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9_01_3A_Algebraic_Considerations

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Outcomes

1. Develop the abstract concept of a vector space through axioms.

2. Deduce basic properties of vector spaces.

3. Use the vector space axioms to determine if a set and its operations constitute a vector space.

In this section we consider the idea of an abstract vector space. A vector space is something which has two operations satisfying the following vector space axioms.

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

A vector space \\V\\ is a set of vectors with two operations defined, addition and scalar multiplication, which satisfy the axioms of addition and scalar multiplication.

In the following definition we define two operations; vector addition, denoted by \\+\\ and scalar multiplication denoted by placing the scalar next to the vector. A vector space need not have usual operations, and for this reason the operations will always be given in the definition of the vector space. The below axioms for addition (written +) and scalar multiplication must hold for however addition and scalar multiplication are defined for the vector space.

It is important to note that we have seen much of this content before, in terms of \\\mathbb{R}^n\\. We will prove in this section that \\\mathbb{R}^n\\ is an example of a vector space and therefore all discussions in this chapter will pertain to \\\mathbb{R}^n\\. While it may be useful to consider all concepts of this chapter in terms of \\\mathbb{R}^n\\, it is also important to understand that these concepts apply to *all* vector spaces.

In the following definition, we will choose scalars \\a,b\\ to be real numbers and are thus dealing with *real* vector spaces. However, we could also choose scalars which are complex numbers. In this case, we would call the vector space \\V\\ complex.

Definition \\\PageIndex{2}\\: Axioms of Addition

Let \\\vec{v}, \vec{w}, \vec{z}\\ be vectors in a vector space \\V\\. Then they satisfy the following axioms of addition:

vectorspaceaxiomsaddition

Definition \\\PageIndex{3}\\: Axioms of Scalar Multiplication

Let \\a, b \in \mathbb{R}\\ and let \\\vec{v}, \vec{w}, \vec{z}\\ be vectors in a vector space \\V\\. Then they satisfy the following axioms of scalar multiplication:

Consider the following example, in which we prove that \\\mathbb{R}^n\\ is in fact a vector space.

Example \\\PageIndex{1}\\: \\\mathbb{R}^n\\

\\\mathbb{R}^n\\, under the usual operations of vector addition and scalar multiplication, is a vector space.

###### Solution

To show that \\\mathbb{R}^n\\ is a vector space, we need to show that the above axioms hold. Let \\\vec{x}, \vec{y}, \vec{z}\\ be vectors in \\\mathbb{R}^n\\. We first prove the axioms for vector addition.

We now need to prove the axioms related to scalar multiplication. Let \\a,b\\ be real numbers and let \\\vec{x}, \vec{y}\\ be vectors in \\\mathbb{R}^n\\.

By the above proofs, it is clear that \\\mathbb{R}^n\\ satisfies the vector space axioms. Hence, \\\mathbb{R}^n\\ is a vector space under the usual operations of vector addition and scalar multiplication.

We now consider some examples of vector spaces.

Example \\\PageIndex{2}\\: Vector Space of Polynomials

Let \\\mathbb{P}\_2\\ be the set of all polynomials of at most degree \\2\\ as well as the zero polynomial. Define addition to be the standard addition of polynomials, and scalar multiplication the usual multiplication of a polynomial by a number. Then \\\mathbb{P}\_2\\ is a vector space.

###### Solution

We can write \\\mathbb{P}\_2\\ explicitly as \\\mathbb{P}\_2 = \left\\ a_2x^2 + a_1x + a_0 \| a_i \in \mathbb{R} \\ \mbox{for all} \\ i \right\\\nonumber \\ To show that \\\mathbb{P}\_2\\ is a vector space, we verify the axioms. Let \\p(x), q(x), r(x)\\ be polynomials in \\\mathbb{P}\_2\\ and let \\a,b,c\\ be real numbers. Write \\p(x)=p_2x^2 + p_1x + p_0\\, \\q(x)=q_2x^2 + q_1x + q_0\\, and \\r(x)=r_2x^2 + r_1x + r_0\\.

We now need to verify the axioms related to scalar multiplication.

Another important example of a vector space is the set of all matrices of the same size.

Example \\\PageIndex{3}\\: Vector Space of Matrices

Let \\\mathbb{M}\_{2,3}\\ be the set of all \\2 \times 3\\ matrices. Using the usual operations of matrix addition and scalar multiplication, show that \\\mathbb{M}\_{2,3}\\ is a vector space.

###### Solution

Let \\A, B\\ be \\2 \times 3\\ matrices in \\\mathbb{M}\_{2,3}\\. We first prove the axioms for addition.

We now turn our attention to the axioms regarding scalar multiplication. Let \\A, B\\ be matrices in \\\mathbb{M}\_{2,3}\\ and let \\c\\ be a real number.

In conclusion, \\\mathbb{M}\_{2,3}\\ satisfies the required axioms and is a vector space.

While here we proved that the set of all \\2 \times 3\\ matrices is a vector space, there is nothing special about this choice of matrix size. In fact if we instead consider \\\mathbb{M}\_{m,n}\\, the set of all \\m \times n\\ matrices, then \\\mathbb{M}\_{m,n}\\ is a vector space under the operations of matrix addition and scalar multiplication.

We now examine an example of a set that does not satisfy all of the above axioms, and is therefore *not* a vector space.

Example \\\PageIndex{4}\\: Not a Vector Space

Let \\V\\ denote the set of \\2 \times 3\\ matrices. Let addition in \\V\\ be defined by \\A + B = A\\ for matrices \\A,B\\ in \\V\\. Let scalar multiplication in \\V\\ be the usual scalar multiplication of matrices. Show that \\V\\ is not a vector space.

###### Solution

In order to show that \\V\\ is not a vector space, it suffices to find only one axiom which is not satisfied. We will begin by examining the axioms for addition until one is found which does not hold. Let \\A,B\\ be matrices in \\V\\.

Consider another example of a vector space.

Example \\\PageIndex{5}\\: Vector Space of Functions

Let \\S\\ be a nonempty set and define \\\mathbb{F}\_S\\ to be the set of *real* functions defined on \\S\\. In other words, we write \\\mathbb{F}\_S: S \mapsto \mathbb{R}\\. Letting \\a,b,c\\ be scalars and \\f,g,h\\ functions, the vector operations are defined as \\\begin{aligned} \left( f+g\right) \left( x\right) &=f\left( x\right) +g\left( x\right) \\ \left( af\right) \left( x\right) &=a\left( f\left( x\right) \right)\end{aligned}\\ Show that \\\mathbb{F}\_S\\ is a vector space.

###### Solution

To verify that \\\mathbb{F}\_S\\ is a vector space, we must prove the axioms beginning with those for addition. Let \\f, g, h\\ be functions in \\\mathbb{F}\_S\\.

Now, check the axioms for scalar multiplication.

It follows that \\V\\ satisfies all the required axioms and is a vector space.

Consider the following important theorem.

Theorem \\\PageIndex{1}\\: Uniqueness

In any vector space, the following are true:

1. \\\vec{0}\\, the additive identity, is unique

2. \\-\vec{x}\\, the additive inverse, is unique

3. \\0\vec{x}=\vec{0}\\ for all vectors \\\vec{x}\\

4. \\\left( -1\right) \vec{x}=-\vec{x}\\ for all vectors \\\vec{x}\\

Proof

1. When we say that the additive identity, \\\vec{0}\\, is unique, we mean that if a vector acts like the additive identity, then it is the additive identity. To prove this uniqueness, we want to show that another vector which acts like the additive identity is actually equal to \\\vec{0}\\. Suppose \\\vec{0}^{\prime }\\ is also an additive identity. Then, \\\vec{0} + \vec{0}^{\prime} = \vec{0}\nonumber \\ Now, for \\\vec{0}\\ the additive identity given above in the axioms, we have that \\\vec{0}^{\prime} + \vec{0} = \vec{0}^{\prime}\nonumber \\ So by the commutative property: \\0 = 0 + 0^{\prime} = 0^{\prime} + 0 = 0^{\prime}\nonumber \\ This says that if a vector acts like an additive identity (such as \\\vec{0}^{\prime}\$, it in fact equals \\\vec{0}\\. This proves the uniqueness of \\\vec{0}\\.

2. When we say that the additive inverse, \\-\vec{x}\\, is unique, we mean that if a vector acts like the additive inverse, then it is the additive inverse. Suppose that \\\vec{y}\\ acts like an additive inverse: \\\vec{x}+\vec{y}=\vec{0}\nonumber \\ Then the following holds: \\\vec{y} = \vec{0} + \vec{y} = (-\vec{x} + \vec{x}) + \vec{y} = -\vec{x} + (\vec{x} + \vec{y}) = -\vec{x} + \vec{0} = -\vec{x}\nonumber \\ Thus if \\\vec{y}\\ acts like the additive inverse, it is equal to the additive inverse \\-\vec{x}\\. This proves the uniqueness of \\-\vec{x}\\.

3. This statement claims that for all vectors \\\vec{x}\\, scalar multiplication by \\0\\ equals the zero vector \\\vec{0}\\. Consider the following, using the fact that we can write \\0=0+0\\: \\0\vec{x}=\left( 0+0\right) \vec{x}=0\vec{x}+0\vec{x}\nonumber \\ We use a small trick here: add \\-0\vec{x}\\ to both sides. This gives \\\begin{aligned} 0\vec{x} + (-0\vec{x})&=0\vec{x}+0\vec{x}+(-\vec{x})\\ \vec{0} + 0 &=0\vec{x} + 0 \\ \vec{0} &= 0\vec{x}\end{aligned}\\ This proves that scalar multiplication of any vector by \\0\\ results in the zero vector \\\vec{0}\\.

4. Finally, we wish to show that scalar multiplication of \\-1\\ and any vector \\\vec{x}\\ results in the additive inverse of that vector, \\-\vec{x}\\. Recall from \\2.\\ above that the additive inverse is unique. Consider the following: \\\begin{aligned} \left( -1\right) \vec{x}+\vec{x} & =\left( -1\right) \vec{x}+1\vec{x}\\ &=\left(-1+1\right) \vec{x} \\ &=0\vec{x} \\ &=\vec{0}\end{aligned}\\ By the uniqueness of the additive inverse shown earlier, any vector which acts like the additive inverse must be equal to the additive inverse. It follows that \\\left( -1\right) \vec{x}=-\vec{x}\\.

An important use of the additive inverse is the following theorem.

Theorem \\\PageIndex{2}\\

Let \\V\\ be a vector space. Then \\\vec{v} + \vec{w} = \vec{v} + \vec{z}\\ implies that \\\vec{w} = \vec{z}\\ for all \\\vec{v}, \vec{w}, \vec{z} \in V\\

Proof

The proof follows from the vector space axioms, in particular the existence of an additive inverse (\\-\vec{u}\$. The proof is left as an exercise to the reader.

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9_02_3A_Spanning_Sets

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Outcomes

1. Determine if a vector is within a given span.

In this section we will examine the concept of spanning introduced earlier in terms of \\\mathbb{R}^n\\. Here, we will discuss these concepts in terms of abstract vector spaces.

Consider the following definition.

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

Let \\X\\ and \\Y\\ be two sets. If all elements of \\X\\ are also elements of \\Y\\ then we say that \\X\\ is a subset of \\Y\\ and we write \\X \subseteq Y\nonumber \\

In particular, we often speak of subsets of a vector space, such as \\X \subseteq V\\. By this we mean that every element in the set \\X\\ is contained in the vector space \\V\\.

Definition \\\PageIndex{2}\\: Linear Combination

Let \\V\\ be a vector space and let \\\vec{v}\_1, \vec{v}\_2, \cdots, \vec{v}\_n \subseteq V\\. A vector \\\vec{v} \in V\\ is called a linear combination of the \\\vec{v}\_i\\ if there exist scalars \\c_i \in \mathbb{R}\\ such that \\\vec{v} = c_1 \vec{v}\_1 + c_2 \vec{v}\_2 + \cdots + c_n \vec{v}\_n\nonumber \\

This definition leads to our next concept of span.

Definition \\\PageIndex{3}\\: Span of Vectors

Let \\\\\vec{v}\_{1},\cdots ,\vec{v}\_{n}\\ \subseteq V\\. Then \\\mathrm{span}\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\ = \left\\ \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}: c\_{i}\in \mathbb{R} \right\\\nonumber \\

When we say that a vector \\\vec{w}\\ is in \\\mathrm{span}\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\\\ we mean that \\\vec{w}\\ can be written as a linear combination of the \\\vec{v}\_1\\. We say that a collection of vectors \\\\\vec{v}\_{1},\cdots ,\vec{v}\_{n}\\\\ is a spanning set for \\V\\ if \\V = \mathrm{span} \\\vec{v}\_{1},\cdots ,\vec{v}\_{n}\\\\.

Consider the following example.

Example \\\PageIndex{1}\\: Matrix Span

Let \\A = \left $$ \begin{array}{rr} 1 & 0 \\ 0 & 2 \end{array}\right $$\\, \\B = \left $$ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array}\right $$\\. Determine if \\A\\ and \\B\\ are in \\\mathrm{span}\left\\ M_1, M_2 \right\\ = \mathrm{span} \left\\ \left $$ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right $$, \left $$ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array}\right $$ \right\\\nonumber \\

###### Solution

First consider \\A\\. We want to see if scalars \\s,t\\ can be found such that \\A = s M_1 + t M_2\\. \\\left $$ \begin{array}{rr} 1 & 0 \\ 0 & 2 \end{array}\right $$ = s \left $$ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right $$ + t \left $$ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array}\right $$\nonumber \\ The solution to this equation is given by \\\begin{aligned} 1 &= s \\ 2 &= t\end{aligned}\\ and it follows that \\A\\ is in \\\mathrm{span} \left\\ M_1, M_2 \right\\\\.

Now consider \\B\\. Again we write \\B = sM_1 + t M_2\\ and see if a solution can be found for \\s, t\\. \\\left $$ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array}\right $$ = s \left $$ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right $$ + t \left $$ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array}\right $$\nonumber \\ Clearly no values of \\s\\ and \\t\\ can be found such that this equation holds. Therefore \\B\\ is not in \\\mathrm{span} \left\\ M_1, M_2 \right\\\\.

Consider another example.

Example \\\PageIndex{2}\\: Polynomial Span

Show that \\p(x) = 7x^2 + 4x - 3\\ is in \\\mathrm{span}\left\\ 4x^2 + x, x^2 -2x + 3 \right\\\\.

###### Solution

To show that \\p(x)\\ is in the given span, we need to show that it can be written as a linear combination of polynomials in the span. Suppose scalars \\a, b\\ existed such that \\7x^2 +4x - 3= a(4x^2+x) + b (x^2-2x+3)\nonumber \\ If this linear combination were to hold, the following would be true: \\\begin{aligned} 4a + b &= 7 \\ a - 2b &= 4 \\ 3b &= -3 \end{aligned}\\

You can verify that \\a = 2, b = -1\\ satisfies this system of equations. This means that we can write \\p(x)\\ as follows: \\7x^2 +4x-3= 2(4x^2+x) - (x^2-2x+3)\nonumber \\

Hence \\p(x)\\ is in the given span.

Consider the following example.

Example \\\PageIndex{3}\\: Spanning Set

Let \\S = \left\\ x^2 + 1, x-2, 2x^2 - x \right\\\\. Show that \\S\\ is a spanning set for \\\mathbb{P}\_2\\, the set of all polynomials of degree at most \\2\\.

###### Solution

Let \\p(x)= ax^2 + bx + c\\ be an arbitrary polynomial in \\\mathbb{P}\_2\\. To show that \\S\\ is a spanning set, it suffices to show that \\p(x)\\ can be written as a linear combination of the elements of \\S\\. In other words, can we find \\r,s,t\\ such that: \\p(x) = ax^2 +bx + c = r(x^2 + 1) + s(x -2) + t(2x^2 - x)\nonumber \\

If a solution \\r,s,t\\ can be found, then this shows that for any such polynomial \\p(x)\\, it can be written as a linear combination of the above polynomials and \\S\\ is a spanning set.

\\\begin{aligned} ax^2 +bx + c &= r(x^2 + 1) + s(x -2) + t(2x^2 - x) \\ &= rx^2 + r + sx - 2s + 2tx^2 - tx \\ &= (r+2t)x^2 + (s-t)x + (r-2s) \end{aligned}\\

For this to be true, the following must hold: \\\begin{aligned} a &= r+2t \\ b &= s-t \\ c &= r-2s\end{aligned}\\

To check that a solution exists, set up the augmented matrix and row reduce: \\\left $$ \begin{array}{rrr\|r} 1 & 0 & 2 & a \\ 0 & 1 & -1 & b \\ 1 & -2 & 0 & c \end{array} \right $$ \rightarrow \cdots \rightarrow \left $$ \begin{array}{rrr\|c} 1 & 0 & 0 & \frac{1}{2} a + 2b + \frac{1}{2}c\\ 0 & 1 & 0 & \frac{1}{4}a - \frac{1}{4}c \\ 0 & 0 & 1 & \frac{1}{4}a - b - \frac{1}{4}c \end{array} \right $$\nonumber \\

Clearly a solution exists for any choice of \\a,b,c\\. Hence \\S\\ is a spanning set for \\\mathbb{P}\_2\\.

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9_03_3A_Linear_Independence

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Outcomes

1. Determine if a set is linearly independent.

In this section, we will again explore concepts introduced earlier in terms of \\\mathbb{R}^n\\ and extend them to apply to abstract vector spaces.

Definition \\\PageIndex{1}\\: Linear Independence

Let \\V\\ be a vector space. If \\\\\vec{v}\_{1},\cdots ,\vec{v}\_{n}\\ \subseteq V,\\ then it is linearly independent if \\\sum\_{i=1}^{n}a\_{i}\vec{v}\_{i}=\vec{0} \\\mbox{implies}\\ a\_{1}=\cdots =a\_{n}=0\nonumber \\ where the \\a_i\\ are real numbers.

The set of vectors is called linearly dependent if it is not linearly independent.

Example \\\PageIndex{1}\\: Linear Independence

Let \\S \subseteq \mathbb{P}\_2\\ be a set of polynomials given by \\S = \left\\ x^2 + 2x - 1, 2x^2 - x + 3 \right\\\nonumber \\ Determine if \\S\\ is linearly independent.

###### Solution

To determine if this set \\S\\ is linearly independent, we write \\a ( x^2 + 2x -1 ) + b(2x^2 - x + 3) = 0x^2 + 0x + 0\nonumber \\ If it is linearly independent, then \\a=b=0\\ will be the only solution. We proceed as follows. \\\begin{aligned} a ( x^2 + 2x -1 ) + b(2x^2 - x + 3) &= 0x^2 + 0x + 0 \\ ax^2 + 2ax - a + 2bx^2 - bx + 3b &= 0x^2 + 0x + 0 \\ (a+2b)x^2 + (2a -b)x - a + 3b &= 0x^2 + 0x + 0\end{aligned}\\

It follows that \\\begin{aligned} a + 2b &= 0 \\ 2a - b &= 0 \\ -a + 3b &= 0\end{aligned}\\

The augmented matrix and resulting reduced row-echelon form are given by \\\left $$ \begin{array}{rr\|r} 1 & 2 & 0 \\ 2 & -1 & 0 \\ -1 & 3 & 0 \end{array} \right $$ \rightarrow \cdots \rightarrow \left $$ \begin{array}{rr\|r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right $$\nonumber \\

Hence the solution is \\a=b=0\\ and the set is linearly independent.

The next example shows us what it means for a set to be dependent.

Example \\\PageIndex{2}\\: Dependent Set

Determine if the set \\S\\ given below is independent. \\S=\left\\ \left $$\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right $$, \left $$\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right $$, \left $$\begin{array}{c} 1 \\ 3 \\ 5 \end{array}\right $$ \right\\\nonumber \\

###### Solution

To determine if \\S\\ is linearly independent, we look for solutions to \\a\left $$\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right $$ +b\left $$\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right $$ +c\left $$\begin{array}{c} 1 \\ 3 \\ 5 \end{array}\right $$ =\left $$\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right $$\nonumber \\ Notice that this equation has nontrivial solutions, for example \\a=2\\, \\b=3\\ and \\c=-1\\. Therefore \\S\\ is dependent.

The following is an important result regarding dependent sets.

Lemma \\\PageIndex{1}\\: Dependent Sets

Let \\V\\ be a vector space and suppose \\W = \left\\ \vec{v}\_1, \vec{v}\_2, \cdots, \vec{v}\_k \right\\\\ is a subset of \\V\\. Then \\W\\ is dependent if and only if \\\vec{v}\_i\\ can be written as a linear combination of \\\left\\ \vec{v}\_1, \vec{v}\_2, \cdots, \vec{v}\_{i-1}, \vec{v}\_{i+1}, \cdots, \vec{v}\_k \right\\\\ for some \\i \leq k\\.

Revisit Example $\PageIndex{2}$ with this in mind. Notice that we can write one of the three vectors as a combination of the others. \\\left $$\begin{array}{c} 1 \\ 3 \\ 5 \end{array}\right $$ = 2\left $$\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right $$ +3\left $$\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right $$\nonumber \\

By Lemma $\PageIndex{1}$ this set is dependent.

If we know that one particular set is linearly independent, we can use this information to determine if a related set is linearly independent. Consider the following example.

Example \\\PageIndex{3}\\: Related Independent Sets

Let \\V\\ be a vector space and suppose \\S \subseteq V\\ is a set of linearly independent vectors given by \\S = \left\\ \vec{u}, \vec{v}, \vec{w} \right\\\\. Let \\R \subseteq V\\ be given by \\R = \left\\ 2\vec{u} - \vec{w}, \vec{w} + \vec{v}, 3\vec{v} + \frac{1}{2} \vec{u} \right\\\\. Show that \\R\\ is also linearly independent.

###### Solution

To determine if \\R\\ is linearly independent, we write \\a(2\vec{u} - \vec{w}) + b(\vec{w} + \vec{v}) + c( 3\vec{v} + \frac{1}{2}\vec{u}) = \vec{0}\nonumber \\ If the set is linearly independent, the only solution will be \\a=b=c=0\\. We proceed as follows. \\\begin{aligned} a(2\vec{u} - \vec{w}) + b(\vec{w} + \vec{v}) + c( 3\vec{v} + \frac{1}{2} \vec{u}) &= \vec{0} \\ 2a\vec{u} - a\vec{w} + b\vec{w} + b\vec{v} + 3c\vec{v} + \frac{1}{2}c\vec{u} &= \vec{0}\\ (2a + \frac{1}{2}c) \vec{u} + (b+3c)\vec{v} + (-a + b) \vec{w} &= \vec{0}\end{aligned}\\

We know that the set \\S = \left\\ \vec{u}, \vec{v}, \vec{w} \right\\\\ is linearly independent, which implies that the coefficients in the last line of this equation must all equal \\0\\. In other words: \\\begin{aligned} 2a + \frac{1}{2} c &= 0 \\ b + 3c &= 0 \\ -a + b &= 0 \end{aligned}\\

The augmented matrix and resulting reduced row-echelon form are given by: \\\left $$ \begin{array}{rrr\|r} 2 & 0 & \frac{1}{2} & 0 \\ 0 & 1 & 3 & 0 \\ -1 & 1 & 0 & 0 \end{array}\right $$ \rightarrow \cdots \rightarrow \left $$ \begin{array}{rrr\|r} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right $$\nonumber \\ Hence the solution is \\a=b=c=0\\ and the set is linearly independent.

The following theorem was discussed in terms in \\\mathbb{R}^n\\. We consider it here in the general case.

Theorem \\\PageIndex{1}\\: Unique Representation

Let \\V\\ be a vector space and let \\U = \left\\ \vec{v}\_1, \cdots, \vec{v}\_k \right\\ \subseteq V\\ be an independent set. If \\\vec{v} \in \mathrm{span} \\U\\, then \\\vec{v}\\ can be written uniquely as a linear combination of the vectors in \\U\\.

Consider the span of a linearly independent set of vectors. Suppose we take a vector which is not in this span and add it to the set. The following lemma claims that the resulting set is still linearly independent.

Lemma \\\PageIndex{2}\\: Adding to a Linearly Independent Set

Suppose \\\vec{v}\notin \mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ and \\\left\\ \vec{u}\_{1},\cdots , \vec{u}\_{k}\right\\\\ is linearly independent. Then the set \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k},\vec{v} \right\\\nonumber \\ is also linearly independent.

Proof

Suppose \\\sum\_{i=1}^{k}c\_{i}\vec{u}\_{i}+d\vec{v}= \vec{0}.\\ It is required to verify that each \\c\_{i}=0\\ and that \\d=0.\\ But if \\d\neq 0,\\ then you can solve for \\\vec{v}\\ as a linear combination of the vectors, \\\left\\ \vec{u}\_{1},\cdots ,\vec{u} \_{k}\right\\\\, \\\vec{v}=-\sum\_{i=1}^{k}\left( \frac{c\_{i}}{d}\right) \vec{u}\_{i}\nonumber \\ contrary to the assumption that \\\vec{v}\\ is not in the span of the \\\vec{u}\_{i}\\. Therefore, \\d=0.\\ But then \\\sum\_{i=1}^{k}c\_{i} \vec{u}\_{i}=\vec{0}\\ and the linear independence of \\\left\\ \vec{u} \_{1},\cdots ,\vec{u}\_{k}\right\\\\ implies each \\c\_{i}=0\\ also.

Consider the following example.

Example \\\PageIndex{4}\\: Adding to a Linearly Independent Set

Let \\S \subseteq M\_{22}\\ be a linearly independent set given by \\S = \left\\ \left $$ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right $$, \left $$ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right $$ \right\\\nonumber \\ Show that the set \\R \subseteq M\_{22}\\ given by \\R = \left\\ \left $$ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right $$, \left $$ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right $$, \left $$ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right $$ \right\\\nonumber \\ is also linearly independent.

###### Solution

Instead of writing a linear combination of the matrices which equals \\0\\ and showing that the coefficients must equal \\0\\, we can instead use Lemma $\PageIndex{2}$.

To do so, we show that \\\left $$ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right $$ \notin \mathrm{span}\left\\ \left $$ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right $$, \left $$ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right $$ \right\\\nonumber \\

Write \\\begin{aligned} \left $$ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right $$ &= a\left $$ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right $$ + b\left $$ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right $$ \\ &= \left $$ \begin{array}{rr} a & 0 \\ 0 & 0 \end{array} \right $$ + \left $$ \begin{array}{rr} 0 & b \\ 0 & 0 \end{array} \right $$ \\ &= \left $$ \begin{array}{rr} a & b \\ 0 & 0 \end{array} \right $$\end{aligned}\\

Clearly there are no possible \\a,b\\ to make this equation true. Hence the new matrix does not lie in the span of the matrices in \\S\\. By Lemma $\PageIndex{2}$, \\R\\ is also linearly independent.

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9_04_3A_Subspaces_and_Basis

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Outcomes

1. Utilize the subspace test to determine if a set is a subspace of a given vector space.

2. Extend a linearly independent set and shrink a spanning set to a basis of a given vector space.

In this section we will examine the concept of subspaces introduced earlier in terms of \\\mathbb{R}^n\\. Here, we will discuss these concepts in terms of abstract vector spaces.

Consider the definition of a subspace.

Definition \\\PageIndex{1}\\: Subspace

Let \\V\\ be a vector space. A subset \\W\subseteq V\\ is said to be a subspace of \\V\\ if \\a\vec{x}+b\vec{y} \in W\\ whenever \\a,b\in \mathbb{R}\\ and \\\vec{x},\vec{y}\in W.\\

The span of a set of vectors as described in Definition 9.2.3 is an example of a subspace. The following fundamental result says that subspaces are subsets of a vector space which are themselves vector spaces.

Theorem \\\PageIndex{1}\\: Subspaces are Vector Spaces

Let \\W\\ be a nonempty collection of vectors in a vector space \\V\\. Then \\W\\ is a subspace if and only if \\W\\ satisfies the vector space axioms, using the same operations as those defined on \\V\\.

Proof

Suppose first that \\W\\ is a subspace. It is obvious that all the algebraic laws hold on \\W\\ because it is a subset of \\V\\ and they hold on \\V\\. Thus \\\vec{u}+\vec{v}=\vec{v}+\vec{u}\\ along with the other axioms. Does \\W\\ contain \\\vec{0}?\\ Yes because it contains \\0\vec{u}=\vec{0}\\. See Theorem 9.1.1.

Are the operations of \\V\\ defined on \\W?\\ That is, when you add vectors of \\W\\ do you get a vector in \\W?\\ When you multiply a vector in \\W\\ by a scalar, do you get a vector in \\W?\\ Yes. This is contained in the definition. Does every vector in \\W\\ have an additive inverse? Yes by Theorem 9.1.1 because \\-\vec{v}=\left( -1\right) \vec{v}\\ which is given to be in \\W\\ provided \\\vec{v}\in W\\.

Next suppose \\W\\ is a vector space. Then by definition, it is closed with respect to linear combinations. Hence it is a subspace.

Consider the following useful Corollary.

Corollary \\\PageIndex{1}\\: Span is a Subspace

Let \\V\\ be a vector space with \\W \subseteq V\\. If \\W = \mathrm{span} \left\\ \vec{v}\_1, \cdots, \vec{v}\_n \right\\\\ then \\W\\ is a subspace of \\V\\.

When determining spanning sets the following theorem proves useful.

Theorem \\\PageIndex{2}\\: Spanning Set

Let \\W \subseteq V\\ for a vector space \\V\\ and suppose \\W = \mathrm{span} \left\\ \vec{v}\_1, \vec{v}\_2, \cdots, \vec{v}\_n \right\\\\.

Let \\U \subseteq V\\ be a subspace such that \\\vec{v}\_1, \vec{v}\_2, \cdots, \vec{v}\_n \in U\\. Then it follows that \\W \subseteq U\\.

In other words, this theorem claims that any subspace that contains a set of vectors must also contain the span of these vectors.

The following example will show that two spans, described differently, can in fact be equal.

Example \\\PageIndex{1}\\: Equal Span

Let \\p(x), q(x)\\ be polynomials and suppose \\U = \mathrm{span}\left\\ 2p(x) - q(x), p(x) + 3q(x)\right\\\\ and \\W = \mathrm{span}\left\\ p(x), q(x) \right\\\\. Show that \\U = W\\.

###### Solution

We will use Theorem $\PageIndex{2}$ to show that \\U \subseteq W\\ and \\W \subseteq U\\. It will then follow that \\U=W\\.

1. \\U \subseteq W\\ Notice that \\2p(x) - q(x)\\ and \\p(x) + 3q(x)\\ are both in \\W=span\\p(x),q(x)\\\\. Then by Theorem $\PageIndex{2}$ \\W\\ must contain the span of these polynomials and so \\U⊆W\\.

2. \\W \subseteq U\\ Notice that \\\begin{aligned}p(x)&=\frac{3}{7}(2p(x)-q(x))+\frac{2}{7}(p(x)+3q(x)) \\ q(x)&=-\frac{1}{7}(2p(x)-q(x))+\frac{2}{7}(p(x)+3q(x))\end{aligned}\\ Hence \\p(x),\\q(x)\\ are in \\span\\2p(x)−q(x),\\ p(x) +3q(x)\\\\. By Theorem $\PageIndex{2}$ \\U\\ must contain the span of these polynomials and so \\W ⊆ U\\.

To prove that a set is a vector space, one must verify each of the axioms given in Definition 9.1.2 and 9.1.3. This is a cumbersome task, and therefore a shorter procedure is used to verify a subspace.

Procedure \\\PageIndex{1}\\: Subspace Test

Suppose \\W\\ is a subset of a vector space \\V\\. To determine if \\W\\ is a subspace of \\V\\, it is sufficient to determine if the following three conditions hold, using the operations of \\V\\:

1. The additive identity \\\vec{0}\\ of \\V\\ is contained in \\W\\.

2. For any vectors \\\vec{w}\_1, \vec{w}\_2\\ in \\W\\, \\\vec{w}\_1 + \vec{w}\_2\\ is also in \\W\\.

3. For any vector \\\vec{w}\_1\\ in \\W\\ and scalar \\a\\, the product \\a\vec{w}\_1\\ is also in \\W\\.

Therefore it suffices to prove these three steps to show that a set is a subspace.

Consider the following example.

Example \\\PageIndex{2}\\: Improper Subspaces

Let \\V\\ be an arbitrary vector space. Then \\V\\ is a subspace of itself. Similarly, the set \\\left\\ \vec{0} \right\\\\ containing only the zero vector is also a subspace.

###### Solution

Using the subspace test in Procedure $\PageIndex{1}$ we can show that \\V\\ and \\\left\\ \vec{0} \right\\\\ are subspaces of \\V\\.

Since \\V\\ satisfies the vector space axioms it also satisfies the three steps of the subspace test. Therefore \\V\\ is a subspace.

Let’s consider the set \\\left\\ \vec{0} \right\\\\.

1. The vector \\\vec{0}\\ is clearly contained in \\\left\\ \vec{0} \right\\\\, so the first condition is satisfied.

2. Let \\\vec{w}\_1, \vec{w}\_2\\ be in \\\left\\ \vec{0} \right\\\\. Then \\\vec{w}\_1 = \vec{0}\\ and \\\vec{w}\_2 = \vec{0}\\ and so \\\vec{w}\_1 + \vec{w}\_2 = \vec{0} + \vec{0} = \vec{0}\nonumber \\ It follows that the sum is contained in \\\left\\ \vec{0} \right\\\\ and the second condition is satisfied.

3. Let \\\vec{w}\_1\\ be in \\\left\\ \vec{0} \right\\\\ and let \\a\\ be an arbitrary scalar. Then \\a\vec{w}\_1 = a\vec{0} = \vec{0}\nonumber \\ Hence the product is contained in \\\left\\ \vec{0} \right\\\\ and the third condition is satisfied.

It follows that \\\left\\ \vec{0} \right\\\\ is a subspace of \\V\\.

The two subspaces described above are called improper subspaces. Any subspace of a vector space \\V\\ which is not equal to \\V\\ or \\\left\\ \vec{0} \right\\\\ is called a proper subspace.

Consider another example.

Example \\\PageIndex{3}\\: Subspace of Polynomials

Let \\\mathbb{P}\_2\\ be the vector space of polynomials of degree two or less. Let \\W \subseteq \mathbb{P}\_2\\ be all polynomials of degree two or less which have \\1\\ as a root. Show that \\W\\ is a subspace of \\\mathbb{P}\_2\\.

###### Solution

First, express \\W\\ as follows: \\W = \left\\ p(x) = ax^2 +bx +c, a,b,c, \in \mathbb{R} \| p(1) = 0 \right\\\nonumber \\

We need to show that \\W\\ satisfies the three conditions of Procedure $\PageIndex{1}$.

1. The zero polynomial of \\\mathbb{P}\_2\\ is given by \\0(x) = 0x^2 + 0x + 0 = 0\\. Clearly \\0(1) = 0\\ so \\0(x)\\ is contained in \\W\\.

2. Let \\p(x), q(x)\\ be polynomials in \\W\\. It follows that \\p(1) = 0\\ and \\q(1) = 0\\. Now consider \\p(x) + q(x)\\. Let \\r(x)\\ represent this sum. \\\begin{aligned} r(1) &= p(1) + q(1) \\ &= 0 + 0 \\ &= 0\end{aligned}\\ Therefore the sum is also in \\W\\ and the second condition is satisfied.

3. Let \\p(x)\\ be a polynomial in \\W\\ and let \\a\\ be a scalar. It follows that \\p(1) = 0\\. Consider the product \\ap(x)\\. \\\begin{aligned} ap(1) &= a(0) \\ &= 0\end{aligned}\\ Therefore the product is in \\W\\ and the third condition is satisfied.

It follows that \\W\\ is a subspace of \\\mathbb{P}\_2\\.

Recall the definition of basis, considered now in the context of vector spaces.

Definition \\\PageIndex{2}\\: Basis

Let \\V\\ be a vector space. Then \\\\\vec{v}\_{1},\cdots ,\vec{v}\_{n}\\\\ is called a basis for \\V\\ if the following conditions hold.

1. \\\mathrm{span}\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\ = V\\

2. \\\\\vec{v}\_{1},\cdots ,\vec{v}\_{n}\\\\ is linearly independent

Consider the following example.

Example \\\PageIndex{4}\\: Polynomials of Degree Two

Let \\\mathbb{P}\_2\\ be the set polynomials of degree no more than 2. We can write \\\mathbb{P}\_2=\mathrm{span}\left\\ x^{2}, x, 1\right\\ .\\ Is \\\left\\ x^{2}, x, 1\right\\\\ a basis for \\\mathbb{P}\_2\\?

###### Solution

It can be verified that \\\mathbb{P}\_2\\ is a vector space defined under the usual addition and scalar multiplication of polynomials.

Now, since \\\mathbb{P}\_2=\mathrm{span}\left\\ x^{2},x, 1\right\\\\, the set \\\left\\ x^{2}, x, 1\right\\\\ is a basis if it is linearly independent. Suppose then that \\ax^{2}+bx+c=0x^2 + 0x + 0\nonumber \\ where \\a,b,c\\ are real numbers. It is clear that this can only occur if \\a=b=c=0\\. Hence the set is linearly independent and forms a basis of \\\mathbb{P}\_2\\.

The next theorem is an essential result in linear algebra and is called the exchange theorem.

Theorem \\\PageIndex{3}\\: Exchange Theorem

Let \\\left\\ \vec{x}\_{1},\cdots ,\vec{x}\_{r}\right\\\\ be a linearly independent set of vectors such that each \\\vec{x}\_{i}\\ is contained in span\\\left\\ \vec{y}\_{1},\cdots ,\vec{y}\_{s}\right\\ .\\ Then \\r\leq s.\\

Proof

The proof will proceed as follows. First, we set up the necessary steps for the proof. Next, we will assume that \\r \> s\\ and show that this leads to a contradiction, thus requiring that \\r \leq s\\.

Define span\\\left\\ \vec{y}\_{1},\cdots ,\vec{y}\_{s}\right\\ = V.\\ Since each \\\vec{x}\_i\\ is in span\\\left\\ \vec{y}\_{1},\cdots ,\vec{y}\_{s}\right\\\\, it follows there exist scalars \\c\_{1},\cdots ,c\_{s}\\ such that \\\vec{x}\_{1}=\sum\_{i=1}^{s}c\_{i}\vec{y}\_{i} \label{lincomb}\\ Note that not all of these scalars \\c_i\\ can equal zero. Suppose that all the \\c_i=0\\. Then it would follow that \\\vec{x}\_{1}=\vec{0}\\ and so \\\left\\ \vec{x} \_{1},\cdots ,\vec{x}\_{r}\right\\\\ would not be linearly independent. Indeed, if \\\vec{x}\_{1}=\vec{0}\\, \\1\vec{x}\_{1}+\sum\_{i=2}^{r}0 \vec{x}\_{i}=\vec{x}\_{1}=\vec{0}\\ and so there would exist a nontrivial linear combination of the vectors \\\left\\ \vec{x}\_{1},\cdots , \vec{x}\_{r}\right\\\\ which equals zero. Therefore at least one \\c_i\\ is nonzero.

Say \\c\_{k}\neq 0.\\ Then solve \\\eqref{lincomb}\\ for \\\vec{y}\_{k}\\ and obtain \\\vec{y}\_{k}\in \mathrm{span}\left\\ \vec{x}\_{1},\overset{\text{s-1 vectors here}}{\overbrace{\vec{y}\_{1},\cdots ,\vec{y}\_{k-1},\vec{y} \_{k+1},\cdots ,\vec{y}\_{s}}}\right\\ .\nonumber \\ Define \\\left\\ \vec{z}\_{1},\cdots ,\vec{z}\_{s-1}\right\\\\ to be \\\left\\ \vec{z}\_{1},\cdots ,\vec{z}\_{s-1}\right\\ = \left\\ \vec{y}\_{1},\cdots ,\vec{y}\_{k-1},\vec{y}\_{k+1},\cdots ,\vec{y} \_{s}\right\\\nonumber \\ Now we can write \\\vec{y}\_{k}\in \mathrm{span}\left\\ \vec{x}\_{1}, \vec{z}\_{1},\cdots, \vec{z}\_{s-1}\right\\\nonumber \\ Therefore, \\\mathrm{span}\left\\ \vec{x}\_{1},\vec{z}\_{1},\cdots ,\vec{z }\_{s-1}\right\\=V\\. To see this, suppose \\\vec{v}\in V\\. Then there exist constants \\c\_{1},\cdots ,c\_{s}\\ such that \\\vec{v}=\sum\_{i=1}^{s-1}c\_{i}\vec{z}\_{i}+c\_{s}\vec{y}\_{k}.\nonumber \\ Replace this \\\vec{y}\_{k}\\ with a linear combination of the vectors \\\left\\ \vec{x}\_{1},\vec{z}\_{1},\cdots ,\vec{z}\_{s-1}\right\\\\ to obtain \\\vec{v}\in \mathrm{span}\left\\ \vec{x}\_{1},\vec{z} \_{1},\cdots ,\vec{z}\_{s-1}\right\\ .\\ The vector \\\vec{y}\_{k},\\ in the list \\\left\\ \vec{y}\_{1},\cdots ,\vec{y}\_{s}\right\\ ,\\ has now been replaced with the vector \\\vec{x}\_{1}\\ and the resulting modified list of vectors has the same span as the original list of vectors, \\\left\\ \vec{y }\_{1},\cdots ,\vec{y}\_{s}\right\\ .\\

We are now ready to move on to the proof. Suppose that \\r\>s\\ and that \\\mathrm{span}\left\\ \vec{x}\_{1},\cdots , \vec{x}\_{l},\vec{z}\_{1},\cdots ,\vec{z}\_{p}\right\\ =V\nonumber \\ where the process established above has continued. In other words, the vectors \\\vec{z}\_{1},\cdots ,\vec{z}\_{p}\\ are each taken from the set \\\left\\ \vec{y}\_{1},\cdots ,\vec{y}\_{s}\right\\\\ and \\l+p=s.\\ This was done for \\l=1\\ above. Then since \\r\>s,\\ it follows that \\l\leq s\

The following corollary follows from the exchange theorem.

Corollary \\\PageIndex{2}\\: Two Bases of the Same Length

Let \\B_1\\, \\B_2\\ be two bases of a vector space \\V\\. Suppose \\B_1\\ contains \\m\\ vectors and \\B_2\\ contains \\n\\ vectors. Then \\m = n\\.

Proof

By Theorem $\PageIndex{3}$, \\m\leq n\\ and \\n\leq m\\. Therefore \\m=n\\.

This corollary is very important so we provide another proof independent of the exchange theorem above.

Proof

Suppose \\n \> m.\\ Then since the vectors \\\left\\ \vec{u} \_{1},\cdots ,\vec{u}\_{m}\right\\\\ span \\V,\\ there exist scalars \\c\_{ij}\\ such that \\\sum\_{i=1}^{m}c\_{ij}\vec{u}\_{i}=\vec{v}\_{j}.\nonumber \\ Therefore, \\\sum\_{j=1}^{n}d\_{j}\vec{v}\_{j}=\vec{0} \text{ if and only if }\sum\_{j=1}^{n}\sum\_{i=1}^{m}c\_{ij}d\_{j}\vec{u}\_{i}= \vec{0}\nonumber \\ if and only if \\\sum\_{i=1}^{m}\left( \sum\_{j=1}^{n}c\_{ij}d\_{j}\right) \vec{u}\_{i}=\vec{ 0}\nonumber \\ Now since \\\\\vec{u}\_{1},\cdots ,\vec{u}\_{n}\\\\ is independent, this happens if and only if \\\sum\_{j=1}^{n}c\_{ij}d\_{j}=0,\\i=1,2,\cdots ,m.\nonumber \\ However, this is a system of \\m\\ equations in \\n\\ variables, \\d\_{1},\cdots ,d\_{n}\\ and \\m\ n\\.

Given the result of the previous corollary, the following definition follows.

Definition \\\PageIndex{3}\\: Dimension

A vector space \\V\\ is of dimension \\n\\ if it has a basis consisting of \\n\\ vectors.

Notice that the dimension is well defined by Corollary $\PageIndex{2}$. It is assumed here that \\n\<\infty\\ and therefore such a vector space is said to be finite dimensional.

Example \\\PageIndex{5}\\: Dimension of a Vector Space

Let \\\mathbb{P}\_2\\ be the set of all polynomials of degree at most \\2\\. Find the dimension of \\\mathbb{P}\_2\\.

###### Solution

If we can find a basis of \\\mathbb{P}\_2\\ then the number of vectors in the basis will give the dimension. Recall from Example $\PageIndex{4}$ that a basis of \\\mathbb{P}\_2\\ is given by \\S = \left\\ x^2, x, 1 \right\\\nonumber \\ There are three polynomials in \\S\\ and hence the dimension of \\\mathbb{P}\_2\\ is three.

It is important to note that a basis for a vector space is not unique. A vector space can have many bases. Consider the following example.

Example \\\PageIndex{6}\\: A Different Basis for Polynomials of Degree Two

Let \\\mathbb{P}\_2\\ be the polynomials of degree no more than 2. Is \\\left\\ x^{2}+x+1,2x+1,3x^{2}+1\right\\\\ a basis for \\\mathbb{P}\_2\\?

###### Solution

Suppose these vectors are linearly independent but do not form a spanning set for \\\mathbb{P}\_2\\. Then by Lemma 9.3.2, we could find a fourth polynomial in \\\mathbb{P}\_2\\ to create a new linearly independent set containing four polynomials. However this would imply that we could find a basis of \\\mathbb{P}\_2\\ of more than three polynomials. This contradicts the result of Example $\PageIndex{5}$ in which we determined the dimension of \\\mathbb{P}\_2\\ is three. Therefore if these vectors are linearly independent they must also form a spanning set and thus a basis for \\\mathbb{P}\_2\\.

Suppose then that \\\begin{aligned} a\left( x^{2}+x+1\right) +b\left( 2x+1\right) +c\left( 3x^{2}+1\right) &= 0\\ \left( a+3c\right) x^{2}+\left( a+2b\right) x+\left( a+b+c\right) &= 0 \end{aligned}\\ We know that \\\left\\ x^2, x, 1 \right\\\\ is linearly independent, and so it follows that \\\begin{aligned} a+3c &= 0 \\ a+2b &= 0 \\ a+b+c &= 0\end{aligned}\\ and there is only one solution to this system of equations, \\a=b=c=0\\. Therefore, these are linearly independent and form a basis for \\\mathbb{P}\_2\\.

Consider the following theorem.

Theorem \\\PageIndex{4}\\: Every Subspace has a Basis

Let \\W\\ be a nonzero subspace of a finite dimensional vector space \\V\\. Suppose \\V\\ has dimension \\n\\. Then \\W\\ has a basis with no more than \\n\\ vectors.

Proof

Let \\\vec{v}\_{1}\in V\\ where \\\vec{v}\_{1}\neq 0.\\ If \\\mathrm{span}\left\\ \vec{v}\_{1}\right\\ =V,\\ then it follows that \\\left\\ \vec{v} \_{1}\right\\\\ is a basis for \\V\\. Otherwise, there exists \\\vec{v} \_{2}\in V\\ which is not in \\\mathrm{span}\left\\ \vec{v}\_{1}\right\\ .\\ By Lemma 9.3.2 \\\left\\ \vec{v}\_{1},\vec{v}\_{2}\right\\\\ is a linearly independent set of vectors. Then \\\left\\ \vec{v}\_{1},\vec{v} \_{2}\right\\\\ is a basis for \\V\\ and we are done. If \\\mathrm{span}\left\\ \vec{v}\_{1}, \vec{v}\_{2}\right\\ \neq V,\\ then there exists \\\vec{v}\_{3}\notin \mathrm{ span}\left\\ \vec{v}\_{1},\vec{v}\_{2}\right\\\\ and \\\left\\ \vec{v} \_{1},\vec{v}\_{2},\vec{v}\_{3}\right\\\\ is a larger linearly independent set of vectors. Continuing this way, the process must stop before \\n+1\\ steps because if not, it would be possible to obtain \\n+1\\ linearly independent vectors contrary to the exchange theorem, Theorem $\PageIndex{3}$.

If in fact \\W\\ has \\n\\ vectors, then it follows that \\W=V\\.

Theorem \\\PageIndex{5}\\: Subspace of Same Dimension

Let \\V\\ be a vector space of dimension \\n\\ and let \\W\\ be a subspace. Then \\W=V\\ if and only if the dimension of \\W\\ is also \\n\\.

Proof

First suppose \\W=V.\\ Then obviously the dimension of \\W=n.\\

Now suppose that the dimension of \\W\\ is \\n\\. Let a basis for \\W\\ be \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{n}\right\\\\. If \\W\\ is not equal to \\V\\, then let \\\vec{v}\\ be a vector of \\V\\ which is not contained in \\W.\\ Thus \\\vec{v}\\ is not in \\\mathrm{span}\left\\ \vec{w}\_{1},\cdots ,\vec{w} \_{n}\right\\\\ and by Lemma 9.7.2, \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{n},\vec{v}\right\\\\ is linearly independent which contradicts Theorem $\PageIndex{3}$ because it would be an independent set of \\n+1\\ vectors even though each of these vectors is in a spanning set of \\n\\ vectors, a basis of \\V\\.

Consider the following example.

Example \\\PageIndex{7}\\: Basis of a Subspace

Let \\U=\left\\ A\in\mathbb{M}\_{22} ~\left\|~ A\left $$\begin{array}{rr} 1 & 0 \\ 1 & -1 \end{array}\right $$\right. = \left $$\begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right $$ A \right\\\\. Then \\U\\ is a subspace of \\\mathbb{M}\_{22}\\ Find a basis of \\U\\, and hence \\\dim(U)\\.

###### Solution

Let \\A=\left $$\begin{array}{rr} a & b \\ c & d \end{array}\right $$ \in\mathbb{M}\_{22}\\. Then \\A\left $$\begin{array}{rr} 1 & 0 \\ 1 & -1 \end{array}\right $$ = \left $$\begin{array}{rr} a & b \\ c & d \end{array}\right $$ \left $$\begin{array}{rr} 1 & 0 \\ 1 & -1 \end{array}\right $$ =\left $$\begin{array}{rr} a+b & -b \\ c+d & -d \end{array}\right $$\nonumber \\ and \\\left $$\begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right $$ A = \left $$\begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right $$ \left $$\begin{array}{rr} a & b \\ c & d \end{array}\right $$ =\left $$\begin{array}{cc} a+c & b+d \\ -c & -d \end{array}\right $$.\nonumber \\ If \\A\in U\\, then \\\left $$\begin{array}{cc} a+b & -b \\ c+d & -d \end{array}\right $$= \left $$\begin{array}{cc} a+c & b+d \\ -c & -d \end{array}\right $$\\.

Equating entries leads to a system of four equations in the four variables \\a,b,c\\ and \\d\\.

\\\begin{array}{ccc} a+b & = & a + c \\ -b & = & b + d \\ c + d & = & -c \\ -d & = & -d \end{array}\nonumber\\

or

\\\begin{array}{rcc} b - c & = & 0 \\ -2b - d & = & 0 \\ 2c + d & = & 0 \end{array}.\nonumber \\

The solution to this system is \\a=s\\, \\b=-\frac{1}{2}t\\, \\c=-\frac{1}{2}t\\, \\d=t\\ for any \\s,t\in\mathbb{R}\\, and thus \\A=\left $$\begin{array}{cc} s & \frac{t}{2} \\ -\frac{t}{2} & t \end{array}\right $$ = s\left $$\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right $$ + t\left $$\begin{array}{rr} 0 & -\frac{1}{2} \\ -\frac{1}{2} & 1 \end{array}\right $$ .\nonumber \\ Let \\B=\left\\ \left $$\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right $$, \left $$\begin{array}{rr} 0 & -\frac{1}{2} \\ -\frac{1}{2} & 1 \end{array}\right $$\right\\.\nonumber \\ Then \\\mathrm{span}(B)=U\\, and it is routine to verify that \\B\\ is an independent subset of \\\mathbb{M}\_{22}\\. Therefore \\B\\ is a basis of \\U\\, and \\\dim(U)=2\\.

The following theorem claims that a spanning set of a vector space \\V\\ can be shrunk down to a basis of \\V\\. Similarly, a linearly independent set within \\V\\ can be enlarged to create a basis of \\V\\.

Theorem \\\PageIndex{6}\\: Basis of \\V\\

If \\V=\mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u} \_{n}\right\\\\ is a vector space, then some subset of \\\\\vec{u}\_{1},\cdots ,\vec{u}\_{n}\\\\ is a basis for \\V.\\ Also, if \\\\\vec{u}\_{1},\cdots ,\vec{u} \_{k}\\\subseteq V\\ is linearly independent and the vector space is finite dimensional, then the set \\\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\\,\\ can be enlarged to obtain a basis of \\V.\\

Proof

Let \\S=\\E\subseteq \\\vec{u}\_{1},\cdots ,\vec{u}\_{n}\\\text{ such that } \mathrm{span}\left\\ E\right\\ =V\\.\nonumber \\ For \\E\in S,\\ let \\\left\vert E\right\vert\\ denote the number of elements of \\E.\\ Let \\m= \min \\\left\vert E\right\vert \text{ such that }E\in S\\.\nonumber \\ Thus there exist vectors \\\\\vec{v}\_{1},\cdots ,\vec{v}\_{m}\\\subseteq \\\vec{u}\_{1},\cdots , \vec{u}\_{n}\\\nonumber \\ such that \\\mathrm{span}\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{m}\right\\ =V\nonumber \\ and \\m\\ is as small as possible for this to happen. If this set is linearly independent, it follows it is a basis for \\V\\ and the theorem is proved. On the other hand, if the set is not linearly independent, then there exist scalars, \\c\_{1},\cdots ,c\_{m}\\ such that \\\vec{0}=\sum\_{i=1}^{m}c\_{i}\vec{v}\_{i}\nonumber \\ and not all the \\c\_{i}\\ are equal to zero. Suppose \\c\_{k}\neq 0.\\ Then solve for the vector \\\vec{v}\_{k}\\ in terms of the other vectors. Consequently, \\V=\mathrm{span}\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{k-1},\vec{v} \_{k+1},\cdots ,\vec{v}\_{m}\right\\\nonumber \\ contradicting the definition of \\m\\. This proves the first part of the theorem.

To obtain the second part, begin with \\\\\vec{u}\_{1},\cdots ,\vec{u} \_{k}\\\\ and suppose a basis for \\V\\ is \\\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\\nonumber \\ If \\\mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\ =V,\nonumber \\ then \\k=n\\. If not, there exists a vector \\\vec{u}\_{k+1}\notin \mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u} \_{k}\right\\\nonumber \\ Then from Lemma 9.3.2, \\\\\vec{u}\_{1},\cdots ,\vec{u}\_{k}, \vec{u}\_{k+1}\\\\ is also linearly independent. Continue adding vectors in this way until \\n\\ linearly independent vectors have been obtained. Then \\\mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\\ =V\nonumber \\ because if it did not do so, there would exist \\\vec{u}\_{n+1}\\ as just described and \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{n+1}\right\\\\ would be a linearly independent set of vectors having \\n+1\\ elements. This contradicts the fact that \\\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\\\ is a basis. In turn this would contradict Theorem $\PageIndex{3}$. Therefore, this list is a basis.

Recall Example 9.3.4 in which we added a matrix to a linearly independent set to create a larger linearly independent set. By Theorem $\PageIndex{6}$ we can extend a linearly independent set to a basis.

Example \\\PageIndex{8}\\: Adding to a Linearly Independent Set

Let \\S \subseteq M\_{22}\\ be a linearly independent set given by \\S = \left\\ \left $$ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right $$, \left $$ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right $$ \right\\\nonumber \\ Enlarge \\S\\ to a basis of \\M\_{22}\\.

###### Solution

Recall from the solution of Example 9.3.4 that the set \\R \subseteq M\_{22}\\ given by \\R = \left\\ \left $$ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right $$, \left $$ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right $$, \left $$ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right $$ \right\\\nonumber \\ is also linearly independent. However this set is still not a basis for \\M\_{22}\\ as it is not a spanning set. In particular, \\\left $$ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array} \right $$\\ is not in \\\mathrm{span} R\\. Therefore, this matrix can be added to the set by Lemma 9.3.2 to obtain a new linearly independent set given by \\T = \left\\ \left $$ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right $$, \left $$ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right $$, \left $$ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right $$, \left $$ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array} \right $$ \right\\\nonumber \\

This set is linearly independent and now spans \\M\_{22}\\. Hence \\T\\ is a basis.

Next we consider the case where you have a spanning set and you want a subset which is a basis. The above discussion involved adding vectors to a set. The next theorem involves removing vectors.

Theorem \\\PageIndex{7}\\: Basis from a Spanning Set

Let \\V\\ be a vector space and let \\W\\ be a subspace. Also suppose that \\W=\mathrm{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 \\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. In any linear combination involving \\\vec{w}\_{j}\\, the linear combination 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\\\\.

Consider the following example of this concept.

Example \\\PageIndex{9}\\: Basis from a Spanning Set

Let \\V\\ be the vector space of polynomials of degree no more than 3, denoted earlier as \\\mathbb{P}\_{3}\\. Consider the following vectors in \\V\\. \\\begin{aligned} &2x^{2}+x+1,x^{3}+4x^{2}+2x+2,2x^{3}+2x^{2}+2x+1, \\ &x^{3}+4x^{2}-3x+2,x^{3}+3x^{2}+2x+1\end{aligned}\\ Then, as mentioned above, \\V\\ has dimension 4 and so clearly these vectors are not linearly independent. A basis for \\V\\ is \\\left\\ 1,x,x^{2},x^{3}\right\\\\. Determine a linearly independent subset of these which has the same span. Determine whether this subset is a basis for \\V\\.

###### Solution

Consider an isomorphism which maps \\\mathbb{R}^{4}\\ to \\V\\ in the obvious way. Thus \\\left $$ \begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array} \right $$\nonumber \\ corresponds to \\2x^{2}+x+1\\ through the use of this isomorphism. Then corresponding to the above vectors in \\V\\ we would have the following vectors in \\\mathbb{R}^{4}.\\ \\\left $$ \begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array} \right $$ ,\left $$ \begin{array}{c} 2 \\ 2 \\ 4 \\ 1 \end{array} \right $$ ,\left $$ \begin{array}{c} 1 \\ 2 \\ 2 \\ 2 \end{array} \right $$ ,\left $$ \begin{array}{r} 2 \\ -3 \\ 4 \\ 1 \end{array} \right $$ ,\left $$ \begin{array}{c} 1 \\ 2 \\ 3 \\ 1 \end{array} \right $$\nonumber \\ Now if we obtain a subset of these which has the same span but which is linearly independent, then the corresponding vectors from \\V\\ will also be linearly independent. If there are four in the list, then the resulting vectors from \\V\\ must be a basis for \\V\\. The reduced row-echelon form for the matrix which has the above vectors as columns is \\\left $$ \begin{array}{rrrrr} 1 & 0 & 0 & -15 & 0 \\ 0 & 1 & 0 & 11 & 0 \\ 0 & 0 & 1 & -5 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right $$\nonumber \\ Therefore, a basis for \\V\\ consists of the vectors \\\begin{aligned} &2x^{2}+x+1,x^{3}+4x^{2}+2x+2,2x^{3}+2x^{2}+2x+1, \\ &x^{3}+3x^{2}+2x+1.\end{aligned}\\ Note how this is a subset of the original set of vectors. If there had been only three pivot columns in this matrix, then we would not have had a basis for \\V\\ but we would at least have obtained a linearly independent subset of the original set of vectors in this way.

Note also that, since all linear relations are preserved by an isomorphism, \\\begin{aligned} &-15\left( 2x^{2}+x+1\right) +11\left( x^{3}+4x^{2}+2x+2\right) +\left( -5\right) \left( 2x^{3}+2x^{2}+2x+1\right) \\ =&x^{3}+4x^{2}-3x+2\end{aligned}\\

Consider the following example.

Example \\\PageIndex{10}\\: Shrinking a Spanning Set

Consider the set \\S \subseteq \mathbb{P}\_2\\ given by \\S = \left\\ 1, x, x^2, x^2 + 1 \right\\\nonumber \\ Show that \\S\\ spans \\\mathbb{P}\_2\\, then remove vectors from \\S\\ until it creates a basis.

###### Solution

First we need to show that \\S\\ spans \\\mathbb{P}\_2\\. Let \\ax^2 + bx + c\\ be an arbitrary polynomial in \\\mathbb{P}\_2\\. Write \\ax^2 + bx + c = r(1) + s(x) + t(x^2) + u (x^2 + 1)\nonumber \\ Then, \\\begin{aligned} ax^2 +bx + c &= r(1) + s(x) + t(x^2) + u (x^2 + 1) \\ &= (t+u) x^2 + s(x) + (r+u) \end{aligned}\\

It follows that \\\begin{aligned} a &= t + u \\ b &= s \\ c &= r + u \end{aligned}\\

Clearly a solution exists for all \\a,b,c\\ and so \\S\\ is a spanning set for \\\mathbb{P}\_2\\. By Theorem $\PageIndex{6}$, some subset of \\S\\ is a basis for \\\mathbb{P}\_2\\.

Recall that a basis must be both a spanning set and a linearly independent set. Therefore we must remove a vector from \\S\\ keeping this in mind. Suppose we remove \\x\\ from \\S\\. The resulting set would be \\\left\\ 1, x^2, x^2 + 1 \right\\\\. This set is clearly linearly dependent (and also does not span \\\mathbb{P}\_2\$ and so is not a basis.

Suppose we remove \\x^2 + 1\\ from \\S\\. The resulting set is \\\left\\ 1, x, x^2 \right\\\\ which is both linearly independent and spans \\\mathbb{P}\_2\\. Hence this is a basis for \\\mathbb{P}\_2\\. Note that removing any one of \\1, x^2\\, or \\x^2 + 1\\ will result in a basis.

Now the following is a fundamental result about subspaces.

Theorem \\\PageIndex{8}\\: Basis of a Vector Space

Let \\V\\ be a finite dimensional vector space and let \\W\\ be a non-zero subspace. Then \\W\\ has a basis. That is, there exists a linearly independent set of vectors \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{r}\right\\\\ such that \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{r}\right\\ =W\nonumber \\ Also if \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{s}\right\\\\ is a linearly independent set of vectors, then \\W\\ has a basis of the form \\\left\\ \vec{w} \_{1},\cdots ,\vec{w}\_{s},\cdots ,\vec{w}\_{r}\right\\\\ for \\r\geq s\\.

Proof

Let the dimension of \\V\\ be \\n\\. Pick \\\vec{w}\_{1}\in W\\ where \\\vec{w}\_{1}\neq \vec{0}.\\ If \\\vec{w}\_{1},\cdots ,\vec{w}\_{s}\\ have been chosen such that \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{s}\right\\\\ is linearly independent, if \\span\left\\ \vec{w}\_{1},\cdots ,\vec{w} \_{r}\right\\ =W,\\ stop. You have the desired basis. Otherwise, there exists \\\vec{w}\_{s+1}\notin span\left\\ \vec{w}\_{1},\cdots ,\vec{w} \_{s}\right\\\\ and \\\left\\ \vec{w}\_{1},\cdots , \vec{w}\_{s},\vec{w}\_{s+1}\right\\\\ is linearly independent. Continue this way until the process stops. It must stop since otherwise, you could obtain a linearly independent set of vectors having more than \\n\\ vectors which is impossible.

The last claim is proved by following the above procedure starting with \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{s}\right\\\\ as above.

This also proves the following corollary. Let \\V\\ play the role of \\W\\ in the above theorem and begin with a basis for \\W\\, enlarging it to form a basis for \\V\\ as discussed above.

Corollary \\\PageIndex{3}\\: Basis Extension

Let \\W\\ be any non-zero subspace of a vector space \\V\\. Then every basis of \\W\\ can be extended to a basis for \\V\\.

Consider the following example.

Example \\\PageIndex{11}\\: Basis Extension

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 \\V\\.

###### 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{vectorspaceeq1}\\ Note how the given vectors were placed as the first two 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{vectorspaceeq2}\\ These 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{vectorspaceeq1}\\ obviously span \\\mathbb{R} ^{4}\\ the span of the first four is the same as the span of all six.

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9_05_3A_Sums_and_Intersections

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> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/09%3A_Vector_Spaces/9.05%3A_Sums_and_Intersections

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Outcomes

1. Show that the sum of two subspaces is a subspace.

2. Show that the intersection of two subspaces is a subspace.

We begin this section with a definition.

Definition \\\PageIndex{1}\\: Sum and Intersection

Let \\V\\ be a vector space, and let \\U\\ and \\W\\ be subspaces of \\V\\. Then

1. \\U+W = \\ \vec{u}+\vec{w} ~\|~ \vec{u}\in U\mbox{ and } \vec{w}\in W\\\\ and is called the sum of \\U\\ and \\W\\.

2. \\U\cap W = \\ \vec{v} ~\|~ \vec{v}\in U\mbox{ and } \vec{v}\in W\\\\ and is called the intersection of \\U\\ and \\W\\.

Therefore the intersection of two subspaces is all the vectors shared by both. If there are no vectors shared by both subspaces, meaning that \\U \cap W = \left\\ \vec{0} \right\\\\, the sum \\U+W\\ takes on a special name.

Definition \\\PageIndex{2}\\: Direct Sum

Let \\V\\ be a vector space and suppose \\U\\ and \\W\\ are subspaces of \\V\\ such that \\U \cap W = \left\\ \vec{0} \right\\\\. Then the sum of \\U\\ and \\W\\ is called the direct sum and is denoted \\U \oplus W\\.

An interesting result is that both the sum \\U + W\\ and the intersection \\U \cap W\\ are subspaces of \\V\\.

Example \\\PageIndex{1}\\: Intersection is a Subspace

Let \\V\\ be a vector space and suppose \\U\\ and \\W\\ are subspaces. Then the intersection \\U \cap W\\ is a subspace of \\V\\.

###### Solution

By the subspace test, we must show three things:

1. \\\vec{0} \in U \cap W\\

2. For vectors \\\vec{v}\_1, \vec{v}\_2 \in U \cap W, \vec{v}\_1+\vec{v}\_2 \in U \cap W\\

3. For scalar \\a\\ and vector \\\vec{v} \in U \cap W, a\vec{v} \in U \cap W\\

We proceed to show each of these three conditions hold.

1. Since \\U\\ and \\W\\ are subspaces of \\V\\, they each contain \\\vec{0}\\. By definition of the intersection, \\\vec{0} \in U \cap W\\.

2. Let \\\vec{v}\_1, \vec{v}\_2 \in U \cap W,\\. Then in particular, \\\vec{v}\_1, \vec{v}\_2 \in U\\. Since \\U\\ is a subspace, it follows that \\\vec{v}\_1+\vec{v}\_2 \in U\\. The same argument holds for \\W\\. Therefore \\\vec{v}\_1+\vec{v}\_2\\ is in both \\U\\ and \\W\\ and by definition is also in \\U \cap W\\.

3. Let \\a\\ be a scalar and \\\vec{v} \in U \cap W\\. Then in particular, \\\vec{v} \in U\\. Since \\U\\ is a subspace, it follows that \\a \vec{v} \in U\\. The same argument holds for \\W\\ so \\a\vec{v}\\ is in both \\U\\ and \\W\\. By definition, it is in \\U \cap W\\.

Therefore \\U \cap W\\ is a subspace of \\V\\.

It can also be shown that \\U + W\\ is a subspace of \\V\\.

We conclude this section with an important theorem on dimension.

Theorem \\\PageIndex{1}\\: Dimension of Sum

Let \\V\\ be a vector space with subspaces \\U\\ and \\W\\. Suppose \\U\\ and \\W\\ each have finite dimension. Then \\U + W\\ also has finite dimension which is given by\\\mathrm{dim} (U+W) = \mathrm{dim}(U) + \mathrm{dim}(W) - \mathrm{dim} (U \cap W)\nonumber \\

Notice that when \\U \cap W = \left\\ \vec{0} \right\\\\, the sum becomes the direct sum and the above equation becomes \\\mathrm{dim} (U \oplus W) = \mathrm{dim}(U) + \mathrm{dim}(W)\nonumber \\

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9_06_3A_Linear_Transformations

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> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/09%3A_Vector_Spaces/9.06%3A_Linear_Transformations

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Outcomes

1. Understand the definition of a linear transformation in the context of vector spaces.

Recall that a function is simply a transformation of a vector to result in a new vector. Consider the following definition.

Definition \\\PageIndex{1}\\: Linear Transformation

Let \\V\\ and \\W\\ be vector spaces. Suppose \\T: V \mapsto W\\ is a function, where for each \\\vec{x} \in V ,T\left(\vec{x}\right)\in W.\\ Then \\T\\ is a linear transformation if whenever \\k ,p\\ are scalars and \\\vec{v}\_1\\ and \\\vec{v}\_2\\ are vectors in \\V\\ \\T\left( k \vec{v}\_1 + p \vec{v}\_2 \right) = kT\left(\vec{v}\_1\right)+ pT\left(\vec{v}\_{2} \right)\nonumber \\

Several important examples of linear transformations include the zero transformation, the identity transformation, and the scalar transformation.

Example \\\PageIndex{1}\\: Linear Transformations

Let \\V\\ and \\W\\ be vector spaces.

1. The zero transformation

\\0:V\to W\\ is defined by \\0(\vec{v})=\vec{0}\\ for all \\\vec{v}\in V\\.

2. The identity transformation

\\1_V:V\to V\\ is defined by \\1_V(\vec{v})=\vec{v}\\ for all \\\vec{v}\in V\\.

3. The scalar transformation Let \\a\in\mathbb{R}\\.

\\s_a:V\to V\\ is defined by \\s_a(\vec{v})=a\vec{v}\text{ for all }\vec{v}\in V\\.

###### Solution

We will show that the scalar transformation \\s_a\\ is linear, the rest are left as an exercise.

By Definition $\PageIndex{1}$ we must show that for all scalars \\k ,p\\ and vectors \\\vec{v}\_1\\ and \\\vec{v}\_2\\ in \\V\\, \\s_a\left( k \vec{v}\_1 + p \vec{v}\_2 \right) = k s_a\left(\vec{v}\_1\right)+ p s_a\left(\vec{v}\_{2} \right)\\. Assume that \\a\\ is also a scalar. \\\begin{aligned} s_a\left( k \vec{v}\_1 + p \vec{v}\_2 \right) &= a \left( k \vec{v}\_1 + p \vec{v}\_2 \right) \\ &= ak \vec{v}\_1 + ap \vec{v}\_2 \\ &= k \left(a \vec{v}\_1\right) + p\left(a \vec{v}\_2\right) \\ &= k s_a\left( \vec{v}\_1 \right) + p s_a \left(\vec{v}\_2 \right)\end{aligned}\\ Therefore \\s_a\\ is a linear transformation.

Consider the following important theorem.

Theorem \\\PageIndex{1}\\: Properties of Linear Transformations

Let \\V\\ and \\W\\ be vector spaces, and \\T:V \mapsto W\\ a linear transformation. Then

1. \\T\\ preserves the zero vector. \\T(\vec{0})=\vec{0}\nonumber \\

2. \\T\\ preserves additive inverses. For all \\\vec{v}\in V\\, \\T(-\vec{v})= -T(\vec{v})\nonumber \\

3. \\T\\ preserves linear combinations. For all \\\vec{v}\_1, \vec{v}\_2, \ldots, \vec{v}\_m \in V\\ and all \\k_1, k_2, \ldots, k_m\in\mathbb{R}\\, \\T(k_1\vec{v}\_1 + k_2\vec{v}\_2 + \cdots + k_m\vec{v}\_m) = k_1T(\vec{v}\_1) + k_2T(\vec{v}\_2) + \cdots + k_mT(\vec{v}\_m).\nonumber \\

Proof

1. Let \\\vec{0}\_V\\ denote the zero vector of \\V\\ and let \\\vec{0}\_W\\ denote the zero vector of \\W\\. We want to prove that \\T(\vec{0}\_V)=\vec{0}\_W\\. Let \\\vec{v}\in V\\. Then \\0\vec{v}=\vec{0}\_V\\ and \\T(\vec{0}\_V)=T(0\vec{v})=0T(\vec{v})=\vec{0}\_W.\nonumber \\

2. Let \\\vec{v}\in V\\; then \\-\vec{v}\in V\\ is the additive inverse of \\\vec{v}\\, so \\\vec{v} + (-\vec{v})=\vec{0}\_V\\. Thus \\\begin{aligned} T(\vec{v} + (-\vec{v})) & = T(\vec{0}\_V) \\ T(\vec{v}) + T(-\vec{v})) & = \vec{0}\_W \\ T(-\vec{v}) & = \vec{0}\_W - T(\vec{v}) = - T(\vec{v}).\end{aligned}\\

3. This result follows from preservation of addition and preservation of scalar multiplication. A formal proof would be by induction on \\m\\.

Consider the following example using the above theorem.

Example \\\PageIndex{2}\\: Linear Combination

Let \\T:\mathbb{P}\_2 \to \mathbb{R}\\ be a linear transformation such that \\T(x^2+x)=-1; T(x^2-x)=1; T(x^2+1)=3.\nonumber \\ Find \\T(4x^2+5x-3)\\.

We provide two solutions to this problem.

Solution 1:

Suppose \\a(x^2+x) + b(x^2-x) + c(x^2+1) = 4x^2+5x-3\\. Then \$a+b+c)x^2 + (a-b)x + c = 4x^2+5x-3.\nonumber \\ Solving for \\a\\, \\b\\, and \\c\\ results in the unique solution \\a=6\\, \\b=1\\, \\c=-3\\. Thus \\\begin{aligned}T(4x^2+5x-3)&=T(6(x^2+x)+(x^2-x)-3(x^2+1)) \\ &=6T(x^2+x)+T(x^2-x)-3T(x^2+1) \\ &=6(-1)+1-3(3)=-14.\end{aligned}\\

Solution 2:

Notice that \\S=\\ x^2+x, x^2-x, x^2+1\\\\ is a basis of \\\mathbb{ P}\_2\\, and thus \\x^2\\, \\x\\, and \\1\\ can each be written as a linear combination of elements of \\S\\.

\\\begin{aligned} x^2 & = \textstyle \frac{1}{2}(x^2+x) + \frac{1}{2}(x^2-x) \\ x & = \textstyle \frac{1}{2}(x^2+x) - \frac{1}{2}(x^2-x) \\ 1 & = (x^2+1)-\textstyle \frac{1}{2}(x^2+x) - \frac{1}{2}(x^2-x).\end{aligned}\\ Then \\\begin{aligned} T(x^2) & = \textstyle T\left(\frac{1}{2}(x^2+x) + \frac{1}{2}(x^2-x)\right) =\frac{1}{2}T(x^2+x) + \frac{1}{2}T(x^2-x)\\ & = \textstyle \frac{1}{2}(-1) + \frac{1}{2}(1) = 0. \\ T(x) & = \textstyle T\left(\frac{1}{2}(x^2+x) - \frac{1}{2}(x^2-x)\right) = \frac{1}{2}T(x^2+x) - \frac{1}{2}T(x^2-x) \\ & = \textstyle \frac{1}{2}(-1) - \frac{1}{2}(1) = -1.\\ T(1) & = \textstyle T\left((x^2+1)-\frac{1}{2}(x^2+x) - \frac{1}{2}(x^2-x)\right)\\ & = \textstyle T(x^2+1)-\frac{1}{2}T(x^2+x) - \frac{1}{2}T(x^2-x) \\ & = \textstyle 3-\frac{1}{2}(-1) - \frac{1}{2}(1) = 3.\end{aligned}\\

Therefore, \\\begin{aligned} T(4x^2+5x-3) & = 4T(x^2) + 5T(x) -3T(1) \\ & = 4(0) + 5(-1) - 3(3)=-14.\end{aligned}\\ The advantage of Solution 2 over Solution 1 is that if you were now asked to find \\T(-6x^2-13x+9)\\, it is easy to use \\T(x^2)=0\\, \\T(x)=-1\\ and \\T(1)= 3\\: \\\begin{aligned} T(-6x^2-13x+9) & = -6T(x^2)-13T(x)+9T(1) \\ & = -6(0)-13(-1)+9(3)=13+27=40.\end{aligned}\\ More generally, \\\begin{aligned} T(ax^2+bx+c) & = aT(x^2)+bT(x)+cT(1) \\ & = a(0)+b(-1)+c(3)=-b+3c.\end{aligned}\\

Suppose two linear transformations act in the same way on \\\vec{v}\\ for all vectors. Then we say that these transformations are equal.

Definition \\\PageIndex{2}\\: Equal Transformations

Let \\S\\ and \\T\\ be linear transformations from \\V\\ to \\W\\. Then \\S = T\\ if and only if for every \\\vec{v} \in V\\, \\S \left( \vec{v} \right) = T \left( \vec{v} \right)\nonumber \\

The definition above requires that two transformations have the same action on every vector in order for them to be equal. The next theorem argues that it is only necessary to check the action of the transformations on basis vectors.

Theorem \\\PageIndex{2}\\: Transformation of a Spanning Set

Let \\V\\ and \\W\\ be vector spaces and suppose that \\S\\ and \\T\\ are linear transformations from \\V\\ to \\W\\. Then in order for \\S\\ and \\T\\ to be equal, it suffices that \\S(\vec{v}\_i) = T(\vec{v}\_i)\\ where \\V = span \\ \vec{v}\_1, \vec{v}\_2, \ldots, \vec{v}\_n\\.\\

This theorem tells us that a linear transformation is completely determined by its actions on a spanning set. We can also examine the effect of a linear transformation on a basis.

Theorem \\\PageIndex{3}\\: Transformation of a Basis

Suppose \\V\\ and \\W\\ are vector spaces and let \\\\ \vec{w}\_1, \vec{w}\_2, \ldots, \vec{w}\_n\\\\ be any given vectors in \\W\\ that may not be distinct. Then there exists a basis \\\\ \vec{v}\_1, \vec{v}\_2, \ldots, \vec{v}\_n\\\\ of \\V\\ and a unique linear transformation \\T: V \mapsto W\\ with \\T (\vec{v}\_i) = \vec{w}\_i\\.

Furthermore, if \\\vec{v} = k_1\vec{v}\_1+k_2\vec{v}\_2+ \cdots+ k_n\vec{v}\_n\nonumber \\ is a vector of \\V\\, then \\T(\vec{v}) = k_1\vec{w}\_1+k_2\vec{w}\_2+ \cdots+ k_n\vec{w}\_n.\nonumber \\

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9_07_3A_Isomorphisms

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##### Outcomes

1. Apply the concepts of one to one and onto to transformations of vector spaces.

2. Determine if a linear transformation of vector spaces is an isomorphism.

3. Determine if two vector spaces are isomorphic.

One to One and Onto Transformations

Recall the following definitions, given here in terms of vector spaces.

##### Definition \\\PageIndex{1}\\: One to One Transformation

Let \\V, W\\ be vector spaces with \\\vec{v}\_1, \vec{v}\_2\\ vectors in \\V\\. Then a linear transformation \\T: V \mapsto W\\ is called one to one if whenever \\\vec{v}\_1 \neq \vec{v}\_2\\ it follows that \\T(\vec{v}\_1) \neq T (\vec{v}\_2)\nonumber \\

##### Definition \\\PageIndex{2}\\: Onto Transformation

Let \\V, W\\ be vector spaces. Then a linear transformation \\T: V \mapsto W\\ is called onto if for all \\\vec{w} \in \vec{W}\\ there exists \\\vec{v} \in V\\ such that \\T(\vec{v}) = \vec{w}\\.

Recall that every linear transformation \\T\\ has the property that \\T(\vec{0})=\vec{0}\\. This will be necessary to prove the following useful lemma.

##### Lemma \\\PageIndex{1}\\: One to One

The assertion that a linear transformation \\T\\ is one to one is equivalent to saying that if \\T(\vec{v})=\vec{0},\\ then \\\vec{v}=0.\\

Proof

Suppose first that \\T\\ is one to one.

\\T(\vec{0})=T\left( \vec{0}+\vec{0}\right) =T(\vec{0})+T(\vec{0})\nonumber \\ and so, adding the additive inverse of \\T(\vec{0})\\ to both sides, one sees that \\T(\vec{0})=\vec{0}\\. Therefore, if \\T(\vec{v})=\vec{0},\\ it must be the case that \\\vec{v}=\vec{0}\\ because it was just shown that \\T(\vec{0})=\vec{0}\\.

Now suppose that if \\T(\vec{v})=\vec{0},\\ then \\\vec{v}=0.\\ If \\T(\vec{v})=T(\vec{u}),\\ then \\T(\vec{v})-T(\vec{u})=T\left( \vec{v}-\vec{u}\right) =\vec{0}\\ which shows that \\\vec{v}-\vec{u}=0\\ or in other words, \\\vec{v}=\vec{u}\\.

Consider the following example.

##### Example \\\PageIndex{1}\\: One to One Transformation

Let \\S:\mathbb{P}\_2\to\mathbb{M}\_{22}\\ be a linear transformation defined by \\S(ax^2+bx+c) = \left $$\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right $$ \nonumber\\ for all \\ax^2+bx+c\in \mathbb{P}\_2.\\

Prove that \\S\\ is one to one but not onto.

###### Solution

By definition, \\\ker(S)=\\ax^2+bx+c\in \mathbb{P}\_2 ~\|~ a+b=0, a+c=0, b-c=0, b+c=0\\. \nonumber\\

Suppose \\p(x)=ax^2+bx+c\in\ker(S)\\. This leads to a homogeneous system of four equations in three variables. Putting the augmented matrix in reduced row-echelon form:

\\\left $$\begin{array}{rrr\|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right $$ \rightarrow \cdots \rightarrow \left $$\begin{array}{ccc\|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right $$.\nonumber\\

The solution is \\a=b=c=0\\. This tells us that if \\S(p(x)) = 0\\, then \\p(x) = ax^2+bx+c = 0x^2 + 0x + 0 = 0\\. Therefore it is one to one.

To show that \\S\\ is not onto, find a matrix \\A\in\mathbb{M}\_{22}\\ such that for every \\p(x)\in \mathbb{P}\_2\\, \\S(p(x))\neq A\\. Let \\A=\left $$\begin{array}{cc} 0 & 1 \\ 0 & 2 \end{array}\right $$, \nonumber\\ and suppose \\p(x)=ax^2+bx+c\in \mathbb{P}\_2\\ is such that \\S(p(x))=A\\. Then \\\begin{array}{ll} a+b=0 & a+c=1 \\ b-c=0 & b+c=2 \end{array}\nonumber \\ Solving this system \\\left $$\begin{array}{ccc\|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 2 \end{array}\right $$ \rightarrow \left $$\begin{array}{rrr\|r} 1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 2 \end{array}\right $$. \nonumber\\

Since the system is inconsistent, there is no \\p(x)\in \mathbb{P}\_2\\ so that \\S(p(x))=A\\, and therefore \\S\\ is not onto.

##### Example \\\PageIndex{2}\\: An Onto Transformation

Let \\T:\mathbb{M}\_{22}\to\mathbb{R}^2\\ be a linear transformation defined by \\T\left $$\begin{array}{cc} a & b \\ c & d \end{array}\right $$ = \left $$\begin{array}{c} a+d \\ b+c \end{array}\right $$ \mbox{ for all } \left $$\begin{array}{cc} a & b \\ c & d \end{array}\right $$ \in\mathbb{M}\_{22}.\nonumber\\

Prove that \\T\\ is onto but not one to one.

###### Solution

Let \\\left $$\begin{array}{c} x \\ y \end{array}\right $$\\ be an arbitrary vector in \\\mathbb{R}^2\\. Since \\T\left $$\begin{array}{cc} x & y \\ 0 & 0 \end{array}\right $$ =\left $$\begin{array}{c} x \\ y \end{array}\right $$\\, \\T\\ is onto.

By Lemma $\PageIndex{1}$ \\T\\ is one to one if and only if \\T(A) = \vec{0}\\ implies that \\A = 0\\ the zero matrix. Observe that \\T \left( \left $$\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right $$ \right) = \left $$ \begin{array}{c} 1 + -1 \\ 0 + 0 \end{array} \right $$ = \left $$ \begin{array}{c} 0 \\ 0 \end{array} \right $$\nonumber\\

There exists a nonzero matrix \\A\\ such that \\T(A) = \vec{0}\\. It follows that \\T\\ is not one to one.

The following example demonstrates that a one to one transformation preserves linear independence.

##### Example \\\PageIndex{3}\\: One to One and Independence

Let \\V\\ and \\W\\ be vector spaces and \\T: V \mapsto W\\ a linear transformation. Prove that if \\T\\ is one to one and \\\\\vec{v}\_1, \vec{v}\_2, \ldots, \vec{v}\_k\\\\ is an independent subset of \\V\\, then \\\\T(\vec{v}\_1), T(\vec{v}\_2), \ldots, T(\vec{v}\_k)\\\\ is an independent subset of \\W\\.

###### Solution

Let \\\vec{0}\_V\\ and \\\vec{0}\_W\\ denote the zero vectors of \\V\\ and \\W\\, respectively. Suppose that \\a_1T(\vec{v}\_1) + a_2T(\vec{v}\_2) +\cdots +a_kT(\vec{v}\_k) =\vec{0}\_W \nonumber\\ for some \\a_1, a_2, \ldots, a_k\in\mathbb{R}\\. Since linear transformations preserve linear combinations (addition and scalar multiplication), \\T(a_1\vec{v}\_1 + a_2\vec{v}\_2 +\cdots +a_k\vec{v}\_k) =\vec{0}\_W. \nonumber\\

Now, since \\T\\ is one to one, \\\ker(T)=\\\vec{0}\_V\\\\, and thus \\a_1\vec{v}\_1 + a_2\vec{v}\_2 +\cdots +a_k\vec{v}\_k =\vec{0}\_V. \nonumber\\

However, \\\\\vec{v}\_1, \vec{v}\_2, \ldots, \vec{v}\_k\\\\ is independent so \\a_1=a_2=\cdots=a_k=0\\. Therefore, \\\\T(\vec{v}\_1), T(\vec{v}\_2), \ldots, T(\vec{v}\_k)\\\\ is independent.

A similar claim can be made regarding onto transformations. In this case, an onto transformation preserves a spanning set.

##### Example \\\PageIndex{4}\\: Onto and Spanning

Let \\V\\ and \\W\\ be vector spaces and \\T:V\to W\\ a linear transformation. Prove that if \\T\\ is onto and \\V=span\\\vec{v}\_1, \vec{v}\_2, \ldots, \vec{v}\_k\\\\, then \\W=span\\T(\vec{v}\_1), T(\vec{v}\_2), \ldots, T(\vec{v}\_k)\\.\nonumber\\

###### Solution

Suppose that \\T\\ is onto and let \\\vec{w}\in W\\. Then there exists \\\vec{v}\in V\\ such that \\T(\vec{v})=\vec{w}\\. Since \\V=span\\\vec{v}\_1, \vec{v}\_2, \ldots, \vec{v}\_k\\\\, there exist \\a_1, a_2, \ldots a_k\in\mathbb{R}\\ such that \\\vec{v} = a_1\vec{v}\_1 + a_2\vec{v}\_2 + \cdots + a_k\vec{v}\_k\\. Using the fact that \\T\\ is a linear transformation, \\\begin{aligned} \vec{w} =T(\vec{v}) & = T(a_1\vec{v}\_1 + a_2\vec{v}\_2 + \cdots + a_k\vec{v}\_k) \\ & = a_1T(\vec{v}\_1) + a_2T(\vec{v}\_2) + \cdots + a_kT(\vec{v}\_k),\end{aligned}\\ i.e., \\\vec{w}\in span\\T(\vec{v}\_1), T(\vec{v}\_2), \ldots, T(\vec{v}\_k)\\\\, and thus \\W\subseteq span\\T(\vec{v}\_1), T(\vec{v}\_2), \ldots, T(\vec{v}\_k)\\. \nonumber\\

Since \\T(\vec{v}\_1), T(\vec{v}\_2), \ldots, T(\vec{v}\_k)\in W\\, it follows from that \\span\\T(\vec{v}\_1), T(\vec{v}\_2), \ldots, T(\vec{v}\_k)\\\subseteq W\\, and therefore \\W=span\\T(\vec{v}\_1), T(\vec{v}\_2), \ldots, T(\vec{v}\_k)\\\\.

Isomorphisms

The focus of this section is on linear transformations which are both one to one and onto. When this is the case, we call the transformation an isomorphism.

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

Let \\V\\ and \\W\\ be two vector spaces and let \\T: V \mapsto W\\ be a linear transformation. Then \\T\\ is called an isomorphism if the following two conditions are satisfied.

##### Definition \\\PageIndex{4}\\: Isomorphic

Let \\V\\ and \\W\\ be two vector spaces and let \\T: V \mapsto W\\ be a linear transformation. Then if \\T\\ is an isomorphism, we say that \\V\\ and \\W\\ are isomorphic.

Consider the following example of an isomorphism.

##### Example \\\PageIndex{5}\\: Isomorphism

Let \\T:\mathbb{M}\_{22}\to\mathbb{R}^4\\ be defined by \\T \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) = \left$$ \begin{array}{c} a\\ b\\ c \\ d \end{array} \right$$ \mbox{ for all } \left$$ \begin{array}{cc} a & b \\ c & d \end{array} \right$$ \in\mathbb{M}\_{22}.\nonumber \\ Show that \\T\\ is an isomorphism.

###### Solution

Notice that if we can prove \\T\\ is an isomorphism, it will mean that \\\mathbb{M}\_{22}\\ and \\\mathbb{R}^4\\ are isomorphic. It remains to prove that

1. \\T\\ is a linear transformation;

2. \\T\\ is one-to-one;

3. \\T\\ is onto.

\\T\\ is linear: Let \\k,p\\ be scalars.

\\\begin{aligned} T \left( k \left$$\begin{array}{cc} a_1 & b_1 \\ c_1 & d_1 \end{array}\right$$ + p \left$$\begin{array}{cc} a_2 & b_2 \\ c_2 & d_2 \end{array}\right$$ \right) &= T \left( \left$$\begin{array}{cc} k a_1 & k b_1 \\ k c_1 & k d_1 \end{array}\right$$ + \left$$\begin{array}{cc} p a_2 & p b_2 \\ p c_2 & p d_2 \end{array}\right$$ \right) \\ &= T \left( \left$$\begin{array}{cc} k a_1 + p a_2 & k b_1 + p b_2 \\ k c_1 + p c_2& k d_1 + p d_2 \end{array}\right$$ \right) \\ &= \left$$ \begin{array}{c} k a_1 + p a_2 \\ k b_1 + p b_2 \\ k c_1 + p c_2 \\ k d_1 + p d_2 \end{array}\right$$ \\ &= \left$$ \begin{array}{c} k a_1 \\ k b_1 \\ k c_1 \\ k d_1 \end{array} \right$$ + \left$$ \begin{array}{c} p a_2 \\ p b_2 \\ p c_2 \\ p d_2 \end{array} \right$$ \\ &= k \left$$ \begin{array}{c} a_1 \\ b_1 \\ c_1 \\ d_1 \end{array} \right$$ + p \left$$ \begin{array}{c} a_2 \\ b_2 \\ c_2 \\ d_2 \end{array} \right$$ \\ &= k T \left(\left$$\begin{array}{cc} a_1 & b_1 \\ c_1 & d_1 \end{array}\right$$ \right) + p T \left(\left$$\begin{array}{cc} a_2 & b_2 \\ c_2 & d_2 \end{array}\right$$ \right)\end{aligned}\\

Therefore \\T\\ is linear.

\\T\\ is one-to-one: By Lemma $\PageIndex{1}$ we need to show that if \\T(A) = 0\\ then \\A = 0\\ for some matrix \\A \in \mathbb{M}\_{22}\\. \\T\left$$\begin{array}{cc} a & b \\ c & d \end{array}\right$$ = \left$$\begin{array}{c} a\\ b\\ c \\ d \end{array}\right$$ = \left$$\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array}\right$$\nonumber \\

This clearly only occurs when \\a=b=c=d=0\\ which means that \\A = \left$$\begin{array}{cc} a & b \\ c & d \end{array}\right$$ = \left$$\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right$$ = 0\nonumber \\

Hence \\T\\ is one-to-one.

\\T\\ is onto: Let

\\\vec{x}=\left$$\begin{array}{c} x_1\\x_2\\x_3\\x_4 \end{array}\right$$\in\mathbb{R}^4, \nonumber\\ and define matrix \\A\in\mathbb{M}\_{22}\\ as follows: \\A=\left$$\begin{array}{cc} x_1 & x_2 \\ x_3 & x_4 \end{array}\right$$. \nonumber\\

Then \\T(A)=\vec{x}\\, and therefore \\T\\ is onto.

Since \\T\\ is a linear transformation which is one-to-one and onto, \\T\\ is an isomorphism. Hence \\\mathbb{M}\_{22}\\ and \\\mathbb{R}^4\\ are isomorphic.

An important property of isomorphisms is that the inverse of an isomorphism is itself an isomorphism and the composition of isomorphisms is an isomorphism. We first recall the definition of composition.

##### Definition \\\PageIndex{5}\\: Composition of Transformations

Let \\V, W, Z\\ be vector spaces and suppose \\T: V \mapsto W\\ and \\S: W \mapsto Z\\ are linear transformations. Then the composite of \\S\\ and \\T\\ is \\S \circ T: V \mapsto Z\nonumber \\ and is defined by \$S \circ T) (\vec{v}) = S(T(\vec{v})) \mbox{ for all } \vec{v} \in V\nonumber \\

Consider now the following proposition.

##### Proposition \\\PageIndex{1}\\: Composite and Inverse Isomorphism

Let \\T:V\rightarrow W\\ be an isomorphism. Then \\T^{-1}:W\rightarrow V\\ is also an isomorphism. Also if \\T:V\rightarrow W\\ is an isomorphism and if \\S:W\rightarrow Z\\ is an isomorphism for the vector spaces \\V,W,Z,\\ then \\S\circ T\\ defined by \\\left( S\circ T\right) \left( v\right) = S\left( T\left( v\right) \right)\\ is also an isomorphism.

Proof

Consider the first claim. Since \\T\\ is onto, a typical vector in \\W\\ is of the form \\T(\vec{v})\\ where \\\vec{v} \in V\\. Consider then for \\a,b\\ scalars, \\T^{-1}\left( aT(\vec{v}\_{1})+bT(\vec{v}\_{2})\right)\nonumber \\ where \\\vec{v}\_{1}, \vec{v}\_2 \in V\\. Consider if this is equal to \\aT^{-1}\left( T(\vec{v}\_{1})\right) +bT^{-1}\left( T(\vec{v}\_{2})\right) =a\vec{v}\_{1}+b\vec{v}\_{2}?\nonumber \\ Since \\T\\ is one to one, this will be so if \\T\left( a\vec{v}\_{1}+b\vec{v}\_{2}\right) =T\left( T^{-1}\left( aT(\vec{v}\_{1})+bT(\vec{v}\_{2})\right) \right) =aT(\vec{v}\_{1})+bT(\vec{v}\_{2})\nonumber \\ However, the above statement is just the condition that \\T\\ is a linear map. Thus \\T^{-1}\\ is indeed a linear map. If \\\vec{v} \in V\\ is given, then \\\vec{v}=T^{-1}\left( T(\vec{v})\right)\\ and so \\T^{-1}\\ is onto. If \\T^{-1}(\vec{v})=\vec{0},\\ then \\\vec{v}=T\left( T^{-1}(\vec{v})\right) =T(\vec{0})=\vec{0}\nonumber \\ and so \\T^{-1}\\ is one to one.

Next suppose \\T\\ and \\S\\ are as described. Why is \\S\circ T\\ a linear map? Let for \\a,b\\ scalars, \\\begin{aligned} S\circ T\left( a\vec{v}\_{1}+b\vec{v}\_{2}\right) &\equiv S\left( T\left( a\vec{v}\_{1}+b\vec{v}\_{2}\right) \right) =S\left( aT(\vec{v}\_{1})+bT(\vec{v}\_{2})\right) \\ &=aS\left( T(\vec{v}\_{1})\right) +bS\left( T(\vec{v}\_{2})\right) \equiv a\left( S\circ T\right) \left( \vec{v}\_{1}\right) +b\left( S\circ T\right) \left( \vec{v}\_{2}\right)\end{aligned}\\ Hence \\S\circ T\\ is a linear map. If \\\left( S\circ T\right) \left( \vec{v}\right) =0,\\ then \\S\left( T\left( \vec{v} \right) \right) =\vec{0}\\ and it follows that \\T(\vec{v})=\vec{0}\\ and hence by this lemma again, \\\vec{v}=\vec{0}\\. Thus \\S\circ T\\ is one to one. It remains to verify that it is onto. Let \\\vec{z}\in Z\\. Then since \\S\\ is onto, there exists \\\vec{w}\in W\\ such that \\S(\vec{w})=\vec{z}.\\ Also, since \\T\\ is onto, there exists \\\vec{v}\in V\\ such that \\T(\vec{v})=\vec{w}.\\ It follows that \\S\left( T\left( \vec{v}\right) \right) =\vec{z}\\ and so \\S\circ T\\ is also onto.

Suppose we say that two vector spaces \\V\\ and \\W\\ are related if there exists an isomorphism of one to the other, written as \\V\sim W\\. Then the above proposition suggests that \\\sim\\ is an equivalence relation. That is: \\\sim\\ satisfies the following conditions:

We leave the proof of these to the reader.

The following fundamental lemma describes the relation between bases and isomorphisms.

##### Lemma \\\PageIndex{2}\\: Bases and Isomorphisms

Let \\T:V\rightarrow W\\ be a linear map where \\V,W\\ are vector spaces. Then a linear transformation \\T\\ which is one to one has the property that if \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ is linearly independent, then so is \\\left\\ T(\vec{u}\_{1}),\cdots ,T(\vec{u}\_{k})\right\\\\. More generally, \\T\\ is an isomorphism if and only if whenever \\\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\\\ is a basis for \\V,\\ it follows that \\\left\\ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\\\\ is a basis for \\W\\.

Proof

First suppose that \\T\\ is a linear map and is one to one and \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\\\\ is linearly independent. It is required to show that \\\left\\ T(\vec{u}\_{1}),\cdots ,T(\vec{u}\_{k})\right\\\\ is also linearly independent. Suppose then that \\\sum\_{i=1}^{k}c\_{i}T(\vec{u}\_{i})=\vec{0}\nonumber \\ Then, since \\T\\ is linear, \\T\left( \sum\_{i=1}^{n}c\_{i}\vec{u}\_{i}\right) =\vec{0}\nonumber \\ Since \\T\\ is one to one, it follows that \\\sum\_{i=1}^{n}c\_{i}\vec{u}\_{i}=0\nonumber \\ Now the fact that \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\\\\ is linearly independent implies that each \\c\_{i}=0\\. Hence \\\left\\ T(\vec{u} \_{1}),\cdots ,T(\vec{u}\_{n})\right\\\\ is linearly independent.

Now suppose that \\T\\ is an isomorphism and \\\left\\ \vec{v}\_{1},\cdots ,\vec{ v}\_{n}\right\\\\ is a basis for \\V\\. It was just shown that \\\left\\ T(\vec{v} \_{1}),\cdots ,T(\vec{v}\_{n})\right\\\\ is linearly independent. It remains to verify that the span of \\\left\\ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\\\\ is all of \\W\\. This is where \\T\\ is onto is used. If \\\vec{w}\in W,\\ there exists \\\vec{v}\in V\\ such that \\T(\vec{v})=\vec{w}\\. Since \\\left\\ \vec{v} \_{1},\cdots ,\vec{v}\_{n}\right\\\\ is a basis, it follows that there exists scalars \\\left\\ c\_{i}\right\\ \_{i=1}^{n}\\ such that \\\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}=\vec{v}.\nonumber \\ Hence, \\\vec{w}=T(\vec{v})=T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) =\sum\_{i=1}^{n}c\_{i}T\vec{v}\_{i}\nonumber \\ which shows that the span of these vectors \\\left\\ T(\vec{v}\_{1}),\cdots ,T (\vec{v}\_{n})\right\\\\ is all of \\W\\ showing that this set of vectors is a basis for \\W\\.

Next suppose that \\T\\ is a linear map which takes a basis to a basis. Then for \\\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\\\ a basis for \\V,\\ it follows \\\left\\ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\\\\ is a basis for \\W.\\ Then if \\w\in W,\\ there exist scalars \\c\_{i}\\ such that \\w=\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i})=T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right)\\ showing that \\T\\ is onto. If \\T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) =0\\ then \\\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i})=\vec{0}\\ and since the vectors \\\left\\ T(\vec{v} \_{1}),\cdots ,T(\vec{v}\_{n})\right\\\\ are linearly independent, it follows that each \\c\_{i}=0.\\ Since \\\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\\ is a typical vector in \\V\\, this has shown that if \\T(\vec{v})=0\\ then \\\vec{v}=\vec{0}\\ and so \\T\\ is also one to one. Thus \\T\\ is an isomorphism.

The following theorem illustrates a very useful idea for defining an isomorphism. Basically, if you know what it does to a basis, then you can construct the isomorphism.

##### Theorem \\\PageIndex{1}\\: Isomorphic Vector Spaces

Suppose \\V\\ and \\W\\ are two vector spaces. Then the two vector spaces are isomorphic if and only if they have the same dimension. In the case that the two vector spaces have the same dimension, then for a linear transformation \\T:V\rightarrow W\\, the following are equivalent.

1. \\T\\ is one to one.

2. \\T\\ is onto.

3. \\T\\ is an isomorphism.

Proof

Suppose first these two vector spaces have the same dimension. Let a basis for \\V\\ be \\\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\\\ and let a basis for \\W\\ be \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{n}\right\\\\. Now define \\T\\ as follows. \\T(\vec{v}\_{i})=\vec{w}\_{i}\nonumber \\ for \\\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\\ an arbitrary vector of \\V,\\ \\T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) = \sum\_{i=1}^{n}c\_{i}T (\vec{v}\_{i})=\sum\_{i=1}^{n}c\_{i}\vec{w}\_{i}.\nonumber \\ It is necessary to verify that this is well defined. Suppose then that \\\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}=\sum\_{i=1}^{n}\hat{c}\_{i}\vec{v}\_{i}\nonumber \\ Then \\\sum\_{i=1}^{n}\left( c\_{i}-\hat{c}\_{i}\right) \vec{v}\_{i}=0\nonumber \\ and since \\\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\\\ is a basis, \\c\_{i}=\hat{c}\_{i}\\ for each \\i\\. Hence \\\sum\_{i=1}^{n}c\_{i}\vec{w}\_{i}=\sum\_{i=1}^{n}\hat{c}\_{i}\vec{w}\_{i}\nonumber \\ and so the mapping is well defined. Also if \\a,b\\ are scalars, \\\begin{aligned} T\left( a\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}+b\sum\_{i=1}^{n}\hat{c}\_{i}\vec{v} \_{i}\right) &=T\left( \sum\_{i=1}^{n}\left( ac\_{i}+b\hat{c}\_{i}\right) \vec{v }\_{i}\right) =\sum\_{i=1}^{n}\left( ac\_{i}+b\hat{c}\_{i}\right) \vec{w}\_{i} \\ &=a\sum\_{i=1}^{n}c\_{i}\vec{w}\_{i}+b\sum\_{i=1}^{n}\hat{c}\_{i}\vec{w}\_{i} \\ &=aT\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) +bT\left( \sum\_{i=1}^{n} \hat{c}\_{i}\vec{v}\_{i}\right)\end{aligned}\\ Thus \\T\\ is a linear map.

Now if \\T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) =\sum\_{i=1}^{n}c\_{i}\vec{w} \_{i}=\vec{0},\nonumber \\ then since the \\\left\\ \vec{w}\_{1},\cdots ,\vec{w}\_{n}\right\\\\ are independent, each \\c\_{i}=0\\ and so \\\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}=\vec{0}\\ also. Hence \\T\\ is one to one. If \\\sum\_{i=1}^{n}c\_{i}\vec{w}\_{i}\\ is a vector in \\W,\\ then it equals \\\sum\_{i=1}^{n}c\_{i}T\vec{v}\_{i}=T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right)\nonumber \\ showing that \\T\\ is also onto. Hence \\T\\ is an isomorphism and so \\V\\ and \\W\\ are isomorphic.

Next suppose these two vector spaces are isomorphic. Let \\T\\ be the name of the isomorphism. Then for \\\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\\\ a basis for \\V\\, it follows that a basis for \\W\\ is \\\left\\ T\vec{v}\_{1},\cdots ,T\vec{v}\_{n}\right\\\\ showing that the two vector spaces have the same dimension.

Now suppose the two vector spaces have the same dimension.

First consider the claim that \\1.)\Rightarrow 2.).\\ If \\T\\ is one to one, then if \\\left\\ \vec{v}\_{1},\cdots ,\vec{v} \_{n}\right\\\\ is a basis for \\V,\\ then \\\left\\ T(\vec{v}\_{1}),\cdots ,T(\vec{v }\_{n})\right\\\\ is linearly independent. If it is not a basis, then it must fail to span \\W\\. But then there would exist \\\vec{w}\notin span \left\\ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\\\\ and it follows that \\\left\\ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n}),\vec{w}\right\\\\ would be linearly independent which is impossible because there exists a basis for \\W\\ of \\n\\ vectors. Hence \\span\left\\ T(\vec{v}\_{1}),\cdots ,T(\vec{v} \_{n})\right\\ =W\nonumber \\ and so \\\left\\ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{n})\right\\\\ is a basis. Hence, if \\\vec{w}\in W,\\ there exist scalars \\c\_{i}\\ such that \\\vec{w}=\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i})=T\left( \sum\_{i=1}^{n}c\_{i}\vec{v} \_{i}\right)\nonumber \\ showing that \\T\\ is onto. This shows that \\1.)\Rightarrow 2.).\\

Next consider the claim that \\2.)\Rightarrow 3.).\\ Since \\2.)\\ holds, it follows that \\T\\ is onto. It remains to verify that \\T\\ is one to one. Since \\T\\ is onto, there exists a basis of the form \\\left\\ T(\vec{v}\_{i}),\cdots ,T (\vec{v}\_{n})\right\\ .\\ If \\\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\\\ is linearly independent, then this set of vectors must also be a basis for \\V\\ because if not, there would exist \\\vec{u}\notin span\left\\ \vec{ v}\_{1},\cdots ,\vec{v}\_{n}\right\\\\ so \\\left\\ \vec{v}\_{1},\cdots ,\vec{v} \_{n},\vec{u}\right\\\\ would be a linearly independent set which is impossible because by assumption, there exists a basis which has \\n\\ vectors. So why is\\\left\\ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\\\\ linearly independent? Suppose \\\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}=\vec{0}\nonumber \\ Then \\\sum\_{i=1}^{n}c\_{i}T\vec{v}\_{i}=\vec{0}\nonumber \\ Hence each \\c\_{i}=0\\ and so, as just discussed, \\\left\\ \vec{v}\_{1},\cdots , \vec{v}\_{n}\right\\\\ is a basis for \\V\\. Now it follows that a typical vector in \\V\\ is of the form \\\sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\\. If \\T\left( \sum\_{i=1}^{n}c\_{i}\vec{v}\_{i}\right) =\vec{0},\\ it follows that \\\sum\_{i=1}^{n}c\_{i}T(\vec{v}\_{i})=\vec{0}\nonumber \\ and so, since \\\left\\ T(\vec{v}\_{i}),\cdots ,T(\vec{v}\_{n})\right\\\\ is independent, it follows each \\c\_{i}=0\\ and hence \\\sum\_{i=1}^{n}c\_{i}\vec{v} \_{i}=\vec{0}\\. Thus \\T\\ is one to one as well as onto and so it is an isomorphism.

If \\T\\ is an isomorphism, it is both one to one and onto by definition so \\3.)\\ implies both \\1.)\\ and \\2.)\\.

Note the interesting way of defining a linear transformation in the first part of the argument by describing what it does to a basis and then “extending it linearly”.

Consider the following example.

##### Example \\\PageIndex{6}\\

Let \\V=\mathbb{R}^{3}\\ and let \\W\\ denote the polynomials of degree at most 2. Show that these two vector spaces are isomorphic.

###### Solution

First, observe that a basis for \\W\\ is \\\left\\ 1,x,x^{2}\right\\\\ and a basis for \\V\\ is \\\left\\ \vec{e}\_{1},\vec{e}\_{2},\vec{e}\_{3}\right\\ .\\ Since these two have the same dimension, the two are isomorphic. An example of an isomorphism is this:

\\T(\vec{e}\_{1})=1,T(\vec{e}\_{2})=x,T(\vec{e}\_{3})=x^{2}\nonumber \\ and extend \\T\\ linearly as in the above proof. Thus \\T\left( a,b,c\right) =a+bx+cx^{2}\nonumber \\

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9_08_3A_The_Kernel_and_Image_of_a_Linear_Map

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> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/09%3A_Vector_Spaces/9.08%3A_The_Kernel_and_Image_of_a_Linear_Map

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Outcomes

1. Describe the kernel and image of a linear transformation.

2. Use the kernel and image to determine if a linear transformation is one to one or onto.

Here we consider the case where the linear map is not necessarily an isomorphism. First here is a definition of what is meant by the image and kernel of a linear transformation.

Definition \\\PageIndex{1}\\: Kernel and Image

Let \\V\\ and \\W\\ be vector spaces and let \\T:V\rightarrow W\\ be a linear transformation. Then the image of \\T\\ denoted as \\\mathrm{im}\left( T\right)\\ is defined to be the set \\\left\\ T(\vec{v}):\vec{v}\in V\right\\\nonumber \\ In words, it consists of all vectors in \\W\\ which equal \\T(\vec{v})\\ for some \\\vec{v}\in V\\. The kernel, \\\ker \left( T\right)\\, consists of all \\\vec{v}\in V\\ such that \\T(\vec{v})=\vec{0}\\. That is, \\\ker \left( T\right) =\left\\ \vec{v}\in V:T(\vec{v})=\vec{0}\right\\\nonumber \\

Then in fact, both \\\mathrm{im}\left( T\right)\\ and \\\ker \left( T\right)\\ are subspaces of \\W\\ and \\V\\ respectively.

Proposition \\\PageIndex{1}\\: Kernel and Image as Subspaces

Let \\V,W\\ be vector spaces and let \\T:V\rightarrow W\\ be a linear transformation. Then \\\ker \left( T\right) \subseteq V\\ and \\\mathrm{im}\left( T\right) \subseteq W\\. In fact, they are both subspaces.

Proof

First consider \\\ker \left( T\right) .\\ It is necessary to show that if \\\vec{v}\_{1},\vec{v}\_{2}\\ are vectors in \\\ker \left( T\right)\\ and if \\a,b\\ are scalars, then \\a\vec{v}\_{1}+b\vec{v}\_{2}\\ is also in \\\ker \left( T\right) .\\ But \\T\left( a\vec{v}\_{1}+b\vec{v}\_{2}\right) =aT(\vec{v}\_{1})+bT(\vec{v}\_{2})=a\vec{0}+b\vec{0}=\vec{0}\nonumber \\

Thus \\\ker \left( T\right)\\ is a subspace of \\V\\.

Next suppose \\T(\vec{v}\_{1}),T(\vec{v}\_{2})\\ are two vectors in \\\mathrm{im}\left( T\right) .\\ Then if \\a,b\\ are scalars, \\aT(\vec{v}\_{2})+bT(\vec{v}\_{2})=T\left( a\vec{v}\_{1}+b\vec{v}\_{2}\right)\nonumber \\ and this last vector is in \\\mathrm{im}\left( T\right)\\ by definition.

Consider the following example.

Example \\\PageIndex{1}\\: Kernel and Image of a Transformation

Let \\T:\mathbb{P}\_1\to\mathbb{R}\\ be the linear transformation defined by \\T(p(x))=p(1)\mbox{ for all } p(x)\in \mathbb{P}\_1.\nonumber \\ Find the kernel and image of \\T\\.

###### Solution

We will first find the kernel of \\T\\. It consists of all polynomials in \\\mathbb{P}\_1\\ that have \\1\\ for a root. \\\begin{aligned} \mathrm{ker}(T) & = \\ p(x)\in \mathbb{P}\_1 ~\|~ p(1)=0\\ \\ & = \\ ax+b ~\|~ a,b\in\mathbb{R} \mbox{ and }a+b=0\\ \\ & = \\ ax-a ~\|~ a\in\mathbb{R} \\\end{aligned}\\ Therefore a basis for \\\mathrm{ker}(T)\\ is \\\left\\ x-1 \right\\\nonumber \\ Notice that this is a subspace of \\\mathbb{P}\_1\\.

Now consider the image. It consists of all numbers which can be obtained by evaluating all polynomials in \\\mathbb{P}\_1\\ at \\1\\. \\\begin{aligned} \mathrm{im}(T) & = \\ p(1) ~\|~ p(x)\in \mathbb{P}\_1 \\ \\ & = \\ a+b ~\|~ ax+b\in \mathbb{P}\_1 \\ \\ & = \\ a+b ~\|~ a,b\in\mathbb{R} \\\\ & = \mathbb{R}\end{aligned}\\ Therefore a basis for \\\mathrm{im}(T)\\ is \\\left\\ 1 \right\\\nonumber \\ Notice that this is a subspace of \\\mathbb{R}\\, and in fact is the space \\\mathbb{R}\\ itself.

Example \\\PageIndex{2}\\: Kernel and Image of a Linear Transformation

Let \\T: \mathbb{M}\_{22} \mapsto \mathbb{R}^2\\ be defined by \\T \left $$ \begin{array}{cc} a & b \\ c & d \end{array} \right $$ = \left $$ \begin{array}{c} a - b \\ c + d \end{array} \right $$\nonumber \\ Then \\T\\ is a linear transformation. Find a basis for \\\mathrm{ker} (T)\\ and \\\mathrm{im}(T)\\.

###### Solution

You can verify that \\T\\ represents a linear transformation.

Now we want to find a way to describe all matrices \\A\\ such that \\T(A) = \vec{0}\\, that is the matrices in \\\mathrm{ker}(T)\\. Suppose \\A = \left $$ \begin{array}{cc} a & b \\ c & d \end{array} \right $$\\ is such a matrix. Then \\T \left $$ \begin{array}{cc} a & b \\ c & d \end{array} \right $$ = \left $$ \begin{array}{c} a - b \\ c + d \end{array} \right $$ = \left $$ \begin{array}{c} 0 \\ 0 \end{array} \right $$\nonumber \\ The values of \\a, b, c, d\\ that make this true are given by solutions to the system \\\begin{aligned} a - b &= 0 \\ c + d &= 0 \end{aligned}\\ The solution is \\a = s, b = s, c = t, d = -t\\ where \\s, t\\ are scalars. We can describe \\\mathrm{ker}(T)\\ as follows. \\\mathrm{ker}(T) = \left\\ \left $$ \begin{array}{cc} s & s \\ t & -t \end{array} \right $$ \right\\ = \mathrm{span} \left\\ \left $$ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right $$, \left $$ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right $$ \right\\\nonumber \\ It is clear that this set is linearly independent and therefore forms a basis for \\\mathrm{ker}(T)\\.

We now wish to find a basis for \\\mathrm{im}(T)\\. We can write the image of \\T\\ as \\\mathrm{im}(T) = \left\\ \left $$ \begin{array}{c} a - b \\ c + d \end{array} \right $$ \right\\\nonumber \\ Notice that this can be written as \\\mathrm{span} \left\\ \left $$ \begin{array}{c} 1 \\ 0 \end{array}\right $$, \left $$ \begin{array}{c} -1 \\ 0 \end{array}\right $$, \left $$ \begin{array}{c} 0 \\ 1 \end{array}\right $$, \left $$ \begin{array}{c} 0 \\ 1 \end{array}\right $$ \right\\\nonumber \\

However this is clearly not linearly independent. By removing vectors from the set to create an independent set gives a basis of \\\mathrm{im}(T)\\. \\\left\\ \left $$ \begin{array}{c} 1 \\ 0 \end{array}\right $$, \left $$ \begin{array}{c} 0 \\ 1 \end{array}\right $$ \right\\\nonumber \\

Notice that these vectors have the same span as the set above but are now linearly independent.

A major result is the relation between the dimension of the kernel and dimension of the image of a linear transformation. A special case was done earlier in the context of matrices. Recall that for an \\m\times n\\ matrix \\% A,\\ it was the case that the dimension of the kernel of \\A\\ added to the rank of \\A\\ equals \\n\\.

Theorem \\\PageIndex{1}\\: Dimension of Kernel + Image

Let \\T:V\rightarrow W\\ be a linear transformation where \\V,W\\ are vector spaces. Suppose the dimension of \\V\\ is \\n\\. Then \\n=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im} \left( T\right) \right)\\.

Proof

From Proposition $\PageIndex{1}$, \\\mathrm{im}\left( T\right)\\ is a subspace of \\W.\\ By Theorem 9.4.8, there exists a basis for \\\mathrm{im}\left( T\right) ,\left\\ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{r})\right\\ .\\ Similarly, there is a basis for \\\ker \left( T\right) ,\left\\ \vec{u} \_{1},\cdots ,\vec{u}\_{s}\right\\\\. Then if \\\vec{v}\in V,\\ there exist scalars \\c\_{i}\\ such that \\T(\vec{v})=\sum\_{i=1}^{r}c\_{i}T(\vec{v}\_{i})\nonumber \\ Hence \\T\left( \vec{v}-\sum\_{i=1}^{r}c\_{i}\vec{v}\_{i}\right) =0.\\ It follows that \\\vec{v}-\sum\_{i=1}^{r}c\_{i}\vec{v}\_{i}\\ is in \\\ker \left( T\right)\\. Hence there are scalars \\a\_{i}\\ such that \\\vec{v}-\sum\_{i=1}^{r}c\_{i}\vec{v}\_{i}=\sum\_{j=1}^{s}a\_{j}\vec{u}\_{j}\nonumber \\ Hence \\\vec{v}=\sum\_{i=1}^{r}c\_{i}\vec{v}\_{i}+\sum\_{j=1}^{s}a\_{j}\vec{u} \_{j}.\\ Since \\\vec{v}\\ is arbitrary, it follows that \\V=\mathrm{span}\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{s},\vec{v}\_{1},\cdots , \vec{v}\_{r}\right\\\nonumber \\ If the vectors \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{s},\vec{v}\_{1},\cdots , \vec{v}\_{r}\right\\\\ are linearly independent, then it will follow that this set is a basis. Suppose then that \\\sum\_{i=1}^{r}c\_{i}\vec{v}\_{i}+\sum\_{j=1}^{s}a\_{j}\vec{u}\_{j}=0\nonumber \\ Apply \\T\\ to both sides to obtain \\\sum\_{i=1}^{r}c\_{i}T(\vec{v}\_{i})+\sum\_{j=1}^{s}a\_{j}T(\vec{u} \_{j})=\sum\_{i=1}^{r}c\_{i}T(\vec{v}\_{i})= \vec{0}\nonumber \\ Since \\\left\\ T(\vec{v}\_{1}),\cdots ,T(\vec{v}\_{r})\right\\\\ is linearly independent, it follows that each \\c\_{i}=0.\\ Hence \\\sum\_{j=1}^{s}a\_{j}\vec{u }\_{j}=0\\ and so, since the \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{s}\right\\\\ are linearly independent, it follows that each \\a\_{j}=0\\ also. It follows that \\\left\\ \vec{u}\_{1},\cdots ,\vec{u}\_{s},\vec{v}\_{1},\cdots ,\vec{v} \_{r}\right\\\\ is a basis for \\V\\ and so \\n=s+r=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right)\nonumber \\

Consider the following definition.

Definition \\\PageIndex{2}\\: Rank of Linear Transformation

Let \\T:V\rightarrow W\\ be a linear transformation and suppose \\V,W\\ are finite dimensional vector spaces. Then the rank of \\T\\ denoted as \\\mathrm{rank}\left( T\right)\\ is defined as the dimension of \\\mathrm{im}\left( T\right) .\\ The nullity of \\T\\ is the dimension of \\\ker \left( T\right) .\\ Thus the above theorem says that \\\mathrm{rank}\left( T\right) +\dim \left( \ker \left( T\right) \right) =\dim \left( V\right) .\\

Recall the following important result.

Theorem \\\PageIndex{2}\\: Subspace of Same Dimension

Let \\V\\ be a vector space of dimension \\n\\ and let \\W\\ be a subspace. Then \\W=V\\ if and only if the dimension of \\W\\ is also \\n\\.

From this theorem follows the next corollary.

Corollary \\\PageIndex{1}\\: One to One and Onto Characterization

Let \\T:V\rightarrow W\\ be a linear map where the dimension of \\V\\ is \\n\\ and the dimension of \\W\\ is \\m\\. Then \\T\\ is one to one if and only if \\\ker \left( T\right) =\left\\ \vec{0}\right\\\\ and \\T\\ is onto if and only if \\\mathrm{rank}\left( T\right) =m\\.

Proof

The statement \\\ker \left( T \right) =\left\\ \vec{0}\right\\\\ is equivalent to saying if \\T \left( \vec{v} \right)=\vec{0},\\ it follows that \\\vec{v}=\vec{0}\\. Thus by Lemma 9.7.1 \\T\\ is one to one. If \\T\\ is onto, then \\\mathrm{im}\left( T\right) =W\\ and so \\\mathrm{rank}\left( T\right)\\ which is defined as the dimension of \\\mathrm{im}\left( T\right)\\ is \\m\\. If \\\mathrm{ rank}\left( T\right) =m,\\ then by Theorem $\PageIndex{2}$, since \\\mathrm{im} \left( T\right)\\ is a subspace of \\W,\\ it follows that \\\mathrm{im}\left( T\right) =W\\.

Example \\\PageIndex{3}\\: One to One Transformation

Let \\S:\mathbb{P}\_2\to\mathbb{M}\_{22}\\ be a linear transformation defined by \\S(ax^2+bx+c) = \left $$\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right $$ \mbox{ for all } ax^2+bx+c\in \mathbb{P}\_2.\nonumber \\ Prove that \\S\\ is one to one but not onto.

###### Solution

You may recall this example from earlier in Example 9.7.1. Here we will determine that \\S\\ is one to one, but not onto, using the method provided in Corollary $\PageIndex{1}$.

By definition, \\\ker(S)=\\ax^2+bx+c\in \mathbb{P}\_2 ~\|~ a+b=0, a+c=0, b-c=0, b+c=0\\.\nonumber \\

Suppose \\p(x)=ax^2+bx+c\in\ker(S)\\. This leads to a homogeneous system of four equations in three variables. Putting the augmented matrix in reduced row-echelon form:

\\\left $$\begin{array}{rrr\|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right $$ \rightarrow \cdots \rightarrow \left $$\begin{array}{ccc\|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right $$.\nonumber \\

Since the unique solution is \\a=b=c=0\\, \\\ker(S)=\\\vec{0}\\\\, and thus \\S\\ is one-to-one by Corollary $\PageIndex{1}$.

Similarly, by Corollary $\PageIndex{1}$, if \\S\\ is onto it will have \\\mathrm{rank}(S) = \mathrm{dim}(\mathbb{M}\_{22}) = 4\\. The image of \\S\\ is given by

\\\mathrm{im}(S) = \left\\ \left $$\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right $$ \right\\ = \mathrm{span} \left\\ \left $$\begin{array}{rr} 1 & 1 \\ 0 & 0 \end{array} \right $$, \left $$\begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right $$, \left $$\begin{array}{rr} 0 & 1 \\ -1 & 1 \end{array} \right $$ \right\\\nonumber \\

These matrices are linearly independent which means this set forms a basis for \\\mathrm{im}(S)\\. Therefore the dimension of \\\mathrm{im}(S)\\, also called \\\mathrm{rank}(S)\\, is equal to \\3\\. It follows that \\S\\ is not onto.

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9_09_3A_The_Matrix_of_a_Linear_Transformation

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> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/09%3A_Vector_Spaces/9.09%3A_The_Matrix_of_a_Linear_Transformation

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Outcomes

1. Find the matrix of a linear transformation with respect to general bases in vector spaces.

You may recall from \\\mathbb{R}^n\\ that the matrix of a linear transformation depends on the bases chosen. This concept is explored in this section, where the linear transformation now maps from one arbitrary vector space to another.

Let \\T: V \mapsto W\\ be an isomorphism where \\V\\ and \\W\\ are vector spaces. Recall from Lemma 9.7.2 that \\T\\ maps a basis in \\V\\ to a basis in \\W\\. When discussing this Lemma, we were not specific on what this basis looked like. In this section we will make such a distinction.

Consider now an important definition.

Definition \\\PageIndex{1}\\: Coordinate Isomorphism

Let \\V\\ be a vector space with \\\mathrm{dim}(V)=n\\, let \\B=\\ \vec{b}\_1, \vec{b}\_2, \ldots, \vec{b}\_n \\\\ be a fixed basis of \\V\\, and let \\\\ \vec{e}\_1, \vec{e}\_2, \ldots, \vec{e}\_n \\\\ denote the standard basis of \\\mathbb{R}^n\\. We define a transformation \\C_B:V\to\mathbb{R}^n\\ by \\C_B(a_1\vec{b}\_1 + a_2\vec{b}\_2 + \cdots + a_n\vec{b}\_n) = a_1\vec{e}\_1 + a_2\vec{e}\_2 + \cdots + a_n\vec{e}\_n = \left $$\begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n \end{array}\right $$.\nonumber \\ Then \\C_B\\ is a linear transformation such that \\C_B(\vec{b}\_i)=\vec{e}\_i\\, \\1\leq i\leq n\\.

\\C_B\\ is an isomorphism, called the coordinate isomorphism corresponding to \\B\\.

We continue with another related definition.

Definition \\\PageIndex{2}\\: Coordinate Vector

Let \\V\\ be a finite dimensional vector space with \\\mathrm{dim}(V)=n\\, and let \\B=\\\vec{b}\_1, \vec{b}\_2, \ldots, \vec{b}\_n\\\\ be an ordered basis of \\V\\ (meaning that the order that the vectors are listed is taken into account). The coordinate vector of \\\vec{v}\\ with respect to \\B\\ is defined as \\C_B(\vec{v})\\.

coordinatevector Consider the following example.

Example \\\PageIndex{1}\\: Coordinate Vector

Let \\V = \mathbb{P}\_2\\ and \\\vec{x} = -x^2 -2x + 4\\. Find \\C_B(\vec{x})\\ for the following bases \\B\\:

1. \\B = \left\\ 1, x, x^2 \right\\\\

2. \\B = \left\\ x^2, x, 1 \right\\\\

3. \\B = \left\\ x + x^2 , x , 4 \right\\\\

###### Solution

1. First, note the order of the basis is important. Now we need to find \\a_1, a_2, a_3\\ such that \\\vec{x} = a_1 (1) + a_2 (x) + a_3(x^2)\\, that is: \\-x^2 -2x + 4 = a_1 (1) + a_2 (x) + a_3(x^2)\nonumber \\ Clearly the solution is \\\begin{aligned} a_1 &= 4 \\ a_2 &= -2 \\ a_3 &= -1\end{aligned}\\ Therefore the coordinate vector is \\C_B(\vec{x}) = \left $$ \begin{array}{r} 4 \\ -2 \\ -1 \end{array} \right $$\nonumber \\

2. Again remember that the order of \\B\\ is important. We proceed as above. We need to find \\a_1, a_2, a_3\\ such that \\\vec{x} = a_1 (x^2) + a_2 (x) + a_3(1)\\, that is: \\-x^2 -2x + 4 = a_1 (x^2) + a_2 (x) + a_3(1)\nonumber \\ Here the solution is \\\begin{aligned} a_1 &= -1 \\ a_2 &= -2 \\ a_3 &= 4\end{aligned}\\ Therefore the coordinate vector is \\C_B(\vec{x}) = \left $$ \begin{array}{r} -1 \\ -2 \\ 4 \end{array} \right $$\nonumber \\

3. Now we need to find \\a_1, a_2, a_3\\ such that \\\vec{x} = a_1 (x + x^2) + a_2 (x) + a_3(4)\\, that is: \\\begin{aligned} -x^2 -2x + 4 &= a_1 (x + x^2 ) + a_2 (x) + a_3(4)\\ &= a_1 (x^2) + (a_1 + a_2) (x) + a_3(4)\end{aligned}\\ The solution is \\\begin{aligned} a_1 &= -1 \\ a_2 &= -1 \\ a_3 &= 1\end{aligned}\\ and the coordinate vector is \\C_B(\vec{x})=\left$$\begin{array}{r}-1\\-1\\1\end{array}\right$$\nonumber\\

Given that the coordinate transformation \\C_B:V\to\mathbb{R}^n\\ is an isomorphism, its inverse exists.

Theorem \\\PageIndex{1}\\: Inverse of the Coordinate Isomorphism

Let \\V\\ be a finite dimensional vector space with dimension \\n\\ and ordered basis \\B=\\\vec{b}\_1, \vec{b}\_2, \ldots, \vec{b}\_n\\\\. Then \\C_B:V\to\mathbb{R}^n\\ is an isomorphism whose inverse, \\C_B^{-1}:\mathbb{R}^n\to V\nonumber \\ is given by \\C_B^{-1} =\left $$\begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n \end{array}\right $$ = a_1\vec{b}\_1 + a_2\vec{b}\_2 + \cdots + a_n\vec{b}\_n ~\mbox{ for all }~ \left $$\begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n \end{array}\right $$ \in\mathbb{R}^n.\nonumber \\

We now discuss the main result of this section, that is how to represent a linear transformation with respect to different bases.

Let \\V\\ and \\W\\ be finite dimensional vector spaces, and suppose

Let \\T:V\to W\\ be a linear transformation. If \\V=\mathbb{R}^n\\ and \\W=\mathbb{R}^m\\, then we can find a matrix \\A\\ so that \\T_A=T\\. For arbitrary vector spaces \\V\\ and \\W\\, our goal is to represent \\T\\ as a matrix., i.e., find a matrix \\A\\ so that \\T_A:\mathbb{R}^n\to\mathbb{R}^m\\ and \\T_A=C\_{B_2}TC\_{B_1}^{-1}\\.

To find the matrix \\A\\:

\\T_A=C\_{B_2}TC\_{B_1}^{-1}~\mbox{ implies that }~ T_AC\_{B_1}=C\_{B_2}T,\nonumber \\ and thus for any \\\vec{v}\in V\\, \\C\_{B_2}$$T(\vec{v})$$ = T_A$$C\_{B_1}(\vec{v})$$ =AC\_{B_1}(\vec{v}).\nonumber \\

Since \\C\_{B_1}(\vec{b}\_j)=\vec{e}\_j\\ for each \\\vec{b}\_j\in B_1\\, \\AC\_{B_1}(\vec{b}\_j)=A\vec{e}\_j\\, which is simply the \\j^{th}\\ column of \\A\\. Therefore, the \\j^{th}\\ column of \\A\\ is equal to \\C\_{B_2}$$T(\vec{b}\_j)$$\\.

The matrix of \\T\\ corresponding to the ordered bases \\B_1\\ and \\B_2\\ is denoted \\M\_{B_2B_1}(T)\\ and is given by \\M\_{B_2B_1}(T)= \left $$\begin{array}{cccc} C\_{B_2} $$ T(\vec{b}\_1)$$ & C\_{B_2}$$T(\vec{b}\_2) $$ & \cdots & C\_{B_2}$$T(\vec{b}\_n) $$ \end{array}\right $$.\nonumber \\ This result is given in the following theorem.

Theorem \\\PageIndex{2}\\

Let \\V\\ and \\W\\ be vectors spaces of dimension \\n\\ and \\m\\ respectively, with \\B_1=\\\vec{b}\_1, \vec{b}\_2, \ldots, \vec{b}\_n\\\\ an ordered basis of \\V\\ and \\B_2\\ an ordered basis of \\W\\. Suppose \\T:V\to W\\ is a linear transformation. Then the unique matrix \\M\_{B_2B_1}(T)\\ of \\T\\ corresponding to \\B_1\\ and \\B_2\\ is given by \\M\_{B_2B_1}(T)= \left $$\begin{array}{cccc} C\_{B_2}$$T(\vec{b}\_1)$$ & C\_{B_2}$$T(\vec{b}\_2)$$ & \cdots & C\_{B_2}$$T(\vec{b}\_n)$$ \end{array}\right $$.\nonumber \\

This matrix satisfies \\C\_{B_2}$$T(\vec{v})$$=M\_{B_2B_1}(T)C\_{B_1}(\vec{v})\\ for all \\\vec{v}\in V\\.

We demonstrate this content in the following examples.

Example \\\PageIndex{2}\\: Matrix of a Linear Transformation

Let \\T: \mathbb{P}\_3 \mapsto \mathbb{R}^4\\ be an isomorphism defined by \\T( ax^3 + bx^2 + cx + d) = \left $$ \begin{array}{c} a + b \\ b - c \\ c + d \\ d + a \end{array} \right $$\nonumber \\

Suppose \\B_1 = \left\\ x^3, x^2, x, 1 \right\\\\ is an ordered basis of \\\mathbb{P}\_3\\ and \\B_2 = \left\\ \left $$ \begin{array}{r} 1 \\ 0 \\ 1 \\ 0 \end{array} \right $$, \left $$ \begin{array}{r} 0 \\ 1 \\ 0 \\ 0 \end{array} \right $$, \left $$ \begin{array}{r} 0 \\ 0 \\ -1 \\ 0 \end{array} \right $$, \left $$ \begin{array}{r} 0 \\ 0 \\ 0 \\ 1 \end{array} \right $$ \right\\\nonumber \\ be an ordered basis of \\\mathbb{R}^4\\. Find the matrix \\M\_{B_2B_1}(T)\\.

###### Solution

To find \\M\_{B_2B_1}(T)\\, we use the following definition. \\M\_{B_2B_1}(T) = \left $$ \begin{array}{cccc} C\_{B_2}$$T(x^3)$$ & C\_{B_2}$$T(x^2)$$ & C\_{B_2}$$T(x)$$ & C\_{B_2}$$T(x^2)$$ \end{array} \right $$\nonumber \\ First we find the result of applying \\T\\ to the basis \\B_1\\. \\T(x^3) = \left $$ \begin{array}{c} 1 \\ 0 \\ 0 \\ 1 \end{array} \right $$, T(x^2) = \left $$ \begin{array}{c} 1 \\ 1 \\ 0 \\ 0 \end{array} \right $$, T(x) = \left $$ \begin{array}{c} 0 \\ -1 \\ 1 \\ 0 \end{array} \right $$, T(1) = \left $$ \begin{array}{c} 0 \\ 0 \\ 1 \\ 1 \end{array} \right $$\nonumber \\

Next we apply the coordinate isomorphism \\C\_{B_2}\\ to each of these vectors. We will show the first in detail. \\C\_{B_2} \left( \left $$ \begin{array}{c} 1 \\ 0 \\ 0 \\ 1 \end{array} \right $$ \right) = a_1 \left $$ \begin{array}{r} 1 \\ 0 \\ 1 \\ 0 \end{array} \right $$ + a_2 \left $$ \begin{array}{r} 0 \\ 1 \\ 0 \\ 0 \end{array} \right $$ + a_3 \left $$ \begin{array}{r} 0 \\ 0 \\ -1 \\ 0 \end{array} \right $$ + a_4 \left $$ \begin{array}{r} 0 \\ 0 \\ 0 \\ 1 \end{array} \right $$\nonumber \\ This implies that \\\begin{aligned} a_1 &= 1 \\ a_2 &= 0 \\ a_1 - a_3 &= 0 \\ a_4 &= 1 \end{aligned}\\ which has a solution given by \\\begin{aligned} a_1 &= 1 \\ a_2 &= 0 \\ a_3 &= 1 \\ a_4 &= 1 \end{aligned}\\

Therefore \\C\_{B_2} $$T(x^3)$$ = \left $$ \begin{array}{r} 1 \\ 0 \\ 1 \\ 1 \end{array} \right $$\\.

You can verify that the following are true. \\C\_{B_2}$$T(x^2)$$ = \left $$ \begin{array}{r} 1 \\ 1 \\ 1 \\ 0 \end{array} \right $$, C\_{B_2}$$T(x)$$ = \left $$ \begin{array}{r} 0 \\ -1 \\ -1 \\ 0 \end{array} \right $$, C\_{B_2}$$T(1)$$ = \left $$ \begin{array}{r} 0 \\ 0 \\ -1 \\ 1 \end{array} \right $$\nonumber \\

Using these vectors as the columns of \\M\_{B_2B_1}(T)\\ we have \\M\_{B_2B_1}(T) = \left $$ \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 0 & 1 \end{array} \right $$\nonumber \\

The next example demonstrates that this method can be used to solve different types of problems. We will examine the above example and see if we can work backwards to determine the action of \\T\\ from the matrix \\M\_{B_2B_1}(T)\\.

Example \\\PageIndex{3}\\: Finding the Action of a Linear Transformation

Let \\T: \mathbb{P}\_3 \mapsto \mathbb{R}^4\\ be an isomorphism with \\M\_{B_2B_1}(T) = \left $$ \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 0 & 1 \end{array} \right $$,\nonumber \\ where \\B_1 = \left\\ x^3, x^2, x, 1 \right\\\\ is an ordered basis of \\\mathbb{P}\_3\\ and \\B_2 = \left\\ \left $$ \begin{array}{r} 1 \\ 0 \\ 1 \\ 0 \end{array} \right $$, \left $$ \begin{array}{r} 0 \\ 1 \\ 0 \\ 0 \end{array} \right $$, \left $$ \begin{array}{r} 0 \\ 0 \\ -1 \\ 0 \end{array} \right $$, \left $$ \begin{array}{r} 0 \\ 0 \\ 0 \\ 1 \end{array} \right $$ \right\\\nonumber \\ is an ordered basis of \\\mathbb{R}^4\\. If \\p(x) = ax^3 + bx^2 + cx + d\\, find \\T(p(x))\\.

###### Solution

Recall that \\C\_{B_2}$$T(p(x))$$ = M\_{B_2B_1}(T) C\_{B_1}(p(x))\\. Then we have \\\begin{aligned} C\_{B_2}$$T(p(x))$$ &= M\_{B_2B_1}(T) C\_{B_1}(p(x)) \\ &= \left $$ \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 0 & 1 \end{array} \right $$ \left $$ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right $$ \\ &= \left $$ \begin{array}{c} a + b \\ b - c \\ a + b - c - d\\ a + d \end{array} \right $$\end{aligned}\\

Therefore \\\begin{aligned} T(p(x)) &= C^{-1}\_D \left $$ \begin{array}{c} a + b \\ b - c \\ a + b - c - d\\ a + d \end{array} \right $$ \\ &= (a+b) \left $$ \begin{array}{r} 1 \\ 0 \\ 1 \\ 0 \end{array} \right $$ + (b-c) \left $$ \begin{array}{r} 0 \\ 1 \\ 0 \\ 0 \end{array} \right $$ + (a+b-c-d) \left $$ \begin{array}{r} 0 \\ 0 \\ -1 \\ 0 \end{array} \right $$ + (a+d) \left $$ \begin{array}{r} 0 \\ 0 \\ 0 \\ 1 \end{array} \right $$ \\ &= \left $$ \begin{array}{c} a + b \\ b - c \\ c + d \\ a +d \end{array} \right $$\end{aligned}\\

You can verify that this was the definition of \\T(p(x))\\ given in the previous example.

We can also find the matrix of the composite of multiple transformations.

Theorem \\\PageIndex{3}\\: Matrix of Composition

Let \\V,W\\ and \\U\\ be finite dimensional vector spaces, and suppose \\T : V \mapsto W\\, \\S: W \mapsto U\\ are linear transformations. Suppose \\V, W\\ and \\U\\ have ordered bases of \\B_1\\, \\B_2\\ and \\B_3\\ respectively. Then the matrix of the composite transformation \\S \circ T\\ (or \\ST\$ is given by \\M\_{B_3B_1}(ST)=M\_{B_3B_2}(S) M\_{B_2B_1}(T).\nonumber \\

The next important theorem gives a condition on when \\T\\ is an isomorphism.

Theorem \\\PageIndex{4}\\: Isomorphism

Let \\V\\ and \\W\\ be vector spaces such that both have dimension \\n\\ and let \\T: V \mapsto W\\ be a linear transformation. Suppose \\B_1\\ is an ordered basis of \\V\\ and \\B_2\\ is an ordered basis of \\W\\.

Then the conditions that \\M\_{B_2B_1}(T)\\ is invertible for all \\B_1\\ and \\B_2\\, and that \\M\_{B_2B_1}(T)\\ is invertible for some \\B_1\\ and \\B_2\\ are equivalent. In fact, these occur if and only if \\T\\ is an isomorphism.

If \\T\\ is an isomorphism, the matrix \\M\_{B_2B_1}(T)\\ is invertible and its inverse is given by \\\left $$ M\_{B_2B_1}(T) \right $$ ^{-1} = M\_{B_1B_2}(T^{-1})\\.

Consider the following example.

Example \\\PageIndex{4}\\

Suppose \\T:\mathbb{P}\_3\to\mathbb{M}\_{22}\\ is a linear transformation defined by \\T(ax^3+bx^2+cx+d)= \left $$\begin{array}{cc} a+d & b-c \\ b+c & a-d \end{array}\right $$\nonumber \\ for all \\ax^3+bx^2+cx+d\in\mathbb{P}\_3\\. Let \\B_1=\\ x^3, x^2, x, 1\\\\ and \\B_2=\left\\ \left $$\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right $$, \left $$\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right $$, \left $$\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right $$, \left $$\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right $$\right\\\nonumber \\ be ordered bases of \\\mathbb{P}\_3\\ and \\\mathbb{M}\_{22}\\, respectively.

1. Find \\M\_{B_2B_1}(T)\\.

2. Verify that \\T\\ is an isomorphism by proving that \\M\_{B_2B_1}(T)\\ is invertible.

3. Find \\M\_{B_1B_2}(T^{-1})\\, and verify that \\M\_{B_1B_2}(T^{-1}) = \left $$ M\_{B_2B_1}(T)\right $$^{-1}\\.

4. Use \\M\_{B_1B_2}(T^{-1})\\ to find \\T^{-1}\\.

###### Solution

1. \\\begin{aligned} M\_{B_2B_1}(T) & = \left $$ \begin{array}{cccc} C\_{B_2}$$T(1)$$ & C\_{B_2}$$T(x)$$ & C\_{B_2}$$T(x^2)$$ & C\_{B_2}$$T(x^3)$$ \end{array}\right $$ \\ & = \left $$ \begin{array}{cccc} C\_{B_2}\left $$\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right $$ & C\_{B_2}\left $$\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right $$ & C\_{B_2}\left $$\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right $$ & C\_{B_2}\left $$\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right $$ \end{array}\right $$ \\ & = \left $$\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & -1 \end{array}\right $$\end{aligned}\\

2. \\\det(M\_{B_2B_1}(T))=4\\, so the matrix is invertible, and hence \\T\\ is an isomorphism.

3. \\T^{-1}\left $$\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right $$ = 1, T^{-1}\left $$\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right $$= x, T^{-1}\left $$\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right $$= x^2, T^{-1}\left $$\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right $$=x^3,\nonumber \\ so \\T^{-1}\left $$\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right $$ = \frac{1+x^3}{2}, T^{-1}\left $$\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right $$= \frac{x-x^2}{2},\nonumber \\ \\T^{-1}\left $$\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right $$ = \frac{x+x^2}{2}, T^{-1}\left $$\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right $$= \frac{1-x^3}{2}.\nonumber \\ Therefore, \\M\_{B_1B_2}(T^{-1})=\frac{1}{2}\left $$\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & -1 \end{array}\right $$\nonumber \\ You should verify that \\M\_{B_2B_1}(T) M\_{B_1B_2}(T^{-1}) = I_4\\. From this it follows that \\$$M\_{B_2B_1}(T)$$^{-1}= M\_{B_1B_2}(T^{-1})\\.

4. \\\begin{aligned} C\_{B_1}\left(T^{-1}\left $$\begin{array}{cc} p & q \\ r & s \end{array}\right $$\right) & = M\_{B_1B_2}(T^{-1}) C\_{B_2}\left( \left $$\begin{array}{cc} p & q \\ r & s \end{array}\right $$\right)\\ T^{-1}\left $$\begin{array}{cc} p & q \\ r & s \end{array}\right $$ & = C\_{B_1}^{-1}\left(M\_{B_1B_2}(T^{-1}) C\_{B_2}\left( \left $$\begin{array}{cc} p & q \\ r & s \end{array}\right $$\right)\right)\\ & = C\_{B_1}^{-1}\left( \frac{1}{2}\left $$\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & -1 \end{array}\right $$ \left $$\begin{array}{c} p \\ q\\ r\\ s\end{array}\right $$\right) \\ & = C\_{B_1}^{-1}\left(\frac{1}{2}\left $$\begin{array}{c} p+s \\ q+r \\ r-q \\ p-s \end{array}\right $$\right) \\ & = \frac{1}{2}(p+s)x^3 +\frac{1}{2}(q+r)x^2 +\frac{1}{2}(r-q)x + \frac{1}{2}(p-s).\end{aligned}\\

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9_E_3A_Exercises

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Exercise \\\PageIndex{1}\\

Suppose you have \\\mathbb{R}^2\\ and the \\+\\ operation is as follows: \$a,b) + (c,d) = (a+d,b+c).\nonumber\\ Scalar multiplication is defined in the usual way. Is this a vector space? Explain why or why not.

Exercise \\\PageIndex{2}\\

Suppose you have \\\mathbb{R}^2\\ and the \\+\\ operation is as follows: \$a,b) + (c,d) = (0,b+d)\nonumber\\ Scalar multiplication is defined in the usual way. Is this a vector space? Explain why or why not.

Exercise \\\PageIndex{3}\\

Suppose you have \\\mathbb{R}^2\\ and scalar multiplication is defined as \\c(a,b) = (a, cb)\\ while vector addition is defined as usual. Is this a vector space? Explain why or why not.

Exercise \\\PageIndex{4}\\

Suppose you have \\\mathbb{R}^2\\ and the \\+\\ operation is defined as follows. \$a,b) + (c,d) = (a−c,b−d)\nonumber\\ Scalar multiplication is same as usual. Is this a vector space? Explain why or why not.

Exercise \\\PageIndex{5}\\

Consider all the functions defined on a non empty set which have values in \\\mathbb{R}\\. Is this a vector space? Explain. The operations are defined as follows. Here \\f ,g\\ signify functions and \\a\\ is a scalar \\\begin{aligned} (f+g)(x)&=f(x)+g(x) \\ (af)(x)&=a(f(x))\end{aligned}\\

Exercise \\\PageIndex{6}\\

Denote by \\\mathbb{R}^{\mathbb{N}}\\ the set of real valued sequences. For \\\vec{a} ≡ \\a_n\\\_{n=1}^∞\\, \\\vec{b} ≡ \\b_n\\\_{n=1}^\infty\\ two of these, define their sum to be given by \\\vec{a}+\vec{b}=\\a_n+b_n\\\_{n=1}^\infty\nonumber\\ and define scalar multiplication by \\c\vec{a}=\\ca_n\\\_{n=1}^\infty\text{ where }\vec{a}=\\a+n\\\_{n=1}^\infty\nonumber\\ Is this a special case of Exercise $\PageIndex{5}$? Is this a vector space?

Exercise \\\PageIndex{7}\\

Let \\\mathbb{C}^2\\ be the set of ordered pairs of complex numbers. Define addition and scalar multiplication in the usual way. \$z,w) + (\hat{z},\hat{w}) = (z+\hat{z},w+\hat{w}), u(z,w) ≡ (uz,uw)\nonumber\\ Here the scalars are from \\\mathbb{C}\\. Show this is a vector space.

Exercise \\\PageIndex{8}\\

Let \\V\\ be the set of functions defined on a nonempty set which have values in a vector space \\W\\. Is this a vector space? Explain.

Exercise \\\PageIndex{9}\\

Consider the space of \\m\times n\\ matrices with operation of addition and scalar multiplication defined the usual way. That is, if \\A,B\\ are two \\m\times n\\ matrices and \\c\\ a scalar, \$A+B)\_{ i j} = A\_{i j} +B\_{i j}, \$cA)\_{ i j} ≡ c (A\_{ij})\nonumber\\

Exercise \\\PageIndex{10}\\

Consider the set of \\n\times n\\ symmetric matrices. That is, \\A = A^T\\. In other words, \\A\_{i j} = A\_{ji}\\. Show that this set of symmetric matrices is a vector space and a subspace of the vector space of \\n\times n\\ matrices.

Exercise \\\PageIndex{11}\\

Consider the set of all vectors in \\\mathbb{R}^2 ,(x, y)\\ such that \\x + y ≥ 0\\. Let the vector space operations be the usual ones. Is this a vector space? Is it a subspace of \\\mathbb{R}^2\\?

Exercise \\\PageIndex{12}\\

Consider the vectors in \\\mathbb{R}^2 ,(x, y)\\ such that \\xy = 0\\. Is this a subspace of \\\mathbb{R}^2\\? Is it a vector space? The addition and scalar multiplication are the usual operations.

Exercise \\\PageIndex{13}\\

Define the operation of vector addition on \\\mathbb{R}^2\\ by \$x, y) + (u, v) = (x+u, y+v+1)\\. Let scalar multiplication be the usual operation. Is this a vector space with these operations? Explain.

Exercise \\\PageIndex{14}\\

Let the vectors be real numbers. Define vector space operations in the usual way. That is \\x+y\\ means to add the two numbers and \\xy\\ means to multiply them. Is \\\mathbb{R}\\ with these operations a vector space? Explain.

Exercise \\\PageIndex{15}\\

Let the scalars be the rational numbers and let the vectors be real numbers which are the form \\a+b\sqrt{2}\\ for \\a,b\\ rational numbers. Show that with the usual operations, this is a vector space.

Exercise \\\PageIndex{16}\\

Let \\\mathbb{P}\_2\\ be the set of all polynomials of degree \\2\\ or less. That is, these are of the form \\a+bx+cx^2\\. Addition is defined as \$a+bx+cx^2)+(\hat{d}+\hat{b}x+\hat{c}x^2)=(a+\hat{a})+(b+\hat{b})x+(c+\hat{c})x^2\nonumber\\ and scalar multiplication is defined as \\d(a+bx+cx^2)=da+dbx+cdx^2\nonumber\\ Show that, with this definition of the vector space operations that \\\mathbb{P}\_2\\ is a vector space. Now let \\V\\ denote those polynomials \\a+bx+cx^2\\ such that \\a+b+c = 0\\. Is \\V\\ a subspace of \\\mathbb{P}\_2\\? Explain.

Exercise \\\PageIndex{17}\\

Let \\M,N\\ be subspaces of a vector space \\V\\ and consider \\M +N\\ defined as the set of all \\m+n\\ where \\m ∈ M\\ and \\n ∈ N\\. Show that \\M +N\\ is a subspace of \\V\\.

Exercise \\\PageIndex{18}\\

Let \\M,N\\ be subspaces of a vector space \\V\\. Then \\M ∩N\\ consists of all vectors which are in both \\M\\ and \\N\\. Show that \\M ∩N\\ is a subspace of \\V\\.

Exercise \\\PageIndex{19}\\

Let \\M,N\\ be subspaces of a vector space \\\mathbb{R}^2\\. Then \\N ∪M\\ consists of all vectors which are in either \\M\\ or \\N\\. Show that \\N ∪M\\ is not necessarily a subspace of \\\mathbb{R}^2\\ by giving an example where \\N ∪M\\ fails to be a subspace.

Exercise \\\PageIndex{20}\\

Let \\X\\ consist of the real valued functions which are defined on an interval \\$$a,b$$\\. For \\f ,g ∈ X, f +g\\ is the name of the function which satisfies \$f +g) (x) = f (x) +g(x)\\. For \\s\\ a real number, \$s f) (x) = s(f (x))\\. Show this is a vector space.

Answer

The axioms of a vector space all hold because they hold for a vector space. The only thing left to verify is the assertions about the things which are supposed to exist. \\0\\ would be the zero function which sends everything to \\0\\. This is an additive identity. Now if \\f\\ is a function, \\−f (x) ≡ (−f (x))\\. Then \$f + (−f)) (x) ≡ f (x) + (−f) (x) ≡ f (x) + (−f (x)) = 0\nonumber\\ Hence \\f + −f = 0\\. For each \\x ∈ $$a,b$$\\, let \\f_x (x) = 1\\ and \\f_x (y) = 0\\ if \\y\neq x\\. Then these vectors are obviously linearly independent.

Exercise \\\PageIndex{21}\\

Consider functions defined on \\\\1, 2,\cdots ,n\\\\ having values in \\\mathbb{R}\\. Explain how, if \\V\\ is the set of all such functions, \\V\\ can be considered as \\\mathbb{R}^n\\.

Answer

Let \\f (i)\\ be the \\i\\th component of a vector \\\vec{x} ∈ \mathbb{R}^n\\. Thus a typical element in \\\mathbb{R}^n\\ is \$f (1),\cdots , f (n))\\.

Exercise \\\PageIndex{22}\\

Let the vectors be polynomials of degree no more than \\3\\. Show that with the usual definitions of scalar multiplication and addition wherein, for \\p(x)\\ a polynomial, \$ap) (x) = ap(x)\\ and for \\p,q\\ polynomials \$p+q) (x) = p(x) +q(x)\\, this is a vector space.

Answer

This is just a subspace of the vector space of functions because it is closed with respect to vector addition and scalar multiplication. Hence this is a vector space.

Exercise \\\PageIndex{23}\\

Let \\V\\ be a vector space and suppose \\\\\vec{x}\_1,\cdots ,\vec{x}\_l\\\\ is a set of vectors in \\V\\. Show that \\\vec{0}\\ is in \\span\\\vec{x}\_1,\cdots ,\vec{x}\_k\\\\.

Answer

\\\sum\limits\_{i=1}^k0\vec{x}\_k=\vec{0}\\

Exercise \\\PageIndex{24}\\

Determine if \\p(x) = 4x^2 −x\\ is in the span given by \\span \\x^2+x,\\x^2-1,\\-x+2\\\nonumber\\

Exercise \\\PageIndex{25}\\

Determine if \\p(x) = −x^2 +x+2\\ is in the span given by \\span\\ x^2 +x+1,\\ 2x^2 +x\\\nonumber\\

Exercise \\\PageIndex{26}\\

Determine if \\A=\left$$\begin{array}{cc}1&3\\0&0\end{array}\right$$\\ is in the span given by \\span\left\\\left$$\begin{array}{cc}1&0\\0&1\end{array}\right$$,\\\left$$\begin{array}{cc}0&1\\1&0\end{array}\right$$,\\\left$$\begin{array}{cc}1&0\\1&1\end{array}\right$$,\\\left$$\begin{array}{cc}0&1\\1&1\end{array}\right$$\right\\\nonumber\\

Exercise \\\PageIndex{27}\\

Show that the spanning set in Exercise $\PageIndex{26}$ is a spanning set for \\M\_{22}\\, the vector space of all \\2\times 2\\ matrices.

Exercise \\\PageIndex{28}\\

Consider the vector space of polynomials of degree at most \\2\\, \\\mathbb{P}\_2\\. Determine whether the following is a basis for \\\mathbb{P}\_2\\. \\\\x^2 +x+1,\\ 2x^2 +2x+1,\\ x+1\\\nonumber\\ Hint: There is a isomorphism from \\\mathbb{R}^3\\ to \\\mathbb{P}\_2\\. It is defined as follows: \\T\vec{e}\_1 = 1,\\ T\vec{e}\_2 = x,\\ T\vec{e}\_3= x^2\nonumber\\ Then extend \\T\\ linearly. Thus \\T\left$$\begin{array}{c}1\\1\\1\end{array}\right$$=x^2+x+1,\\T\left$$\begin{array}{c}1\\2\\2\end{array}\right$$=2x^2+2x+1,\\T\left$$\begin{array}{c}1\\1\\0\end{array}\right$$=1+x\nonumber\\ It follows that if \\\left\\\left$$\begin{array}{c}1\\1\\1\end{array}\right$$,\\\left$$\begin{array}{c}1\\2\\2\end{array}\right$$,\\\left$$\begin{array}{c}1\\1\\0\end{array}\right$$\right\\\nonumber\\ is a basis for \\\mathbb{R}^3\\, then the polynomials will be a basis for \\\mathbb{P}\_2\\ because they will be independent. Recall that an isomorphism takes a linearly independent set to a linearly independent set. Also, since \\T\\ is an isomorphism, it preserves all linear relations.

Exercise \\\PageIndex{29}\\

Find a basis in \\\mathbb{P}\_2\\ for the subspace \\span\\ 1+x+x^2 ,\\ 1+2x,\\ 1+5x−3x^2\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others. Hint: This is the situation in which you have a spanning set and you want to cut it down to form a linearly independent set which is also a spanning set. Use the same isomorphism above. Since \\T\\ is an isomorphism, it preserves all linear relations so if such can be found in \\\mathbb{R}^3\\, the same linear relations will be present in \\\mathbb{P}\_2\\.

Exercise \\\PageIndex{30}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ 1+x−x^2 +x^3 ,\\ 1+2x+3x^3 ,\\−1+3x+5x^2 +7x^3 ,\\ 1+6x+4x^2 +11x^3\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{31}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ 1+x−x^2 +x^3 ,\\ 1+2x+3x^3 ,\\−1+3x+5x^2 +7x^3 ,\\ 1+6x+4x^2 +11x^3\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{32}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 −2x^2 +x+2,\\ 3x^3 −x^2 +2x+2,\\ 7x^3 +x^2 +4x+2,\\ 5x^3 +3x+2\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{33}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 +2x^2 +x−2,\\ 3x^3 +3x^2 +2x−2,\\ 3x^3 +x+2,\\ 3x^3 +x+2\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{34}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 −5x^2 +x+5,\\ 3x^3 −4x^2 +2x+5,\\ 5x^3 +8x^2 +2x−5,\\ 11x^3 +6x+5\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{35}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\x^3 −3x^2 +x+3,\\ 3x^3 −2x^2 +2x+3,\\ 7x^3 +7x^2 +3x−3,\\ 7x^3 +4x+3\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{36}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 −x^2 +x+1,\\ 3x^3 +2x+1,\\ 4x^3 +x^2 +2x+1,\\ 3x^3 +2x−1\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{37}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 −x^2 +x+1,\\ 3x^3 +2x+1,\\ 13x^3 +x^2 +8x+4,\\ 3x^3 +2x−1\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{38}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 −3x^2 +x+3,\\ 3x^3 −2x^2 +2x+3,\\−5x^3 +5x^2 −4x−6,\\ 7x^3 +4x−3\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{39}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 −2x^2 +x+2,\\ 3x^3 −x^2 +2x+2,\\ 7x^3 −x^2 +4x+4,\\ 5x^3 +3x−2\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{40}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 −2x^2 +x+2,\\ 3x^3 −x^2 +2x+2,\\ 3x^3 +4x^2 +x−2,\\ 7x^3 −x^2 +4x+4\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{41}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 −4x^2 +x+4,\\ 3x^3 −3x^2 +2x+4,\\−3x^3 +3x^2 −2x−4,\\−2x^3 +4x^2 −2x−4\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{42}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 +2x^2 +x−2,\\ 3x^3 +3x^2 +2x−2,\\ 5x^3 +x^2 +2x+2,\\ 10x^3 +10x^2 +6x−6\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{43}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 +x^2 +x−1,\\ 3x^3 +2x^2 +2x−1,\\ x^3 +1,\\ 4x^3 +3x^2 +2x−1\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{44}\\

Find a basis in \\\mathbb{P}\_3\\ for the subspace \\span\\ x^3 −x^2 +x+1,\\ 3x^3 +2x+1,\\ x^3 +2x^2 −1,\\ 4x^3 +x^2 +2x+1\\\nonumber\\ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise \\\PageIndex{45}\\

Here are some vectors. \\\\ x^3 +x^2 −x−1,\\ 3x^3 +2x^2 +2x−1\\\nonumber\\ If these are linearly independent, extend to a basis for all of \\\mathbb{P}\_3\\.

Exercise \\\PageIndex{46}\\

Here are some vectors. \\\\ x^3 −2x^2 −x+2,\\ 3x^3 −x^2 +2x+2\\\nonumber\\ If these are linearly independent, extend to a basis for all of \\\mathbb{P}\_3\\.

Exercise \\\PageIndex{47}\\

Here are some vectors. \\\\ x^3 −3x^2 −x+3,\\ 3x^3 −2x^2 +2x+3\\\nonumber\\ If these are linearly independent, extend to a basis for all of \\\mathbb{P}\_3\\.

Exercise \\\PageIndex{48}\\

Here are some vectors. \\\\ x^3 −2x^2 −3x+2,\\ 3x^3 −x^2 −6x+2,\\−8x^3 +18x+10\\\nonumber\\ If these are linearly independent, extend to a basis for all of \\\mathbb{P}\_3\\.

Exercise \\\PageIndex{49}\\

Here are some vectors. \\\\ x^3 −3x^2 −3x+3,\\ 3x^3 −2x^2 −6x+3,\\−8x^3 +18x+40\\\nonumber\\ If these are linearly independent, extend to a basis for all of \\\mathbb{P}\_3\\.

Exercise \\\PageIndex{50}\\

Here are some vectors. \\\\ x^3 −x^2 +x+1,\\ 3x^3 +2x+1,\\ 4x^3 +2x+2\\\nonumber\\ If these are linearly independent, extend to a basis for all of \\\mathbb{P}\_3\\.

Exercise \\\PageIndex{51}\\

Here are some vectors. \\\\ x^3 +x^2 +2x−1,\\ 3x^3 +2x^2 +4x−1,\\ 7x^3 +8x+23\\\nonumber\\ If these are linearly independent, extend to a basis for all of \\\mathbb{P}\_3\\.

Exercise \\\PageIndex{52}\\

Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. \\\\ x+1,\\ x^2 +2,\\ x^2 −x−3\\\nonumber\\

Exercise \\\PageIndex{53}\\

Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. \\\\ x^2 +x,\\−2x^2 −4x−6,\\ 2x−2\\\nonumber\\

Exercise \\\PageIndex{54}\\

Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. \\\left\\\left$$\begin{array}{cc}1&2\\0&1\end{array}\right$$,\\\left$$\begin{array}{rr}-7&2\\-2&-3\end{array}\right$$,\\\left$$\begin{array}{cc}4&0\\1&2\end{array}\right$$\right\\\nonumber\\

Exercise \\\PageIndex{55}\\

Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. \\\left\\\left$$\begin{array}{cc}1&0\\0&1\end{array}\right$$,\\\left$$\begin{array}{cc}0&1\\0&1\end{array}\right$$,\\\left$$\begin{array}{cc}1&0\\1&0\end{array}\right$$,\\\left$$\begin{array}{cc}0&0\\1&1\end{array}\right$$\right\\\nonumber\\

Exercise \\\PageIndex{56}\\

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\\?

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{57}\\

If you have \\6\\ vectors in \\\mathbb{R}^5\\, is it possible they are linearly independent? Explain.

Answer

No. They can't be.

Exercise \\\PageIndex{58}\\

Let \\\mathbb{P}\_3\\ be the polynomials of degree no more than \\3\\. Determine which of the following are bases for this vector space.

1. \\\\ x+1,\\ x^3 +x^2 +2x,\\ x^2 +x,\\ x^3 +x^2 +x\\\\

2. \\\\ x^3 +1,\\ x^2 +x,\\ 2x^3 +x^2 ,\\ 2x^3 −x^2 −3x+1\\\\

Answer

1.

2. Suppose \\c_1(x^3 +1)+c_2 (x^2 +x) +c_3( 2x^3 +x^2) +c_4 (2x^3 −x^2 −3x+1) = 0\nonumber\\ Then combine the terms according to power of \\x\\. \$c_1 +2c_3 +2c_4) x^3 + (c_2 +c_3 −c_4) x^2 + (c_2 −3c_4) x+ (c_1 +c_4) = 0\nonumber\\ Is there a non zero solution to the system \\\begin{aligned}c_1 +2c_3 +2c_4 &= 0\\ c_2 +c_3 −c_4 &= 0\\ c_2 −3c_4 &= 0\\ c_1 +c_4 &= 0\end{aligned}\\, Solution is: \\$$c_1 = 0,\\ c_2 = 0,\\ c_3 = 0,\\ c_4 = 0$$\nonumber\\ Therefore, these are linearly independent.

Exercise \\\PageIndex{59}\\

In the context of the above problem, consider polynomials \\\\a_ix^3 +b_ix^2 +c_ix+d_i ,\\ i = 1, 2, 3, 4\\\nonumber\\ Show that this collection of polynomials is linearly independent on an interval \\$$s,t$$\\ if and only if \\\left$$\begin{array}{cccc}a_1&b_1&c_1&d_1 \\ a_2&b_2&c_2&d_2 \\ a_3&b_3&c_3&d_3\\ a_4&b_4&c_4&d_4\end{array}\right$$\nonumber\\ is an invertible matrix.

Answer

Let \\p_i(x)\\ denote the \\i\\th of these polynomials. Suppose \\\sum_i C_ip_i(x) = 0\\. Then collecting terms according to the exponent of \\x\\, you need to have \\\begin{aligned}C_1a_1 +C_2a_2 +C_3a_3 +C_4a_4 &= 0\\ C_1b_1 +C_2b_2 +C_3b_3 +C_4b_4 &= 0\\ C_1c_1 +C_2c_2 +C_3c_3 +C_4c_4 &= 0\\ C_1d_1 +C_2d_2 +C_3d_3 +C_4d_4 &= 0\end{aligned}\\ The matrix of coefficients is just the transpose of the above matrix. There exists a non trivial solution if and only if the determinant of this matrix equals \\0\\.

Exercise \\\PageIndex{60}\\

Let the field of scalars be \\\mathbb{Q}\\, the rational numbers and let the vectors be of the form \\a+b\sqrt{2}\\ where \\a,b\\ are rational numbers. Show that this collection of vectors is a vector space with field of scalars \\\mathbb{Q}\\ and give a basis for this vector space.

Answer

When you add two of these you get one and when you multiply one of these by a scalar, you get another one. A basis is \\\\1,\sqrt{2}\\\\. By definition, the span of these gives the collection of vectors. Are they independent? Say \\a + b\sqrt{2} = 0\\ where \\a,b\\ are rational numbers. If \\a\neq 0\\, then \\b\sqrt{2} = −a\\ which can’t happen since a is rational. If \\b\neq 0\\, then \\−a = b\sqrt{2}\\ which again can’t happen because on the left is a rational number and on the right is an irrational. Hence both \\a,b = 0\\ and so this is a basis.

Exercise \\\PageIndex{61}\\

Suppose \\V\\ is a finite dimensional vector space. Based on the exchange theorem above, it was shown that any two bases have the same number of vectors in them. Give a different proof of this fact using the earlier material in the book. Hint: Suppose \\\\\vec{x}\_1,\cdots ,\vec{x}\_n\\\\ and \\\\\vec{y}\_1,\cdots , \vec{y}\_m\\\\ are two bases with \\m \< n\\. Then define \\φ : \mathbb{R}^n \mapsto V,\\ ψ :\mathbb{R}^m\mapsto V\nonumber\\ by \\φ (\vec{a}) = \sum\limits\_{k=1}^n a_k\vec{x}\_k ,\\ ψ(\vec{b}) =\sum\limits\_{j=1}^m b_j\vec{y}\_j\nonumber\\ Consider the linear transformation, \\ψ^{−1}\circ φ\\. Argue it is a one to one and onto mapping from \\\mathbb{R}^n\\ to \\\mathbb{R}^m\\. Now consider a matrix of this linear transformation and its reduced row-echelon form.

Answer

This is obvious because when you add two of these you get one and when you multiply one of these by a scalar, you get another one. A basis is \\\\1,\sqrt{2}\\\\. By definition, the span of these gives the collection of vectors. Are they independent? Say \\a+b\sqrt{2} = 0\\ where \\a,b\\ are rational numbers. If \\a\neq 0\\, then \\b\sqrt{2} = −a\\ which can’t happen since \\a\\ is rational. If \\b\neq 0\\, then \\−a = b\sqrt{2}\\ which again can’t happen because on the left is a rational number and on the right is an irrational. Hence both \\a,b = 0\\ and so this is a basis.

Exercise \\\PageIndex{62}\\

Let \\M =\\\vec{u} = (u_1,\\u_2,\\u_3,\\u_4)\in \mathbb{R}^4\\ :\\ \|u_1\| ≤ 4\\\\. Is \\M\\ a subspace of \\\mathbb{R}^4\\?

Answer

This is not a subspace. \\\left$$\begin{array}{c}1\\1\\1\\1\end{array}\right$$\\ is in it, but \\20\left$$\begin{array}{c}1\\1\\1\\1\end{array}\right$$\\ is not.

Exercise \\\PageIndex{63}\\

Let \\M =\\\vec{u} = (u_1,\\u_2,\\u_3,\\u_4)\in \mathbb{R}^4\\ :\\ \sin(u_1) = 1\\\\. Is \\M\\ a subspace of \\\mathbb{R}^4\\?

Answer

This is not a subspace.

Exercise \\\PageIndex{64}\\

Let \\W\\ be a subset of \\M\_{22}\\ given by \\W = \\ A\|A\in M\_{22},A^T = A\\\nonumber\\ In words, \\W\\ is the set of all symmetric \\2\times 2\\ matrices. Is \\W\\ a subspace of \\M\_{22}\\?

Exercise \\\PageIndex{65}\\

Let \\W\\ be a subset of \\M\_{22}\\ given by \\W=\left\\\left$$\begin{array}{cc}a&b\\c&d\end{array}\right$$ \\ \|a,b,c,d\in\mathbb{R},\\a+b=c+d\right\\\nonumber\\ Is \\W\\ a subspace of \\M\_{22}\\?

Exercise \\\PageIndex{66}\\

Let \\W\\ be a subset of \\P_3\\ given by \\W = \\ ax^3 +bx^2 +cx+d\|\\ a,b, c,d\in\mathbb{R},d = 0\\\nonumber\\ Is \\W\\ a subspace of \\P_3\\?

Exercise \\\PageIndex{67}\\

Let \\W\\ be a subset of \\P_3\\ given by \\W = \\ p(x) = ax^3 +bx^2 +cx+d\|\\ a,b, c,d\in\mathbb{R}, p(2) = 1\\\nonumber\\ Is \\W\\ a subspace of \\P_3\\?

Exercise \\\PageIndex{68}\\

Let \\T\\: \\\mathbb{P}\_2\to\mathbb{R}\\ be a linear transformation such that \\T(x^2)=1;\\ T(x^2+x)=5;\\ T(x^2+x+1)=-1.\nonumber\\ Find \\T(ax^2+bx+c)\\.

Answer

By linearity we have \\T(x^2 ) = 1,\\ T(x) = T(x^2 +x−x^2 ) = T(x^2 +x)−T (x^2 ) = 5−1 = 5,\\ and \\T(1) = T(x^2 +x+1−(x^2 +x)) = T(x^2 +x+1)−T(x^2 +x)) = −1−5 = −6\\. Thus \\T(ax^2 +bx+c) = aT(x^2 ) +bT(x) +cT(1) = a+5b−6c\\.

Exercise \\\PageIndex{69}\\

Consider the following functions \\T\\: \\\mathbb{R}^3\to\mathbb{R}^2\\. Explain why each of these functions \\T\\ is not linear.

1. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}x+2y+3z+1 \\ 2y-3x+z\end{array}\right$$\\

2. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}x+2y^2+3z \\ 2y+3z+z\end{array}\right$$\\

3. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}\sin x+2y+3z \\ 2y+3z+z\end{array}\right$$\\

4. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}x+2y+3z \\ 2y+3z-\ln z\end{array}\right$$\\

Exercise \\\PageIndex{70}\\

Suppose \\T\\ is a linear transformation such that \\\begin{aligned} T\left$$\begin{array}{r}1\\1\\-7\end{array}\right$$&=\left$$\begin{array}{c}3\\3\\3\end{array}\right$$ \\ T\left$$\begin{array}{r}-1\\0\\6\end{array}\right$$&=\left$$\begin{array}{c}1\\2\\3\end{array}\right$$ \\ T\left$$\begin{array}{r}0\\-1\\2\end{array}\right$$&=\left$$\begin{array}{r}1\\3\\-1\end{array}\right$$\end{aligned}\\ Find the matrix of \\T\\. That is find \\A\\ such that \\T(\vec{x})=A\vec{x}\\.

Answer

\\\left$$\begin{array}{rrr}3&1&1\\3&2&3\\3&3&-1\end{array}\right$$\left$$\begin{array}{ccc}6&2&1\\5&2&1\\6&1&1\end{array}\right$$=\left$$\begin{array}{ccc}29&9&5\\46&13&8\\27&11&5\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{71}\\

Suppose \\T\\ is a linear transformation such that \\\begin{aligned} T\left$$\begin{array}{r}1\\2\\-18\end{array}\right$$&=\left$$\begin{array}{c}5\\2\\5\end{array}\right$$ \\ T\left$$\begin{array}{r}-1\\-1\\15\end{array}\right$$&=\left$$\begin{array}{c}3\\3\\5\end{array}\right$$ \\ T\left$$\begin{array}{r}0\\-1\\4\end{array}\right$$&=\left$$\begin{array}{r}2\\5\\-2\end{array}\right$$\end{aligned}\\ Find the matrix of \\T\\. That is find \\A\\ such that \\T(\vec{x})=A\vec{x}\\.

Answer

\\\left$$\begin{array}{rrr}5&3&2\\2&3&5\\5&5&-2\end{array}\right$$\left$$\begin{array}{ccc}11&4&1\\10&4&1\\12&3&1\end{array}\right$$=\left$$\begin{array}{ccc}109&38&10\\112&35&10\\81&34&8\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{72}\\

Consider the following functions \\T\\: \\\mathbb{R}^3\to\mathbb{R}^2\\. Show that each is a linear transformation and determine for each the matrix \\A\\ such that \\T(\vec{x}) = A\vec{x}\\.

1. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}x+2y+3z \\ 2y-3x+z\end{array}\right$$\\

2. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}7x+2y+z \\ 3x-11y+2z\end{array}\right$$\\

3. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}3x+2y+z \\ x+2y+6z\end{array}\right$$\\

4. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}2y-5x+z \\ x+y+z\end{array}\right$$\\

Exercise \\\PageIndex{73}\\

Suppose \\$$A_1\cdots A_n$$^{-1}\nonumber\\ exists where each \\A_j\in\mathbb{R}^n\\ and let vectors \\\\B_1,\cdots ,B_n\\\\ in \\\mathbb{R}^m\\ be given. Show that there always exists a linear transformation \\T\\ such that \\T(A_i)=B_i\\.

Exercise \\\PageIndex{74}\\

Let \\V\\ and \\W\\ be subspaces of \\\mathbb{R}^n\\ and \\\mathbb{R}^m\\ respectively and let \\T\\: \\V → W\\ be a linear transformation. Suppose that \\\\T\vec{v}\_1,\cdots ,T\vec{v}\_r\\\\ is linearly independent. Show that it must be the case that \\\\\vec{v}\_1,\cdots ,\vec{v}\_r\\\\ is also linearly independent.

Answer

If \\\sum\limits_i^ra_i\vec{v}\_r=0\\, then using linearity properties of \\T\\ we get \\0=T(0)=T\left(\sum\limits_i^ra_i\vec{v}\_r\right)=\sum\limits_i^ra_iT(\vec{v}\_r).\nonumber\\ Since we assume that \\\\T\vec{v}\_1,\cdots ,T\vec{v}\_r\\\\ is linearly independent, we must have all \\a_i = 0\\, and therefore we conclude that \\\\\vec{v}\_1,\cdots ,\vec{v}\_r\\\\ is also linearly independent.

Exercise \\\PageIndex{75}\\

Let \\V=span\left\\\left$$\begin{array}{c}1\\1\\2\\0\end{array}\right$$,\\\left$$\begin{array}{c}0\\1\\1\\1\end{array}\right$$,\\\left$$\begin{array}{c}1\\1\\0\\1\end{array}\right$$\right\\\nonumber\\ Let \\T\vec{x}=A\vec{x}\\ where \\A\\ is the matrix \\\left$$\begin{array}{cccc}1&1&1&1\\0&1&1&0\\0&1&2&1\\1&1&1&2\end{array}\right$$\nonumber\\ Give a basis for \\Im(T)\\.

Exercise \\\PageIndex{76}\\

Let \\V=span\left\\\left$$\begin{array}{c}1\\0\\0\\1\end{array}\right$$,\\\left$$\begin{array}{c}1\\1\\1\\1\end{array}\right$$,\\\left$$\begin{array}{c}1\\4\\4\\1\end{array}\right$$\right\\\nonumber\\ Let \\T\vec{x}=A\vec{x}\\ where \\A\\ is the matrix \\\left$$\begin{array}{cccc}1&1&1&1\\0&1&1&0\\0&1&2&1\\1&1&1&2\end{array}\right$$\nonumber\\ Find a basis for \\Im(T)\\. In this case, the original vectors do not form an independent set.

Answer

Since the third vector is a linear combinations of the first two, then the image of the third vector will also be a linear combinations of the image of the first two. However the image of the first two vectors are linearly independent (check!), and hence form a basis of the image. Thus a basis for \\Im(T)\\ is: \\V=span\left\\\left$$\begin{array}{c}2\\0\\1\\3\end{array}\right$$,\\\left$$\begin{array}{c}4\\2\\4\\5\end{array}\right$$\right\\\nonumber\\

Exercise \\\PageIndex{77}\\

If \\\\\vec{v}\_1,\cdots ,\vec{v}\_r\\\\ is linearly independent and \\T\\ is a one to one linear transformation, show that \\\\T\vec{v}\_1,\cdots ,T\vec{v}\_r\\\\ is also linearly independent. Give an example which shows that if \\T\\ is only linear, it can happen that, although \\\\\vec{v}\_1,\cdots ,\vec{v}\_r\\\\ is linearly independent, \\\\T\vec{v}\_1,\cdots ,T\vec{v}\_r\\\\ is not. In fact, show that it can happen that each of the \\T\vec{v}\_j\\ equals \\0\\.

Exercise \\\PageIndex{78}\\

Let \\V\\ and \\W\\ be subspaces of \\\mathbb{R}^n\\ and \\\mathbb{R}^m\\ respectively and let \\T\\: \\V → W\\ be a linear transformation. Show that if \\T\\ is onto \\W\\ and if \\\\\vec{v}\_1,\cdots ,\vec{v}\_r\\\\ is a basis for \\V\\, then \\span\\T\vec{v}\_1,\cdots ,T\vec{v}\_r\\ = W\\.

Exercise \\\PageIndex{79}\\

Define \\T\\: \\\mathbb{R}^4 → \mathbb{R}^3\\ as follows. \\T\vec{x}=\left$$\begin{array}{rrrr}3&2&1&8\\2&2&-2&6\\1&1&-1&3\end{array}\right$$\vec{x}\nonumber\\ Find a basis for \\Im(T)\\. Also find a basis for \\\text{ker}(T)\\.

Exercise \\\PageIndex{80}\\

Define \\T\\: \\\mathbb{R}^4 → \mathbb{R}^3\\ as follows. \\T\vec{x}=\left$$\begin{array}{rrr}1&2&0\\1&1&1\\0&1&1\end{array}\right$$\vec{x}\nonumber\\ where on the right, it is just matrix multiplication of the vector \\\vec{x}\\ which is meant. Explain why \\T\\ is an isomorphism of \\\mathbb{R}^3\\ to \\\mathbb{R}^3\\.

Exercise \\\PageIndex{81}\\

Suppose \\T\\: \\\mathbb{R}^3 → \mathbb{R}^3\\ is a linear transformation given by \\T\vec{x}=A\vec{x}\nonumber\\ where \\A\\ is a \\3\times 3\\ matrix. Show that \\T\\ is an isomorphism if and only if \\A\\ is invertible.

Exercise \\\PageIndex{82}\\

Suppose \\T\\: \\\mathbb{R}^3 → \mathbb{R}^3\\ is a linear transformation given by \\T\vec{x}=A\vec{x}\nonumber\\ where \\A\\ is a \\m\times n\\ matrix. Show that \\T\\ is never an isomorphism if \\m\neq n\\. In particular, show that if \\m\>n\\, \\T\\ cannot be onto and if \\m\

Exercise \\\PageIndex{83}\\

Define \\T\\: \\\mathbb{R}^2 → \mathbb{R}^3\\ as follows. \\T\vec{x}=\left$$\begin{array}{cc}1&0\\1&1\\0&1\end{array}\right$$\vec{x}\nonumber\\ where on the right, it is just matrix multiplication of the vector \\\vec{x}\\ which is meant. Show that \\T\\ is one to one. Next let \\W = Im(T)\\. Show that \\T\\ is an isomorphism of \\\mathbb{R}^2\\ and \\Im (T)\\.

Exercise \\\PageIndex{84}\\

In the above problem, find a \\2\times 3\\ matrix \\A\\ such that the restriction of \\A\\ to \\Im(T)\\ gives the same result as \\T^{−1}\\ on \\Im(T)\\. Hint: You might let \\A\\ be such that \\A\left$$\begin{array}{c}1\\1\\0\end{array}\right$$=\left$$\begin{array}{c}1\\0\end{array}\right$$,\\A\left$$\begin{array}{c}0\\1\\1\end{array}\right$$=\left$$\begin{array}{c}0\\1\end{array}\right$$\nonumber\\ now find another vector \\\vec{v} ∈ \mathbb{R}^3\\ such that \\\left\\\left$$\begin{array}{c}1\\1\\0\end{array}\right$$,\\\left$$\begin{array}{c}0\\1\\1\end{array}\right$$,\\\vec{v}\right\\\nonumber\\ is a basis. You could pick \\\vec{v}=\left$$\begin{array}{c}0\\0\\1\end{array}\right$$\nonumber\\ for example. Explain why this one works or one of your choice works. Then you could define \\A\vec{v}\\ to equal some vector in \\\mathbb{R}^2\\. Explain why there will be more than one such matrix \\A\\ which will deliver the inverse isomorphism \\T^{−1}\\ on \\Im(T)\\.

Exercise \\\PageIndex{85}\\

Now let \\V\\ equal \\span\left\\\left$$\begin{array}{c}1\\0\\1\end{array}\right$$,\\\left$$\begin{array}{c}0\\1\\1\end{array}\right$$\right\\\\ and let \\T\\: \\V\to W\\ be a linear transformation where \\W=span\left\\\left$$\begin{array}{c}1\\0\\1\\0\end{array}\right$$,\\\left$$\begin{array}{c}0\\1\\1\\1\end{array}\right$$\right\\\nonumber\\ and \\T\left$$\begin{array}{c}1\\0\\1\end{array}\right$$=\left$$\begin{array}{c}1\\0\\1\\0\end{array}\right$$,\\T\left$$\begin{array}{c}0\\1\\1\end{array}\right$$=\left$$\begin{array}{c}0\\1\\1\\1\end{array}\right$$\nonumber\\

Explain why \\T\\ is an isomorphism. Determine a matrix \\A\\ which, when multiplied on the left gives the same result as \\T\\ on \\V\\ and a matrix \\B\\ which delivers \\T^{−1}\\ on \\W\\. Hint: You need to have \\A\left$$\begin{array}{cc}1&0\\0&1\\1&1\end{array}\right$$=\left$$\begin{array}{cc}1&0\\0&1\\1&1\\0&1\end{array}\right$$\nonumber\\

Now enlarge \\\left$$\begin{array}{c}1\\0\\1\end{array}\right$$\\, \\\left$$\begin{array}{c}0\\1\\1\end{array}\right$$\\ to obtain a basis for \\\mathbb{R}^3\\. You could add in \\\left$$\begin{array}{c}0\\0\\1\end{array}\right$$\\ for example, and then pick another vector in \\\mathbb{R}^4\\ and let \\A\left$$\begin{array}{c}0\\0\\1\end{array}\right$$\\ equal this other vector. Then you would have \\A\left$$\begin{array}{ccc}1&0&0\\0&1&0\\1&1&1\end{array}\right$$=\left$$\begin{array}{ccc}1&0&0\\0&1&0\\1&1&0\\0&1&1\end{array}\right$$\nonumber\\

This would involve picking for the new vector in \\\mathbb{R}^4\\ the vector \\\left$$\begin{array}{cccc}0&0&0&1\end{array}\right$$^T\\. Then you could find \\A\\. You can do something similar to find a matrix for \\T^{-1}\\ denoted as \\B\\.

Exercise \\\PageIndex{86}\\

Let \\V=\mathbb{R}^3\\ and let \\W=span(S),\text{ where }S=\left\\\left$$\begin{array}{r}1\\-1\\1\end{array}\right$$,\\\left$$\begin{array}{r}-2\\2\\-2\end{array}\right$$,\\\left$$\begin{array}{r}-1\\1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\-1\\3\end{array}\right$$\right\\\nonumber\\ Find a basis of \\W\\ consisting of vectors in \\S\\.

Answer

In this case \\\text{dim}(W) = 1\\ and a basis for \\W\\ consisting of vectors in \\S\\ can be obtained by taking any (nonzero) vector from \\S\\.

Exercise \\\PageIndex{87}\\

Let \\T\\ be a linear transformation given by \\T\left$$\begin{array}{c}x\\y\end{array}\right$$=\left$$\begin{array}{cc}1&1\\1&1\end{array}\right$$\left$$\begin{array}{c}x\\y\end{array}\right$$\nonumber\\ Find a basis for \\\text{ker}(T)\\ and \\Im(T)\\.

Answer

A basis for \\\text{ker}(T)\\ is \\\left\\\left$$\begin{array}{r}1\\-1\end{array}\right$$\right\\\\ and a basis for \\Im(T)\\ is \\\left\\\left$$\begin{array}{r}1\\1\end{array}\right$$\right\\\\. There are many other possibilities for the specific bases, but in this case \\\text{dim}(\text{ker}(T)) = 1\\ and \\\text{dim}(Im(T)) = 1\\.

Exercise \\\PageIndex{88}\\

Let \\T\\ be a linear transformation given by \\T\left$$\begin{array}{c}x\\y\end{array}\right$$=\left$$\begin{array}{cc}1&0\\1&1\end{array}\right$$\left$$\begin{array}{c}x\\y\end{array}\right$$\nonumber\\ Find a basis for \\\text{ker}(T)\\ and \\Im(T)\\.

Answer

In this case \\\text{ker}(T) = \\0\\\\ and \\Im(T) = \mathbb{R}^2\\ (pick any basis of \\\mathbb{R}^2\$.

Exercise \\\PageIndex{89}\\

Let \\V=\mathbb{R}^3\\ and let \\W=span\left\\\left$$\begin{array}{c}1\\1\\1\end{array}\right$$,\\\left$$\begin{array}{r}-1\\2\\-1\end{array}\right$$\right\\\nonumber\\ Extend this basis of \\W\\ to a basis of \\V\\.

Answer

There are many possible such extensions, one is (how do we know?): \\\left\\\left$$\begin{array}{r}1\\1\\1\end{array}\right$$,\\\left$$\begin{array}{r}-1\\2\\-1\end{array}\right$$,\\\left$$\begin{array}{c}0\\0\\1\end{array}\right$$\right\\\nonumber\\

Exercise \\\PageIndex{90}\\

Let \\T\\ be a linear transformation given by \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{ccc}1&1&1\\1&1&1\end{array}\right$$\left$$\begin{array}{c}x\\y\\z\end{array}\right$$\nonumber\\ What is \\\text{dim}(\text{ker}(T))\\?

Answer

We can easily see that \\\text{dim}(Im(T)) = 1\\, and thus \\\text{dim}(\text{ker}(T)) = 3−\text{dim}(Im(T)) = 3−1 = 2\\.

Exercise \\\PageIndex{91}\\

Consider the following functions which map \\\mathbb{R}^n\\ to \\\mathbb{R}^n\\.

1. \\T\\ multiplies the \\j\\th component of \\\vec{x}\\ by a nonzero number \\b\\.

2. \\T\\ replaces the \\i\\th component of \\\vec{x}\\ with \\b\\ times the \\j\\th component added to the \\i\\h component.

3. \\T\\ switches the \\i\\th and \\j\\th components.

Show these functions are linear transformations and describe their matrices \\A\\ such that \\T (\vec{x}) = A\vec{x}\\.

Answer

1. The matrix of \\T\\ is the elementary matrix which multiplies the \\j\\th diagonal entry of the identity matrix by \\b\\.

2. The matrix of \\T\\ is the elementary matrix which takes \\b\\ times the \\j\\th row and adds to the \\i\\th row.

3. The matrix of \\T\\ is the elementary matrix which switches the \\i\\th and the \\j\\th rows where the two components are in the \\i\\th and \\j\\th positions.

Exercise \\\PageIndex{92}\\

You are given a linear transformation \\T\\: \\\mathbb{R}^n → \mathbb{R}^m\\ and you know that \\T(A_i)=B_i\nonumber\\ where \\\left$$\begin{array}{ccc}A_1&\cdots&A_n\end{array}\right$$^{-1}\\ exists. Show that the matrix of \\T\\ is of the form \\\left$$\begin{array}{ccc}B_1&\cdots&B_n\end{array}\right$$\\\left$$\begin{array}{ccc}A_1&\cdots&A_n\end{array}\right$$^{-1}\nonumber\\

Answer

Suppose \\\left$$\begin{array}{c}\vec{c}\_1^T \\ \vdots \\ \vec{c}\_n^T\end{array}\right$$=\left$$\begin{array}{ccc}\vec{a}\_1&\cdots&\vec{a}\_n\end{array}\right$$^{-1}\nonumber\\ Thus \\\vec{c}\_i^T\vec{a}\_j=\delta\_{ij}\\. Therefore \\\begin{aligned} \left$$\begin{array}{ccc}\vec{b}\_1&\cdots&\vec{b}\_n\end{array}\right$$\\ \left$$\begin{array}{ccc}\vec{a}\_1&\cdots&\vec{a}\_n\end{array}\right$$^{-1}\vec{a}\_i &=\left$$\begin{array}{ccc}\vec{b}\_1&\cdots&\vec{b}\_n\end{array}\right$$\\\left$$\begin{array}{c}\vec{c}\_1^T \\ \vdots \\ \vec{c}\_n^T\end{array}\right$$ \vec{a}\_i \\ &=\left$$\begin{array}{ccc}\vec{b}\_1&\cdots&\vec{b}\_n\end{array}\right$$ \vec{e}\_i \\ &=\vec{b}\_i\end{aligned}\\ Thus \\T\vec{a}\_i=\left$$\begin{array}{ccc}\vec{b}\_1&\cdots&\vec{b}\_n\end{array}\right$$\\ \left$$\begin{array}{ccc}\vec{a}\_1&\cdots&\vec{a}\_n\end{array}\right$$^{-1}\vec{a}\_i=A\vec{a}\_i\\. If \\\vec{x}\\ is arbitrary, then since the matrix \\\left$$\begin{array}{ccc}\vec{a}\_1&\cdots&\vec{a}\_n\end{array}\right$$\\ is invertible, there exists a unique \\\vec{y}\\ such that \\\left$$\begin{array}{ccc}\vec{a}\_1&\cdots&\vec{a}\_n\end{array}\right$$\vec{y}=\vec{x}\\ Hence \\T\vec{x}=T\left(\sum\limits\_{i=1}^ny_i\vec{a}\_i\right)=\sum\limits\_{i=1}^ny_iT\vec{a}\_i=\sum\limits\_{i=1}^ny_1A\vec{a}\_i=A\left(\sum\limits\_{i=1}^ny_i\vec{a}\_i\right)=A\vec{x}\nonumber\\

Exercise \\\PageIndex{93}\\

Suppose \\T\\ is a linear transformation such that \\\begin{aligned}T\left$$\begin{array}{r}1\\2\\-6\end{array}\right$$&=\left$$\begin{array}{c}5\\1\\3\end{array}\right$$ \\ T\left$$\begin{array}{r}-1\\-1\\5\end{array}\right$$&=\left$$\begin{array}{c}1\\1\\5\end{array}\right$$ \\ T\left$$\begin{array}{r}0\\-1\\2\end{array}\right$$&=\left$$\begin{array}{r}5\\3\\-2\end{array}\right$$\end{aligned}\\ Find the matrix of \\T\\. That is find \\A\\ such that \\T(\vec{x}) = A\vec{x}\\.

Answer

\\\left$$\begin{array}{rrr}5&1&5\\1&1&3\\3&5&-2\end{array}\right$$\left$$\begin{array}{ccc}3&2&1\\2&2&1\\4&1&1\end{array}\right$$=\left$$\begin{array}{ccc}37&17&11\\17&7&5\\11&14&6\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{94}\\

Suppose \\T\\ is a linear transformation such that \\\begin{aligned}T\left$$\begin{array}{r}1\\1\\-8\end{array}\right$$&=\left$$\begin{array}{c}1\\3\\1\end{array}\right$$ \\ T\left$$\begin{array}{r}-1\\0\\6\end{array}\right$$&=\left$$\begin{array}{c}2\\4\\1\end{array}\right$$ \\ T\left$$\begin{array}{r}0\\-1\\3\end{array}\right$$&=\left$$\begin{array}{r}6\\1\\-1\end{array}\right$$\end{aligned}\\ Find the matrix of \\T\\. That is find \\A\\ such that \\T(\vec{x}) = A\vec{x}\\.

Answer

\\\left$$\begin{array}{rrr}1&2&6\\3&4&1\\1&1&-1\end{array}\right$$\left$$\begin{array}{ccc}6&3&1\\5&3&1\\6&2&1\end{array}\right$$=\left$$\begin{array}{ccc}52&21&9\\44&23&8\\5&4&1\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{95}\\

Suppose \\T\\ is a linear transformation such that \\\begin{aligned}T\left$$\begin{array}{r}1\\3\\-7\end{array}\right$$&=\left$$\begin{array}{c}-3\\1\\3\end{array}\right$$ \\ T\left$$\begin{array}{r}-1\\-2\\6\end{array}\right$$&=\left$$\begin{array}{4}1\\3\\-3\end{array}\right$$ \\ T\left$$\begin{array}{r}0\\-1\\2\end{array}\right$$&=\left$$\begin{array}{r}5\\3\\-3\end{array}\right$$\end{aligned}\\ Find the matrix of \\T\\. That is find \\A\\ such that \\T(\vec{x}) = A\vec{x}\\.

Answer

\\\left$$\begin{array}{rrr}-3&1&5\\1&3&3\\3&-3&-3\end{array}\right$$\left$$\begin{array}{ccc}2&2&1\\1&2&1\\4&1&1\end{array}\right$$=\left$$\begin{array}{rrr}15&1&3\\17&11&7\\-9&-3&-3\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{96}\\

Suppose \\T\\ is a linear transformation such that \\\begin{aligned}T\left$$\begin{array}{r}1\\1\\-7\end{array}\right$$&=\left$$\begin{array}{c}3\\3\\3\end{array}\right$$ \\ T\left$$\begin{array}{r}-1\\0\\6\end{array}\right$$&=\left$$\begin{array}{c}1\\2\\3\end{array}\right$$ \\ T\left$$\begin{array}{r}0\\-1\\2\end{array}\right$$&=\left$$\begin{array}{r}1\\3\\-1\end{array}\right$$\end{aligned}\\ Find the matrix of \\T\\. That is find \\A\\ such that \\T(\vec{x}) = A\vec{x}\\.

Answer

\\\left$$\begin{array}{rrr}3&1&1\\3&2&3\\3&3&-1\end{array}\right$$\left$$\begin{array}{ccc}6&2&1\\5&2&1\\6&1&1\end{array}\right$$=\left$$\begin{array}{ccc}29&9&5\\46&13&8\\27&11&5\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{97}\\

Suppose \\T\\ is a linear transformation such that \\\begin{aligned}T\left$$\begin{array}{r}1\\2\\-18\end{array}\right$$&=\left$$\begin{array}{c}5\\2\\5\end{array}\right$$ \\ T\left$$\begin{array}{r}-1\\-1\\15\end{array}\right$$&=\left$$\begin{array}{c}3\\3\\5\end{array}\right$$ \\ T\left$$\begin{array}{r}0\\-1\\4\end{array}\right$$&=\left$$\begin{array}{r}2\\5\\-2\end{array}\right$$\end{aligned}\\ Find the matrix of \\T\\. That is find \\A\\ such that \\T(\vec{x}) = A\vec{x}\\.

Answer

\\\left$$\begin{array}{rrr}5&3&2\\2&3&5\\5&5&-2\end{array}\right$$\left$$\begin{array}{ccc}11&4&1\\10&4&1\\12&3&1\end{array}\right$$=\left$$\begin{array}{ccc}109&38&10 \\112&35&10\\81&34&8\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{98}\\

Consider the following functions \\T\\: \\\mathbb{R}^3 → \mathbb{R}^2\\. Show that each is a linear transformation and determine for each the matrix \\A\\ such that \\T(\vec{x}) = A\vec{x}\\.

1. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}x+2y+3z \\ 2y-3x+z\end{array}\right$$\\

2. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}7x+2y+z \\ 3x-11y+2z\end{array}\right$$\\

3. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}3x+2y+z \\ x+2y+6z\end{array}\right$$\\

4. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}2y-5x+z \\ x+y+z\end{array}\right$$\\

Exercise \\\PageIndex{99}\\

Consider the following functions \\T\\: \\\mathbb{R}^3 → \mathbb{R}^2\\. Explain why each of these functions \\T\\ is not linear.

1. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}x+2y+3z+1 \\ 2y-3x+z\end{array}\right$$\\

2. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}x+2y^2+3z \\ 2y+3x+z\end{array}\right$$\\

3. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}\sin x+2y+3z \\ 2y+3x+z\end{array}\right$$\\

4. \\T\left$$\begin{array}{c}x\\y\\z\end{array}\right$$=\left$$\begin{array}{c}x+2y+3z \\ 2y+3x-\ln z\end{array}\right$$\\

Exercise \\\PageIndex{100}\\

Suppose \\\left$$\begin{array}{ccc}A_1&\cdots&A_n\end{array}\right$$^{-1}\nonumber\\ exists where each \\A_j ∈ \mathbb{R}^n\\ and let vectors \\\\B_1,\cdots ,B_n\\\\ in \\\mathbb{R}^m\\ be given. Show that there always exists a linear transformation \\T\\ such that \\T(A_i) = B_i\\.

Exercise \\\PageIndex{101}\\

Find the matrix for \\T (\vec{w}) = \text{proj}\_{\vec{v}} (\vec{w})\\ where \\\vec{v}=\left$$\begin{array}{ccc}1&-2&3\end{array}\right$$^T\\.

Answer

Recall that \\\text{proj}\_{\vec{u}}(\vec{v}) = \frac{\vec{v}\bullet\vec{u}}{\|\|\vec{u}\|\|^2}\vec{u}\\ and so the desired matrix has \\i\\th column equal to \\\text{proj}\_{\vec{u}} (\vec{e}\_i)\\. Therefore, the matrix desired is \\\frac{1}{14}\left$$\begin{array}{rrr}1&-2&3\\-2&4&-6\\3&-6&9\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{102}\\

Find the matrix for \\T (\vec{w}) = \text{proj}\_{\vec{v}} (\vec{w})\\ where \\\vec{v}=\left$$\begin{array}{ccc}1&5&3\end{array}\right$$^T\\.

Answer

\\\frac{1}{35}\left$$\begin{array}{ccc}1&5&3\\5&25&15\\3&15&9\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{103}\\

Find the matrix for \\T (\vec{w}) = \text{proj}\_{\vec{v}} (\vec{w})\\ where \\\vec{v}=\left$$\begin{array}{ccc}1&0&3\end{array}\right$$^T\\.

Answer

\\\frac{1}{10}\left$$\begin{array}{ccc}1&0&3\\0&0&0\\3&0&9\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{104}\\

Let \\B=\left\\\left$$\begin{array}{r}2\\-1\end{array}\right$$,\\\left$$\begin{array}{c}3\\2\end{array}\right$$\right\\\\ be a basis of \\\mathbb{R}^2\\ and let \\\vec{x}=\left$$\begin{array}{r}5\\-7\end{array}\right$$\\ be a vector in \\\mathbb{R}^2\\. Find \\C_B(\vec{x})\\.

Exercise \\\PageIndex{105}\\

Let \\B=\left\\\left$$\begin{array}{r}1\\-1\\2\end{array}\right$$,\\\left$$\begin{array}{c}2\\1\\2\end{array}\right$$,\\\left$$\begin{array}{r}-1\\0\\2\end{array}\right$$\right\\\\ be a basis of \\\mathbb{R}^3\\ and let \\\vec{x}=\left$$\begin{array}{r}5\\-1\\4\end{array}\right$$\\ be a vector in \\\mathbb{R}^2\\. Find \\C_B(\vec{x})\\.

Answer

\\C_B(\vec{x})=\left$$\begin{array}{r}2\\1\\-1\end{array}\right$$\\.

Exercise \\\PageIndex{106}\\

Let \\T\\: \\\mathbb{R}^2\mapsto \mathbb{R}^2\\ be a linear transformation defined by \\T\left(\left$$\begin{array}{c}a\\b\end{array}\right$$\right)=\left$$\begin{array}{c}a+b\\a-b\end{array}\right$$\\.

Consider the two bases \\B_1=\\\vec{v}\_1,\vec{v}\_2\\=\left\\\left$$\begin{array}{c}1\\0\end{array}\right$$,\\\left$$\begin{array}{r}-1\\1\end{array}\right$$\right\\\nonumber\\ and \\B_2=\left\\\left$$\begin{array}{c}1\\1\end{array}\right$$,\\\left$$\begin{array}{r}1\\-1\end{array}\right$$\right\\\nonumber\\ Find the matrix \\M\_{B_2,B_1}\\ of \\T\\ with respect to the bases \\B_1\\ and \\B_2\\.

Answer

\\M\_{B_2B_1}=\left$$\begin{array}{rr}1&0\\-1&1\end{array}\right$$\\

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