonumber\\ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Exercise \\\PageIndex{122}\\

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \\\left$$\begin{array}{r}1\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}2\\1\\-1\end{array}\right$$,\\\left$$\begin{array}{r}0\\1\\1\end{array}\right$$\nonumber\\ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Exercise \\\PageIndex{123}\\

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \\\left$$\begin{array}{r}1\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}2\\1\\-1\end{array}\right$$,\\\left$$\begin{array}{r}1\\2\\1\end{array}\right$$\nonumber\\ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Exercise \\\PageIndex{124}\\

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. \\\left$$\begin{array}{r}1\\0\\0\\0\end{array}\right$$,\\\left$$\begin{array}{r}0\\1\\-1\\0\end{array}\right$$,\\\left$$\begin{array}{r}0\\0\\0\\1\end{array}\right$$\nonumber\\ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Exercise \\\PageIndex{125}\\

Here are some matrices. Label according to whether they are symmetric, skew symmetric, or orthogonal.

1. \\\left$$\begin{array}{ccc}1&0&0 \\ 0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \\ 0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{array}\right$$\\

2. \\\left$$\begin{array}{ccc}1&2&-3 \\ 2&1&4 \\ -3&4&7\end{array}\right$$\\

3. \\\left$$\begin{array}{ccc}0&-2&-3 \\ 2&0&-4 \\ 3&4&0\end{array}\right$$\\

Answer

1. Orthogonal

2. Symmetric

3. Skew Symmetric

Exercise \\\PageIndex{126}\\

For \\U\\ an orthogonal matrix, explain why \\\|\|U\vec{x}\|\| =\|\|\vec{x}\|\|\\ for any vector \\\vec{x}\\. Next explain why if \\U\\ is an \\n\times n\\ matrix with the property that \\\|\|U\vec{x}\|\| =\|\|\vec{x}\|\|\\ for all vectors, \\\vec{x}\\, then \\U\\ must be orthogonal. Thus the orthogonal matrices are exactly those which preserve length.

Answer

\\\|\|U\vec{x}\|\|^2=U\vec{x}\bullet U\vec{x}=U^TU\vec{x}\bullet\vec{x}=I\vec{x}\bullet\vec{x}=\|\|\vec{x}\|\|^2\\. Next suppose distance is preserved by \\U\\. Then \\\begin{aligned} (U(\vec{x}+\vec{y}))\bullet (U(\vec{x}+\vec{y}))&=\|\|Ux\|\|^2+\|\|Uy\|\|^2+2(Ux\bullet Uy) \\ &=\|\|\vec{x}\|\|^2+\|\|\vec{y}\|\|^2+2(U^TU\vec{x}\bullet\vec{y})\end{aligned}\\ But since \\U\\ preserves distances, it is also the case that \$U(\vec{x}+\vec{y})\bullet U(\vec{x}+\vec{y}))=\|\|\vec{x}\|\|^2+\|\|\vec{y}\|\|^2+2(\vec{x}\bullet\vec{y})\nonumber\\ Hence \\\vec{x}\bullet\vec{y}=U^TU\vec{x}\bullet\vec{y}\nonumber\\ and so \$(U^TU-I)\vec{x})\bullet\vec{y}=0\nonumber\\ Since \\y\\ is arbitrary, it follows that \\U^TU-I=0\\. Thus \\U\\ is orthogonal.

Exercise \\\PageIndex{127}\\

Suppose \\U\\ is an orthogonal \\n\times n\\ matrix. Explain why \\rank(U) = n\\.

Answer

You could observe that \\\text{det}(UU^T)=(\text{det}(U))^2-1\\ so \\\text{det}(U)\neq 0\\.

Exercise \\\PageIndex{128}\\

Fill in the missing entries to make the matrix orthogonal. \\\left$$\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\underline{\\}&\underline{\\} \\ \underline{\\}&\frac{\sqrt{6}}{3}&\underline{\\}\end{array}\right$$.\nonumber\\

Answer

\\\begin{aligned} &\left$$\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&a \\ 0&\frac{\sqrt{6}}{3}&b\end{array}\right$$\left$$\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&a \\ 0&\frac{\sqrt{6}}{3}&b\end{array}\right$$^T \\ =&\left$$\begin{array}{ccc} 1&\frac{1}{3}\sqrt{3}a-\frac{1}{3} &\frac{1}{3}\sqrt{3}b-\frac{1}{3} \\ \frac{1}{3}\sqrt{3}a-\frac{1}{3}&a^2+\frac{2}{3}&ab-\frac{1}{3} \\ \frac{1}{3}\sqrt{3}b-\frac{1}{3}&ab-\frac{1}{3}&b^2+\frac{2}{3}\end{array}\right$$\end{aligned}\\ This requires, \\a=1/\sqrt{3},b=1/\sqrt{3}\\. \\\left$$\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&1/\sqrt{3} \\ 0&\frac{\sqrt{6}}{3}&1/\sqrt{3}\end{array}\right$$\left$$\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&1/\sqrt{3} \\ 0&\frac{\sqrt{6}}{3}&1/\sqrt{3}\end{array}\right$$^T =\left$$\begin{array}{ccc}1&0&0 \\ 0&1&0 \\ 0&0&1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{129}\\

Fill in the missing entries to make the matrix orthogonal. \\\left$$\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\underline{\\}&\underline{\\} \\ \underline{\\}&0&\underline{\\}\end{array}\right$$\nonumber\\

Answer

\\\left$$\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2}\\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&a \\ -\frac{1}{3}&0&b\end{array}\right$$\left$$\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&a \\ -\frac{1}{3}&0&b\end{array}\right$$^T=\left$$\begin{array}{ccc}1&\frac{1}{6}\sqrt{2}a-\frac{1}{18}&\frac{1}{6}\sqrt{2}b-\frac{2}{9} \\ \frac{1}{6}\sqrt{2}a-\frac{1}{18}&a^2+\frac{17}{18} &ab-\frac{2}{9} \\ \frac{1}{6}\sqrt{2}b-\frac{2}{9}&ab-\frac{2}{9}&b^2+\frac{1}{9}\end{array}\right$$\nonumber\\ This requires \\a=\frac{1}{3\sqrt{2}},\\b=\frac{4}{3\sqrt{2}}\\. \\\left$$\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&\frac{1}{3\sqrt{2}} \\ -\frac{1}{3}&0&\frac{4}{3\sqrt{2}}\end{array}\right$$\left$$\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2} \\ \frac{2}{3}&\frac{-\sqrt{2}}{2}&\frac{1}{3\sqrt{2}} \\ -\frac{1}{3}&0&\frac{4}{3\sqrt{2}}\end{array}\right$$^T=\left$$\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{130}\\

Fill in the missing entries to make the matrix orthogonal. \\\left$$\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\underline{\\} \\ \frac{2}{3}&0&\underline{\\} \\ \underline{\\}&\underline{\\}&\frac{4}{15}\sqrt{5}\end{array}\right$$\nonumber\\

Answer

Try \\\begin{aligned}&\left$$\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&c \\ \frac{2}{3}&0&d \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right$$\left$$\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&c \\ \frac{2}{3}&0&d \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right$$^T \\ =&\left$$\begin{array}{ccc}c^2+\frac{41}{45} &cd+\frac{2}{9}&\frac{4}{15}\sqrt{5}c-\frac{8}{45} \\ cd+\frac{2}{9}&d^2+\frac{4}{9} &\frac{4}{15}\sqrt{5}d+\frac{4}{9} \\ \frac{4}{15}\sqrt{5}c-\frac{8}{45}&\frac{4}{15}\sqrt{5}d+\frac{4}{9}&1\end{array}\right$$\end{aligned}\\ This requires that \\c=\frac{2}{3\sqrt{5}},d=\frac{-5}{3\sqrt{5}}\\. \\\left$$\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\frac{2}{3\sqrt{5}} \\ \frac{2}{3}&0&\frac{-5}{3\sqrt{5}} \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right$$\left$$\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\frac{2}{3\sqrt{5}} \\ \frac{2}{3}&0&\frac{-5}{3\sqrt{5}} \\ \frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right$$^T=\left$$\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{131}\\

Find an orthonormal basis for the span of each of the following sets of vectors.

1. \\\left$$\begin{array}{r}3\\-4\\0\end{array}\right$$,\\\left$$\begin{array}{r}7\\-1\\0\end{array}\right$$,\\\left$$\begin{array}{r}1\\7\\1\end{array}\right$$\\

2. \\\left$$\begin{array}{r}3\\0\\-4\end{array}\right$$,\\\left$$\begin{array}{r}11\\0\\2\end{array}\right$$,\\\left$$\begin{array}{r}1\\1\\7\end{array}\right$$\\

3. \\\left$$\begin{array}{r}3\\0\\-4\end{array}\right$$,\\\left$$\begin{array}{r}5\\0\\10\end{array}\right$$,\\\left$$\begin{array}{r}-7\\1\\1\end{array}\right$$\\

Answer

1. \\\left$$\begin{array}{c}\frac{3}{5} \\ -\frac{4}{5} \\ 0\end{array}\right$$,\\\left$$\begin{array}{c}\frac{4}{5}\\ \frac{3}{5} \\ 0\end{array}\right$$,\\\left$$\begin{array}{c}0\\0\\1\end{array}\right$$\\

2. \\\left$$\begin{array}{c}\frac{3}{5}\\ 0\\ -\frac{4}{5}\end{array}\right$$,\\\left$$\begin{array}{c}\frac{4}{5} \\ 0\\ \frac{3}{5}\end{array}\right$$,\\\left$$\begin{array}{c}0\\1\\0\end{array}\right$$\\

3. \\\left$$\begin{array}{c}\frac{3}{5}\\0\\-\frac{4}{5}\end{array}\right$$,\\\left$$\begin{array}{c}\frac{4}{5}\\0\\ \frac{3}{5}\end{array}\right$$,\\\left$$\begin{array}{c}0\\1\\0\end{array}\right$$\\

Exercise \\\PageIndex{132}\\

Using the Gram Schmidt process find an orthonormal basis for the following span: \\span\left\\\left$$\begin{array}{r}1\\2\\1\end{array}\right$$,\\\left$$\begin{array}{r}2\\-1\\3\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\0\end{array}\right$$\right\\\nonumber\\

Answer

A solution is \\\left$$\begin{array}{c}\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{6} \\ \frac{1}{6}\sqrt{6}\end{array}\right$$,\\\left$$\begin{array}{c}\frac{3}{10}\sqrt{2} \\ -\frac{2}{5}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right$$,\\\left$$\begin{array}{c}\frac{7}{15}\sqrt{3} \\ -\frac{1}{15}\sqrt{3} \\ -\frac{1}{3}\sqrt{3}\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{133}\\

Using the Gram Schmidt process find an orthonormal basis for the following span: \\span\left\\\left$$\begin{array}{r}1\\2\\1\\0\end{array}\right$$,\\\left$$\begin{array}{r}2\\-1\\3\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\0\\0\\1\end{array}\right$$\right\\\nonumber\\

Answer

Then a solution is \\\left$$\begin{array}{c}\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{6} \\ \frac{1}{6}\sqrt{6} \\ 0\end{array}\right$$,\\\left$$\begin{array}{c}\frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{2}{9}\sqrt{2}\sqrt{3} \\ \frac{5}{18}\sqrt{2}\sqrt{3} \\ \frac{1}{9}\sqrt{2}\sqrt{3}\end{array}\right$$,\\\left$$\begin{array}{c}\frac{5}{111}\sqrt{3}\sqrt{37} \\ \frac{1}{133}\sqrt{3}\sqrt{37} \\ -\frac{17}{333}\sqrt{3}\sqrt{37} \\ \frac{22}{333}\sqrt{3}\sqrt{37}\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{134}\\

The set \\V=\left\\\left$$\begin{array}{c}x\\y\\z\end{array}\right$$ :2x+3y-z=0\right\\\\ is a subspace of \\\mathbb{R}^3\\. Find an orthonormal basis for this subspace.

Answer

The subspace is of the form \\\left$$\begin{array}{c}x\\y\\2x+3y\end{array}\right$$\nonumber\\ and a basis is \\\left$$\begin{array}{c}1\\0\\2\end{array}\right$$,\\\left$$\begin{array}{c}0\\1\\3\end{array}\right$$\\. Therefore, an orthonormal basis is \\\left$$\begin{array}{c}\frac{1}{5}\sqrt{5} \\ 0\\ \frac{2}{5}\sqrt{5}\end{array}\right$$,\\\left$$\begin{array}{c}-\frac{3}{35}\sqrt{5}\sqrt{14} \\ \frac{1}{14}\sqrt{5}\sqrt{14} \\ \frac{3}{70}\sqrt{5}\sqrt{14}\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{135}\\

Consider the following scalar equation of a plane. \\2x-3y+z=0\nonumber\\ Find the orthogonal complement of the vector \\\vec{v}=\left$$\begin{array}{c}3\\4\\1\end{array}\right$$\\. Also find the point on the plane which is closest to \$3,4,1)\\.

Exercise \\\PageIndex{136}\\

Consider the following scalar equation of a plane. \\x+3y+z=0\nonumber\\ Find the orthogonal complement of the vector \\\vec{v}=\left$$\begin{array}{c}1\\2\\1\end{array}\right$$\\. Also find the point on the plane which is closest to \$3,4,1)\\.

Exercise \\\PageIndex{137}\\

Let \\\vec{v}\\ be a vector and let \\\vec{n}\\ be a normal vector for a plane through the origin. Find the equation of the line through the point determined by \\\vec{v}\\ which has direction vector \\\vec{n}\\. Show that it intersects the plane at the point determined by \\\vec{v}−proj\_{\vec{n}}\vec{v}\\. Hint: The line:\\\vec{v}+t\vec{n}\\. It is in the plane if \\\vec{n}•(\vec{v}+t\vec{n}) = 0\\. Determine \\t\\. Then substitute in to the equation of the line.

Exercise \\\PageIndex{138}\\

As shown in the above problem, one can find the closest point to~v in a plane through the origin by finding the intersection of the line through \\\vec{v}\\ having direction vector equal to the normal vector to the plane with the plane. If the plane does not pass through the origin, this will still work to find the point on the plane closest to the point determined by \\\vec{v}\\. Here is a relation which defines a plane \\2x+y+z=11\nonumber\\ and here is a point: \$1, 1, 2)\\. Find the point on the plane which is closest to this point. Then determine the distance from the point to the plane by taking the distance between these two points. Hint: Line: \$x, y,z) = (1, 1, 2) +t(2, 1, 1)\\. Now require that it intersect the plane.

Exercise \\\PageIndex{139}\\

In general, you have a point \$x_0, y_0,z_0)\\ and a scalar equation for a plane \\ax+by+cz = d\\ where \\a^2 +b^2 +c^2 \> 0\\. Determine a formula for the closest point on the plane to the given point. Then use this point to get a formula for the distance from the given point to the plane. Hint: Find the line perpendicular to the plane which goes through the given point: \$x, y,z) = (x_0, y_0,z_0) + t(a,b, c)\\. Now require that this point satisfy the equation for the plane to determine \\t\\.

Exercise \\\PageIndex{140}\\

Find the least squares solution to the following system. \\\begin{aligned}x+2y&=1 \\ 2x+3y&=2 \\ 3x+5y&=4\end{aligned}\\

Answer

\\\begin{aligned}\left$$\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right$$^T\left$$\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right$$&=\left$$\begin{array}{cc}14&23\\23&38\end{array}\right$$\left$$\begin{array}{cc}14&23\\23&38\end{array}\right$$\left$$\begin{array}{c}x\\y\end{array}\right$$ \\ &=\left$$\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right$$^T\left$$\begin{array}{c}1\\2\\4\end{array}\right$$=\left$$\begin{array}{c}17\\28\end{array}\right$$\end{aligned}\\ \\\begin{aligned}\left$$\begin{array}{cc}14&23\\23&38\end{array}\right$$\left$$\begin{array}{c}x\\y\end{array}\right$$&=\left$$\begin{array}{c}17\\28\end{array}\right$$ \\ \left$$\begin{array}{cc}14&23\\23&38\end{array}\right$$\left$$\begin{array}{c}x\\y\end{array}\right$$&=\left$$\begin{array}{c}17\\28\end{array}\right$$,\end{aligned}\\ Solution is: \\\left$$\begin{array}{c}\frac{2}{3}\\ \frac{1}{3}\end{array}\right$$\\

Exercise \\\PageIndex{141}\\

You are doing experiments and have obtained the ordered pairs, \$0, 1),(1, 2),(2, 3.5),(3, 4)\nonumber\\ Find \\m\\ and \\b\\ such that \\\vec{y} = m\vec{x}+b\\ approximates these four points as well as possible.

Exercise \\\PageIndex{142}\\

Suppose you have several ordered triples, \$x_i , y_i ,z_i)\\. Describe how to find a polynomial such as \\z = a+bx+cy+dxy+ex^2 + fy^2\nonumber\\ giving the best fit to the given ordered triples.

Exercise \\\PageIndex{143}\\

The wind blows from the South at \\20\\ kilometers per hour and an airplane which flies at \\600\\ kilometers per hour in still air is heading East. Find the velocity of the airplane and its location after two hours.

Exercise \\\PageIndex{144}\\

The wind blows from the West at \\30\\ kilometers per hour and an airplane which flies at \\400\\ kilometers per hour in still air is heading North East. Find the velocity of the airplane and its position after two hours.

Exercise \\\PageIndex{145}\\

The wind blows from the North at \\10\\ kilometers per hour. An airplane which flies at \\300\\ kilometers per hour in still air is supposed to go to the point whose coordinates are at \\\left( 100, 100 \right).\\ In what direction should the airplane fly?

Exercise \\\PageIndex{146}\\

Three forces act on an object. Two are \\\left $$ \begin{array}{r} 3 \\ -1 \\ -1 \end{array} \right $$\\ and \\\left $$ \begin{array}{r} 1 \\ -3 \\ 4 \end{array} \right $$\\ Newtons. Find the third force if the object is not to move.

Exercise \\\PageIndex{147}\\

Three forces act on an object. Two are \\\left $$ \begin{array}{r} 6 \\ -3 \\ 3 \end{array} \right $$\\ and \\\left $$ \begin{array}{r} 2 \\ 1 \\ 3 \end{array} \right $$\\ Newtons. Find the third force if the total force on the object is to be \\\left $$ \begin{array}{r} 7 \\ 1 \\ 3 \end{array} \right $$ .\\

Exercise \\\PageIndex{148}\\

A river flows West at the rate of \\b\\ miles per hour. A boat can move at the rate of \\8\\ miles per hour. Find the smallest value of \\b\\ such that it is not possible for the boat to proceed directly across the river.

Exercise \\\PageIndex{149}\\

The wind blows from West to East at a speed of \\50\\ miles per hour and an airplane which travels at \\400\\ miles per hour in still air is heading North West. What is the velocity of the airplane relative to the ground? What is the component of this velocity in the direction North?

Answer

The velocity is the sum of two vectors. \\50\vec{i}+\frac{ 300}{\sqrt{2}} \left( \vec{i}+\vec{j}\right) =\left( 50+\frac{300}{\sqrt{2}} \right) \vec{i}+ \frac{300}{\sqrt{2}}\vec{j}.\\ The component in the direction of North is then \\\frac{300}{\sqrt{2}}= 150\sqrt{2}\\ and the velocity relative to the ground is \\\left( 50+\frac{300}{\sqrt{2}}\right) \vec{i}+\frac{300}{\sqrt{2}}\vec{j}\nonumber \\

Exercise \\\PageIndex{150}\\

The wind blows from West to East at a speed of \\60\\ miles per hour and an airplane can travel travels at \\100\\ miles per hour in still air. How many degrees West of North should the airplane head in order to travel exactly North?

Exercise \\\PageIndex{151}\\

The wind blows from West to East at a speed of \\50\\ miles per hour and an airplane which travels at \\400\\ miles per hour in still air heading somewhat West of North so that, with the wind, it is flying due North. It uses \\30.0\\ gallons of gas every hour. If it has to travel \\600.0\\ miles due North, how much gas will it use in flying to its destination?

Exercise \\\PageIndex{152}\\

An airplane is flying due north at \\150.0\\ miles per hour but it is not actually going due North because there is a wind which is pushing the airplane due east at \\40.0\\ miles per hour. After one hour, the plane starts flying \\30^{\circ }\\ East of North. Assuming the plane starts at \\\left( 0,0\right) ,\\ where is it after \\2\\ hours? Let North be the direction of the positive \\y\\ axis and let East be the direction of the positive \\x\\ axis.

Answer

Velocity of plane for the first hour: \\\left $$ \begin{array}{cc} 0 & 150 \end{array} \right $$ + \left $$ \begin{array}{cc} 40 & 0 \end{array} \right $$ =\left $$ \begin{array}{cc} 40 & 150 \end{array} \right $$ .\\ After one hour it is at \\\left( 40,150\right) .\\ Next the velocity of the plane is \\150\left $$ \begin{array}{cc} \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array} \right $$ +\left $$ \begin{array}{cc} 40 & 0 \end{array} \right $$\\ in miles per hour. After two hours it is then at \\\left( 40,150\right) + 150\left $$ \begin{array}{cc} \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array} \right $$ +\left $$ \begin{array}{cc} 40 & 0 \end{array} \right $$ = \left $$ \begin{array}{cc} 155 & 75\sqrt{3}+150 \end{array} \right $$ = \left $$ \begin{array}{cc} 155.0 & 279.\\ 9 \end{array} \right $$\\

Exercise \\\PageIndex{153}\\

City A is located at the origin \\\left( 0,0 \right)\\ while city B is located at \\\left(300,500 \right)\\ where distances are in miles. An airplane flies at \\250\\ miles per hour in still air. This airplane wants to fly from city A to city B but the wind is blowing in the direction of the positive \\y\\ axis at a speed of \\50\\ miles per hour. Find a unit vector such that if the plane heads in this direction, it will end up at city B having flown the shortest possible distance. How long will it take to get there?

Answer

Wind: \\\left $$ \begin{array}{cc} 0 & 50 \end{array} \right $$ .\\ Direction it needs to travel: \\\left( 3,5 \right) \frac{1}{\sqrt{34}}.\\ Then you need \\250 \left $$ \begin{array}{cc} a & b \end{array} \right $$ + \left $$ \begin{array}{cc} 0 & 50 \end{array} \right $$\\ to have this direction where \\\left $$ \begin{array}{cc} a & b \end{array} \right $$\\ is an appropriate unit vector. Thus you need \\\begin{aligned} a^{2}+b^{2} &=1 \\ \frac{250b+50}{250a} &=\frac{5}{3}\end{aligned}\\ Thus \\a=\frac{3}{5},b=\frac{4}{5}.\\ The velocity of the plane relative to the ground is \\\left $$ \begin{array}{cc} 150 & 250 \end{array} \right $$ .\\ The speed of the plane relative to the ground is given by \\\sqrt{\left( 150\right) ^{2}+\left( 250\right) ^{2}}= 291.55 \text{ miles per hour }\nonumber\\ It has to go a distance of \\\sqrt{\left( 300\right) ^{2}+\left( 500\right) ^{2}}= 583.\\ 10\\ miles. Therefore, it takes \\\frac{ 583.\\ 1}{ 291.\\ 55}=2 \text{ hours}\nonumber \\

Exercise \\\PageIndex{154}\\

A certain river is one half mile wide with a current flowing at \\2\\ miles per hour from East to West. A man swims directly toward the opposite shore from the South bank of the river at a speed of \\3\\ miles per hour. How far down the river does he find himself when he has swam across? How far does he end up traveling?

Answer

Water:\\\left $$ \begin{array}{rr} -2 & 0 \end{array} \right $$\\ Swimmer:\\\left $$ \begin{array}{rr} 0 & 3 \end{array} \right $$\\ Speed relative to earth: \\\left $$ \begin{array}{rr} -2 & 3 \end{array} \right $$ .\\ It takes him \\1/6\\ of an hour to get across. Therefore, he ends up traveling \\\frac{1}{6}\sqrt{4+9}= \frac{1}{6}\sqrt{13}\\ miles. He ends up \\1/3\\ mile down stream.

Exercise \\\PageIndex{155}\\

A certain river is one half mile wide with a current flowing at 2 miles per hour from East to West. A man can swim at \\3\\ miles per hour in still water. In what direction should he swim in order to travel directly across the river? What would the answer to this problem be if the river flowed at 3 miles per hour and the man could swim only at the rate of 2 miles per hour?

Answer

Man: \\3\left $$ \begin{array}{rr} a & b \end{array} \right $$\\ Water: \\\left $$ \begin{array}{rr} -2 & 0 \end{array} \right $$\\ Then you need \\3a=2\\ and so \\a=2/3\\ and hence \\b=\sqrt{5}/3\\. The vector is then \\\left $$ \begin{array}{cc} \frac{2}{3} & \frac{\sqrt{5}}{3} \end{array} \right $$ .\\

In the second case, he could not do it. You would need to have a unit vector \\\left $$ \begin{array}{rr} a & b \end{array} \right $$\\ such that \\2a=3\\ which is not possible.

Exercise \\\PageIndex{156}\\

Three forces are applied to a point which does not move. Two of the forces are \\2 \vec{i}+2 \vec{j} -6 \vec{k}\\ Newtons and \\8 \vec{i}+ 8 \vec{j}+ 3 \vec{k}\\ Newtons. Find the third force.

Exercise \\\PageIndex{157}\\

The total force acting on an object is to be \\4 \vec{i}+ 2 \vec{j} -3 \vec{k}\\ Newtons. A force of \\-3 \vec{i} -1 \vec{j}+ 8 \vec{k}\\ Newtons is being applied. What other force should be applied to achieve the desired total force?

Exercise \\\PageIndex{158}\\

A bird flies from its nest \\8\\ km in the direction \\\frac{5}{6}\pi\\ north of east where it stops to rest on a tree. It then flies \\1\\ km in the direction due southeast and lands atop a telephone pole. Place an \\xy\\ coordinate system so that the origin is the bird’s nest, and the positive \\x\\ axis points east and the positive \\y\\ axis points north. Find the displacement vector from the nest to the telephone pole.

Exercise \\\PageIndex{159}\\

If \\\vec{F}\\ is a force and \\\vec{D}\\ is a vector, show \\\mathrm{proj}\_{\vec{D}}\left( \vec{F}\right) =\left( \\ \vec{F} \\ \cos \theta \right) \vec{u}\\ where \\\vec{u}\\ is the unit vector in the direction of \\\vec{D}\\, where \\\vec{u}=\vec{D}/ \\ \vec{D} \\\\ and \\\theta\\ is the included angle between the two vectors, \\\vec{F}\\ and \\\vec{D}\\. \\ \\ \vec{F} \\ \cos \theta\\ is sometimes called the component of the force, \\\vec{F}\\ in the direction, \\\vec{D}\\.

Answer

\\\mathrm{proj}\_{\vec{D}}\left( \vec{F}\right) = \frac{\vec{F}\bullet \vec{D}}{ \\ \vec{D} \\ }\frac{\vec{D}}{ \\ \vec{D} \\ }=\left( \\ \vec{F} \\ \cos \theta \right) \frac{\vec{D}}{ \\ \vec{D} \\ }=\left( \\ \vec{F} \\ \cos \theta \right) \vec{u}\\

Exercise \\\PageIndex{160}\\

A boy drags a sled for \\100\\ feet along the ground by pulling on a rope which is \\20\\ degrees from the horizontal with a force of \\40\\ pounds. How much work does this force do?

Answer

\\40\cos \left( \frac{20}{180}\pi \right)100=3758.8\\

Exercise \\\PageIndex{161}\\

A girl drags a sled for \\200\\ feet along the ground by pulling on a rope which is \\30\\ degrees from the horizontal with a force of \\20\\ pounds. How much work does this force do?

Answer

\\20\cos \left( \frac{\pi }{6}\right)200= 3464.1\\

Exercise \\\PageIndex{162}\\

A large dog drags a sled for \\300\\ feet along the ground by pulling on a rope which is \\45\\ degrees from the horizontal with a force of \\20\\ pounds. How much work does this force do?

Answer

\\20\left( \cos \frac{\pi }{4}\right)300=4242.6\\

Exercise \\\PageIndex{163}\\

How much work does it take to slide a crate \\20\\ meters along a loading dock by pulling on it with a \\200\\ Newton force at an angle of \\30^{\circ }\\ from the horizontal? Express your answer in Newton meters.

Answer

\\200\left( \cos \left( \frac{\pi }{6}\right) \right) 20= 3464.1\\

Exercise \\\PageIndex{164}\\

An object moves \\10\\ meters in the direction of \\\vec{j}\\. There are two forces acting on this object, \\\vec{F}\_{1}=\vec{i}+\vec{j}+ 2\vec{k}\\, and \\\vec{F}\_{2}=-5\vec{i}+2\vec{j}-6\vec{k}\\. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?

Answer

\\\left $$ \begin{array}{r} -4 \\ 3 \\ -4 \end{array} \right $$ \bullet \left $$ \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right $$ \times 10= 30\\ You can consider the resultant of the two forces because of the properties of the dot product.

Exercise \\\PageIndex{165}\\

An object moves \\10\\ meters in the direction of \\\vec{j}+\vec{i}\\. There are two forces acting on this object, \\\vec{F}\_{1}=\vec{i}+2\vec{j} +2\vec{k}\\, and \\\vec{F}\_{2}=5\vec{i}+2\vec{j}-6\vec{k}\\. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?

Answer

\\\begin{aligned} \vec{F}\_{1}\bullet \left $$ \begin{array}{r} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \right $$ 10+\vec{F}\_{2}\bullet \left $$ \begin{array}{r} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \right $$ 10 &=\left( \vec{F}\_{1}+\vec{F}\_{2}\right) \bullet \left $$ \begin{array}{r} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \right $$ 10 \\ &= \left $$ \begin{array}{r} 6 \\ 4 \\ -4 \end{array} \right $$ \bullet \left $$ \begin{array}{r} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \right $$ 10 \\ &= 50\sqrt{2}\end{aligned}\\

Exercise \\\PageIndex{166}\\

An object moves \\20\\ meters in the direction of \\\vec{k}+\vec{j}\\. There are two forces acting on this object, \\\vec{F}\_{1}=\vec{i}+\vec{j}+ 2\vec{k}\\, and \\\vec{F}\_{2}=\vec{i}+2\vec{j}-6\vec{k}\\. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force.

Answer

\\\left $$ \begin{array}{r} 2 \\ 3 \\ -4 \end{array} \right $$ \bullet \left $$ \begin{array}{r} 0 \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right $$ 20= -10\sqrt{2}\\

---