Vector Spaces 向量空间

The following corollary follows from the exchange theorem.

Corollary $\PageIndex{2}$: Two Bases of the Same Length 推论 2：等长两基

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 $\sum\_{i=1}^{m}c\_{ij}\vec{u}\_{i}=\vec{v}\_{j}$. Therefore, $\sum\_{j=1}^{n}d\_{j}\vec{v}\_{j}=\vec{0} \text{ if and only if }\sum\_{j=1}^{n}\sum\_{i=1}^{m}c\_{ij}d\_{j}\vec{u}\_{i}= \vec{0}$ if and only if $\sum\_{i=1}^{m}\left( \sum\_{j=1}^{n}c\_{ij}d\_{j}\right) \vec{u}\_{i}=\vec{0}$. Now since $$\vec{u}\_{1},\cdots ,\vec{u}\_{n}$$ is independent, this happens if and only if $\sum\_{j=1}^{n}c\_{ij}d\_{j}=0, i=1,2,\cdots ,m$. 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 定义 3：维数

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 示例 5：向量空间的维数

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 $S = \left\{ x^2, x, 1 \right\}$. 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 示例 6：二次多项式的另一个基

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 定理 4：每个子空间都有基

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 定理 5：同维子空间

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 示例 7：子空间的基

Let $U=\left\{ A\in\mathbb{M}\_{22} ~\left\|~ A\left[ \begin{array}{rr} 1 & 0 \\ 1 & -1 \end{array}\right] = \left[ \begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right] A \right\}$. 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 $A\left[ \begin{array}{rr} 1 & 0 \\ 1 & -1 \end{array}\right] = \left[ \begin{array}{rr} a & b \\ c & d \end{array}\right] \left[ \begin{array}{rr} 1 & 0 \\ 1 & -1 \end{array}\right] =\left[ \begin{array}{rr} a+b & -b \\ c+d & -d \end{array}\right]$ and $\left[ \begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right] A = \left[ \begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right] \left[ \begin{array}{rr} a & b \\ c & d \end{array}\right] =\left[ \begin{array}{rr} a+c & b+d \\ -c & -d \end{array}\right]$. 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 $d$.

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

or

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

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 $A=\left[ \begin{array}{cc} s & \frac{t}{2} \\ -\frac{t}{2} & t \end{array}\right] = s\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right] + t\left[ \begin{array}{rr} 0 & -\frac{1}{2} \\ -\frac{1}{2} & 1 \end{array}\right]$. Let $B=\left\{ \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right], \left[ \begin{array}{rr} 0 & -\frac{1}{2} \\ -\frac{1}{2} & 1 \end{array}\right] \right\}$. 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 $V$ 定理 6：向量空间 V 的基

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 $\left\{ \vec{u}\_{1},\cdots ,\vec{u} \_{k}\right\} \subseteq V$ is linearly independent and the vector space is finite dimensional, then the set $\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\}$ can be enlarged to obtain a basis of $V$.

Proof.

Let $S=\left\{ E\subseteq \left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\} \text{ such that } \mathrm{span}\left\{ E\right\} =V\right\}$. For $E\in S$, let $\left\vert E\right\vert$ denote the number of elements of $E$. Let $m= \min \left\{ \left\vert E\right\vert \mid E\in S\right\}$. Thus there exist vectors $\left\{ \vec{v}\_{1},\cdots ,\vec{v}\_{m}\right\} \subseteq \left\{ \vec{u}\_{1},\cdots , \vec{u}\_{n}\right\}$ such that $\mathrm{span}\left\{ \vec{v}\_{1},\cdots ,\vec{v}\_{m}\right\} =V$ 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 $\vec{0}=\sum\_{i=1}^{m}c\_{i}\vec{v}\_{i}$ 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, $V=\mathrm{span}\left\{ \vec{v}\_{1},\cdots ,\vec{v}\_{k-1},\vec{v} \_{k+1},\cdots ,\vec{v}\_{m}\right\}$ 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 $\left\{ \vec{v}\_{1},\cdots ,\vec{v}\_{n}\right\}$. If $\mathrm{span}\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\} =V$, then $k=n$. If not, there exists a vector $\vec{u}\_{k+1}\notin \mathrm{span}\left\{ \vec{u}\_{1},\cdots ,\vec{u} \_{k}\right\}$. 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 $\mathrm{span}\left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{n}\right\} =V$ 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 示例 8：扩充线性无关集

Let $S \subseteq M\_{22}$ be a linearly independent set given by $S = \left\{ \left[ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right], \left[ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array}\right] \right\}$. 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 $R = \left\{ \left[ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right], \left[ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array}\right], \left[ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array}\right] \right\}$ 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 $T = \left\{ \left[ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right], \left[ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array}\right], \left[ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array}\right], \left[ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array}\right] \right\}$

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 定理 7：由生成集得到基

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 $\mathrm{span} \left\{ \vec{u}\_{1},\cdots ,\vec{u}\_{k}\right\} =W$. If $\sum\_{i=1}^{k}c\_{i}\vec{w}\_{i}=\vec{0}$ 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. $\vec{w}\_{j}=\sum\_{i\neq j}\left( -\frac{c\_{i}}{c\_{j}}\right) \vec{w}\_{i}$ 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 示例 9：由生成集得到基

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 $\left[ \begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array}\right]$ 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 $\mathbb{R}^{4}$. $\left[ \begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array}\right] ,\left[ \begin{array}{c} 2 \\ 2 \\ 4 \\ 1 \end{array}\right] ,\left[ \begin{array}{c} 1 \\ 2 \\ 2 \\ 2 \end{array}\right] ,\left[ \begin{array}{r} 2 \\ -3 \\ 4 \\ 1 \end{array}\right] ,\left[ \begin{array}{c} 1 \\ 2 \\ 3 \\ 1 \end{array}\right]$ 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 $\left[ \begin{array}{rrrrr} 1 & 0 & 0 & -15 & 0 \\ 0 & 1 & 0 & 11 & 0 \\ 0 & 0 & 1 & -5 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right]$ 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 示例 10：缩减生成集

Consider the set $S \subseteq \mathbb{P}\_2$ given by $S = \left\{ 1, x, x^2, x^2 + 1 \right\}$. 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 $ax^2 + bx + c = r(1) + s(x) + t(x^2) + u (x^2 + 1)$. 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 定理 8：向量空间的基

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 $\left\{ \vec{w}\_{1},\cdots ,\vec{w}\_{r}\right\} =W$. 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 $\mathrm{span}\left\{ \vec{w}\_{1},\cdots ,\vec{w} \_{r}\right\} =W$, stop. You have the desired basis. Otherwise, there exists $\vec{w}\_{s+1}\notin \mathrm{span}\left\{ \vec{w}\_{1},\cdots ,\vec{w} \_{s}\right\}$ 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 推论 3：基的扩充

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 示例 11：基的扩充

Let $V=\mathbb{R}^{4}$ and let $W=\mathrm{span}\left\{ \left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array}\right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\}$. 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 $\left[ \begin{array}{cccccc} 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \end{array}\right]$. 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 $\left[ \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 1 & -1 \end{array}\right]$. These are $\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array}\right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array}\right] ,\left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array}\right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right]$ 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 上述矩阵 obviously span $\mathbb{R} ^{4}$ and the span of the first four is the same as the span of all six.

Outcomes 学习目标

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

1. 证明两个子空间的和是子空间。
2. 证明两个子空间的交是子空间。

We begin this section with a definition.

Definition $\PageIndex{1}$: Sum and Intersection 定义 1：子空间的和与交

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

1. $U+W = \left\{ \vec{u}+\vec{w} ~\|~ \vec{u}\in U\mbox{ and } \vec{w}\in W\right\}$ and is called the sum of $U$ and $W$.
2. $U\cap W = \left\{ \vec{v} ~\|~ \vec{v}\in U\mbox{ and } \vec{v}\in W\right\}$ and is called the intersection of $U$ and $W$.

1. $U+W = \left\{ \vec{u}+\vec{w} ~\|~ \vec{u}\in U\mbox{ 且 } \vec{w}\in W\right\}$ 称为 $U$ 与 $W$ 的和。
2. $U\cap W = \left\{ \vec{v} ~\|~ \vec{v}\in U\mbox{ 且 } \vec{v}\in W\right\}$ 称为 $U$ 与 $W$ 的交。

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 定义 2：直和

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 示例 1：交是子空间

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 $\vec{v} \in U \cap W$, $a\vec{v} \in U \cap W$

1. $\vec{0} \in U \cap W$
2. 对任意向量 $\vec{v}_1, \vec{v}_2 \in U \cap W$，有 $\vec{v}_1+\vec{v}_2 \in U \cap W$
3. 对任意标量 $a$ 与向量 $\vec{v} \in U \cap W$，有 $a\vec{v} \in U \cap W$

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 $U \cap W$.

1. 由于 $U$ 与 $W$ 都是 $V$ 的子空间，它们都含有 $\vec{0}$。由交的定义，$\vec{0} \in U \cap W$。
2. 设 $\vec{v}_1, \vec{v}_2 \in U \cap W$。特别地，$\vec{v}_1, \vec{v}_2 \in U$。由于 $U$ 是子空间，故 $\vec{v}_1+\vec{v}_2 \in U$。同理对 $W$ 也成立。因此 $\vec{v}_1+\vec{v}_2$ 既属于 $U$ 又属于 $W$，由定义也属于 $U \cap W$。
3. 设 $a$ 为标量且 $\vec{v} \in U \cap W$。特别地，$\vec{v} \in U$。由于 $U$ 是子空间，故 $a \vec{v} \in U$。同理对 $W$ 也成立，所以 $a\vec{v}$ 既属于 $U$ 又属于 $W$。由定义，它属于 $U \cap W$。

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 定理 1：和的维数

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 $\mathrm{dim} (U+W) = \mathrm{dim}(U) + \mathrm{dim}(W) - \mathrm{dim} (U \cap W)$

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

Outcomes 学习目标

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

1. 理解向量空间意义下线性变换的定义。

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

Definition 1: Linear Transformation 定义 1：线性变换

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) $

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

Example 1: Linear Transformations 例 1：线性变换

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 $s_a(\vec{v})=a\vec{v}\text{ for all }\vec{v}\in V$.

1. 零变换 $0:V\to W$ 定义为：对所有 $\vec{v}\in V$，$0(\vec{v})=\vec{0}$。
2. 恒等变换 $1_V:V\to V$ 定义为：对所有 $\vec{v}\in V$，$1_V(\vec{v})=\vec{v}$。
3. 标量变换 设 $a\in\mathbb{R}$。$s_a:V\to V$ 定义为 $s_a(\vec{v})=a\vec{v}\text{ for all }\vec{v}\in V$，即对所有 $\vec{v}\in V$ 取 $a\vec{v}$。

Solution.

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

By Definition 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 1: Properties of Linear Transformations 定理 1：线性变换的性质

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

1. $T$ 保持零向量：$T(\vec{0})=\vec{0}$。
2. $T$ 保持加法逆元：对所有 $\vec{v}\in V$，$T(-\vec{v})= -T(\vec{v})$。
3. $T$ 保持线性组合：对所有 $\vec{v}\_1, \vec{v}\_2, \ldots, \vec{v}\_m \in V$ 与所有 $k_1, k_2, \ldots, k_m\in\mathbb{R}$，$T(k_1\vec{v}\_1 + k_2\vec{v}\_2 + \cdots + k_m\vec{v}\_m) = k_1T(\vec{v}\_1) + k_2T(\vec{v}\_2) + \cdots + k_mT(\vec{v}\_m)$。

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. $
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 $m$.

1. 以 $\vec{0}\_V$ 记 $V$ 的零向量，以 $\vec{0}\_W$ 记 $W$ 的零向量。要证 $T(\vec{0}\_V)=\vec{0}\_W$。取 $\vec{v}\in V$，则 $0\vec{v}=\vec{0}\_V$，于是 $T(\vec{0}\_V)=T(0\vec{v})=0T(\vec{v})=\vec{0}\_W. $
2. 取 $\vec{v}\in V$；则 $-\vec{v}\in V$ 是 $\vec{v}$ 的加法逆元，故 $\vec{v} + (-\vec{v})=\vec{0}\_V$。于是 $\begin{aligned} T(\vec{v} + (-\vec{v})) & = T(\vec{0}\_V) \\ T(\vec{v}) + T(-\vec{v})) & = \vec{0}\_W \\ T(-\vec{v}) & = \vec{0}\_W - T(\vec{v}) = - T(\vec{v}).\end{aligned}$
3. 这一结论由保持加法与保持标量乘法直接得到。严格的证明可对 $m$ 作归纳。

Consider the following example using the above theorem.

Example 2: Linear Combination 例 2：线性组合

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. $ 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 $(a+b+c)x^2 + (a-b)x + c = 4x^2+5x-3.$ 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 2: Equal Transformations 定义 2：相等的变换

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) $

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 2: Transformation of a Spanning Set 定理 2：张成集上的变换

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 3: Transformation of a Basis 定理 3：基上的变换

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 $ 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. $

Outcomes 学习目标

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.

1. 把单射与满射的概念应用于向量空间之间的变换。
2. 判定向量空间之间的线性变换是否为同构。
3. 判定两个向量空间是否同构。

One to One and Onto Transformations 单射与满射变换

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

Definition 1: One to One Transformation 定义 1：单射变换

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) $

Definition 2: Onto Transformation 定义 2：满射变换

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 1: One to One 引理 1：单射

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}) $ 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 1: One to One Transformation 例 1：单射变换

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

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

Solution.

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

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:

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

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 $A=\left[\begin{array}{cc} 0 & 1 \\ 0 & 2 \end{array}\right],$ and suppose $p(x)=ax^2+bx+c\in \mathbb{P}\_2$ is such that $S(p(x))=A$. Then $\begin{array}{ll} a+b=0 & a+c=1 \\ b-c=0 & b+c=2 \end{array}$ Solving this system $\left[\begin{array}{ccc\|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 2 \end{array}\right] \rightarrow \left[\begin{array}{rrr\|r} 1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 2 \end{array}\right].$

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 2: An Onto Transformation 例 2：一个满射变换

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

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

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 3: One to One and Independence 例 3：单射与线性无关性

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 $a_1T(\vec{v}\_1) + a_2T(\vec{v}\_2) +\cdots +a_kT(\vec{v}\_k) =\vec{0}\_W$ for some $a_1, a_2, \ldots, a_k\in\mathbb{R}$. Since linear transformations preserve linear combinations (addition and scalar multiplication), $T(a_1\vec{v}\_1 + a_2\vec{v}\_2 +\cdots +a_k\vec{v}\_k) =\vec{0}\_W.$

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

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 4: Onto and Spanning 例 4：满射与张成

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 $W=span\{T(\vec{v}\_1), T(\vec{v}\_2), \ldots, T(\vec{v}\_k)\}.$

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 $W\subseteq span\{T(\vec{v}\_1), T(\vec{v}\_2), \ldots, T(\vec{v}\_k)\}.$

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 同构

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 3: Isomorphism 定义 3：同构

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.

• $T$ is one to one.
• $T$ is onto.

• $T$ 是单射。
• $T$ 是满射。

Definition 4: Isomorphic 定义 4：同构的

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 5: Isomorphism 例 5：同构

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}. $ 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.

1. $T$ 是线性变换；
2. $T$ 是单射；
3. $T$ 是满射。

$T$ 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 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] $

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 $

Hence $T$ is one-to-one.

$T$ is onto: Let

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

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 5: Composition of Transformations 定义 5：变换的复合

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 $ and is defined by $(S \circ T) (\vec{v}) = S(T(\vec{v})) \mbox{ for all } \vec{v} \in V $

Consider now the following proposition.

Proposition 1: Composite and Inverse Isomorphism 命题 1：复合同构与逆同构

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) $ 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}? $ 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}) $ 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} $ 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:

• $V\sim V$
• If $V\sim W,$ it follows that $W\sim V$
• If $V\sim W$ and $W\sim Z,$ then $V\sim Z$

• $V\sim V$
• 若 $V\sim W,$ 则 $W\sim V$
• 若 $V\sim W$ 且 $W\sim Z,$ 则 $V\sim Z$

We leave the proof of these to the reader.

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

Lemma 2: Bases and Isomorphisms 引理 2：基与同构

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} $ Then, since $T$ is linear, $T\left( \sum\_{i=1}^{n}c\_{i}\vec{u}\_{i}\right) =\vec{0} $ Since $T$ is one to one, it follows that $\sum\_{i=1}^{n}c\_{i}\vec{u}\_{i}=0 $ 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}. $ 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} $ 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 1: Isomorphic Vector Spaces 定理 1：同构的向量空间

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.

1. $T$ 是单射。
2. $T$ 是满射。
3. $T$ 是同构。

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} $ 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}. $ 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} $ Then $\sum\_{i=1}^{n}\left( c\_{i}-\hat{c}\_{i}\right) \vec{v}\_{i}=0 $ 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} $ 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}, $ 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) $ 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 $ 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) $ 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} $ Then $\sum\_{i=1}^{n}c\_{i}T\vec{v}\_{i}=\vec{0} $ 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} $ 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 6 例 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:

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

Outcomes 学习目标

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.

1. 描述线性变换的核与像。
2. 利用核与像判断线性变换是否单射或满射。

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 定义 \PageIndex{1}：核与像

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 $\left\{ T(\vec{v}):\vec{v}\in V\right\}$. 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, $\ker \left( T\right) =\left\{ \vec{v}\in V:T(\vec{v})=\vec{0}\right\}$.

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 命题 \PageIndex{1}：核与像作为子空间

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 $T\left( a\vec{v}_{1}+b\vec{v}_{2}\right) =aT(\vec{v}_{1})+bT(\vec{v}_{2})=a\vec{0}+b\vec{0}=\vec{0}$.

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, $aT(\vec{v}_{2})+bT(\vec{v}_{2})=T\left( a\vec{v}_{1}+b\vec{v}_{2}\right)$ 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 示例 \PageIndex{1}：变换的核与像

Let $T:\mathbb{P}_1\to\mathbb{R}$ be the linear transformation defined by $T(p(x))=p(1)\mbox{ for all } p(x)\in \mathbb{P}_1$. 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. $\begin{aligned} \mathrm{ker}(T) & = \{ p(x)\in \mathbb{P}_1 \mid p(1)=0\} \\ & = \{ ax+b \mid a,b\in\mathbb{R} \mbox{ and }a+b=0\} \\ & = \{ ax-a \mid a\in\mathbb{R} \}\end{aligned}$ Therefore a basis for $\mathrm{ker}(T)$ is $\left\{ x-1 \right\}$. 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 $1$. $\begin{aligned} \mathrm{im}(T) & = \{ p(1) \mid p(x)\in \mathbb{P}_1 \} \\ & = \{ a+b \mid ax+b\in \mathbb{P}_1 \} \\ & = \{ a+b \mid a,b\in\mathbb{R} \} \\ & = \mathbb{R}\end{aligned}$ Therefore a basis for $\mathrm{im}(T)$ is $\left\{ 1 \right\}$. 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 示例 \PageIndex{2}：线性变换的核与像

Let $T: \mathbb{M}_{22} \mapsto \mathbb{R}^2$ be defined by $T \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{c} a - b \\ c + d \end{array} \right]$. 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 $T \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{c} a - b \\ c + d \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \end{array} \right]$. 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. $\mathrm{ker}(T) = \left\{ \left[ \begin{array}{cc} s & s \\ t & -t \end{array} \right] \right\} = \mathrm{span} \left\{ \left[ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right] \right\}$. 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 $\mathrm{im}(T) = \left\{ \left[ \begin{array}{c} a - b \\ c + d \end{array} \right] \right\}$. Notice that this can be written as $\mathrm{span} \left\{ \left[ \begin{array}{c} 1 \\ 0 \end{array}\right], \left[ \begin{array}{c} -1 \\ 0 \end{array}\right], \left[ \begin{array}{c} 0 \\ 1 \end{array}\right], \left[ \begin{array}{c} 0 \\ 1 \end{array}\right] \right\}$.

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

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 定理 \PageIndex{1}：核的维数 + 像的维数

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 $T(\vec{v})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})$. 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 $\vec{v}-\sum_{i=1}^{r}c_{i}\vec{v}_{i}=\sum_{j=1}^{s}a_{j}\vec{u}_{j}$. 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 $V=\mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots , \vec{v}_{r}\right\}$. 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 $\sum_{i=1}^{r}c_{i}\vec{v}_{i}+\sum_{j=1}^{s}a_{j}\vec{u}_{j}=0$. Apply $T$ to both sides to obtain $\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})+\sum_{j=1}^{s}a_{j}T(\vec{u}_{j})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})= \vec{0}$. 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

Consider the following definition.

Definition $\PageIndex{2}$: Rank of Linear Transformation 定义 \PageIndex{2}：线性变换的秩

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 定理 \PageIndex{2}：同维数的子空间

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 推论 \PageIndex{1}：单射与满射的刻画

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 示例 \PageIndex{3}：单射变换

Let $S:\mathbb{P}_2\to\mathbb{M}_{22}$ be a linear transformation defined by $S(ax^2+bx+c) = \left[ \begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right] \mbox{ for all } ax^2+bx+c\in \mathbb{P}_2$. 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, $\ker(S)=\{ ax^2+bx+c\in \mathbb{P}_2 \mid a+b=0, a+c=0, b-c=0, b+c=0\}$.

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:

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

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

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.

Outcomes 学习目标

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

1. 求线性变换关于向量空间中一般基的矩阵。

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 定义 \PageIndex{1}：坐标同构

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 $C_B(a_1\vec{b}_1 + a_2\vec{b}_2 + \cdots + a_n\vec{b}_n) = a_1\vec{e}_1 + a_2\vec{e}_2 + \cdots + a_n\vec{e}_n = \left[ \begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n \end{array}\right].$ 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 定义 \PageIndex{2}：坐标向量

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 $C_B(\vec{v})$.

Consider the following example.

Example $\PageIndex{1}$: Coordinate Vector 示例 \PageIndex{1}：坐标向量

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. $B = \left\{ x + x^2 , x , 4 \right\}$

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: $-x^2 -2x + 4 = a_1 (1) + a_2 (x) + a_3(x^2)$ Clearly the solution is $\begin{aligned} a_1 &= 4 \\ a_2 &= -2 \\ a_3 &= -1\end{aligned}$ Therefore the coordinate vector is $C_B(\vec{x}) = \left[ \begin{array}{r} 4 \\ -2 \\ -1 \end{array} \right]$
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: $-x^2 -2x + 4 = a_1 (x^2) + a_2 (x) + a_3(1)$ Here the solution is $\begin{aligned} a_1 &= -1 \\ a_2 &= -2 \\ a_3 &= 4\end{aligned}$ Therefore the coordinate vector is $C_B(\vec{x}) = \left[ \begin{array}{r} -1 \\ -2 \\ 4 \end{array} \right]$
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 $C_B(\vec{x})=\left[\begin{array}{r}-1\\-1\\1\end{array}\right]$

1. 首先注意基的顺序很重要。现需找到 $a_1, a_2, a_3$ 使得 $\vec{x} = a_1 (1) + a_2 (x) + a_3(x^2)$，即：$-x^2 -2x + 4 = a_1 (1) + a_2 (x) + a_3(x^2)$。显然解为 $\begin{aligned} a_1 &= 4 \\ a_2 &= -2 \\ a_3 &= -1\end{aligned}$，因此坐标向量为 $C_B(\vec{x}) = \left[ \begin{array}{r} 4 \\ -2 \\ -1 \end{array} \right]$。
2. 再次注意 $B$ 的顺序很重要。按上面同样方法进行。需找到 $a_1, a_2, a_3$ 使得 $\vec{x} = a_1 (x^2) + a_2 (x) + a_3(1)$，即：$-x^2 -2x + 4 = a_1 (x^2) + a_2 (x) + a_3(1)$。此处解为 $\begin{aligned} a_1 &= -1 \\ a_2 &= -2 \\ a_3 &= 4\end{aligned}$，因此坐标向量为 $C_B(\vec{x}) = \left[ \begin{array}{r} -1 \\ -2 \\ 4 \end{array} \right]$。
3. 现需找到 $a_1, a_2, a_3$ 使得 $\vec{x} = a_1 (x + x^2) + a_2 (x) + a_3(4)$，即：$\begin{aligned} -x^2 -2x + 4 &= a_1 (x + x^2 ) + a_2 (x) + a_3(4)\\ &= a_1 (x^2) + (a_1 + a_2) (x) + a_3(4)\end{aligned}$。解为 $\begin{aligned} a_1 &= -1 \\ a_2 &= -1 \\ a_3 &= 1\end{aligned}$，坐标向量为 $C_B(\vec{x})=\left[\begin{array}{r}-1\\-1\\1\end{array}\right]$。

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 定理 \PageIndex{1}：坐标同构的逆

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, $C_B^{-1}:\mathbb{R}^n\to V$ is given by $C_B^{-1} =\left[ \begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n \end{array}\right] = a_1\vec{b}_1 + a_2\vec{b}_2 + \cdots + a_n\vec{b}_n ~\mbox{ for all }~ \left[ \begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n \end{array}\right] \in\mathbb{R}^n$.

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

• $\dim(V)=n$ and $B_1=\{\vec{b}_1, \vec{b}_2, \ldots, \vec{b}_n\}$ is an ordered basis of $V$;
• $\dim(W)=m$ and $B_2$ is an ordered basis of $W$.

• $\dim(V)=n$，且 $B_1=\{\vec{b}_1, \vec{b}_2, \ldots, \vec{b}_n\}$ 是 $V$ 的有序基；
• $\dim(W)=m$，且 $B_2$ 是 $W$ 的有序基。

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 $A$:

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 $M_{B_2B_1}(T)= \left[ \begin{array}{cccc} C_{B_2}(T(\vec{b}_1)) & C_{B_2}(T(\vec{b}_2)) & \cdots & C_{B_2}(T(\vec{b}_n)) \end{array}\right].$ This result is given in the following theorem.

Theorem $\PageIndex{2}$ 定理 \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 $M_{B_2B_1}(T)= \left[ \begin{array}{cccc} C_{B_2}(T(\vec{b}_1)) & C_{B_2}(T(\vec{b}_2)) & \cdots & C_{B_2}(T(\vec{b}_n)) \end{array}\right].$

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 示例 \PageIndex{2}：线性变换的矩阵

Let $T: \mathbb{P}_3 \mapsto \mathbb{R}^4$ be an isomorphism defined by $T( ax^3 + bx^2 + cx + d) = \left[ \begin{array}{c} a + b \\ b - c \\ c + d \\ d + a \end{array} \right]$. Suppose $B_1 = \left\{ x^3, x^2, x, 1 \right\}$ is an ordered basis of $\mathbb{P}_3$ and $B_2 = \left\{ \left[ \begin{array}{r} 1 \\ 0 \\ 1 \\ 0 \end{array}\right], \left[ \begin{array}{r} 0 \\ 1 \\ 0 \\ 0 \end{array}\right], \left[ \begin{array}{r} 0 \\ 0 \\ -1 \\ 0 \end{array}\right], \left[ \begin{array}{r} 0 \\ 0 \\ 0 \\ 1 \end{array}\right] \right\}$ is 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. $M_{B_2B_1}(T) = \left[ \begin{array}{cccc} C_{B_2}(T(x^3)) & C_{B_2}(T(x^2)) & C_{B_2}(T(x)) & C_{B_2}(T(x^2)) \end{array} \right]$ First we find the result of applying $T$ to the basis $B_1$. $T(x^3) = \left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ 1 \end{array}\right]$, $T(x^2) = \left[ \begin{array}{c} 1 \\ 1 \\ 0 \\ 0 \end{array}\right]$, $T(x) = \left[ \begin{array}{c} 0 \\ -1 \\ 1 \\ 0 \end{array}\right]$, $T(1) = \left[ \begin{array}{c} 0 \\ 0 \\ 1 \\ 1 \end{array}\right]$ Next we apply the coordinate isomorphism $C_{B_2}$ to each of these vectors. We will show the first in detail. $C_{B_2} \left( \left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ 1 \end{array}\right] \right) = a_1 \left[ \begin{array}{r} 1 \\ 0 \\ 1 \\ 0 \end{array}\right] + a_2 \left[ \begin{array}{r} 0 \\ 1 \\ 0 \\ 0 \end{array}\right] + a_3 \left[ \begin{array}{r} 0 \\ 0 \\ -1 \\ 0 \end{array}\right] + a_4 \left[ \begin{array}{r} 0 \\ 0 \\ 0 \\ 1 \end{array}\right]$ 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. $C_{B_2}(T(x^2)) = \left[ \begin{array}{r} 1 \\ 1 \\ 1 \\ 0 \end{array}\right]$, $C_{B_2}(T(x)) = \left[ \begin{array}{r} 0 \\ -1 \\ -1 \\ 0 \end{array}\right]$, $C_{B_2}(T(1)) = \left[ \begin{array}{r} 0 \\ 0 \\ -1 \\ 1 \end{array}\right]$ Using these vectors as the columns of $M_{B_2B_1}(T)$ we have $M_{B_2B_1}(T) = \left[ \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 0 & 1 \end{array} \right]$

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 示例 \PageIndex{3}：求线性变换的作用

Let $T: \mathbb{P}_3 \mapsto \mathbb{R}^4$ be an isomorphism with $M_{B_2B_1}(T) = \left[ \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 0 & 1 \end{array} \right]$, where $B_1 = \left\{ x^3, x^2, x, 1 \right\}$ is an ordered basis of $\mathbb{P}_3$ and $B_2 = \left\{ \left[ \begin{array}{r} 1 \\ 0 \\ 1 \\ 0 \end{array}\right], \left[ \begin{array}{r} 0 \\ 1 \\ 0 \\ 0 \end{array}\right], \left[ \begin{array}{r} 0 \\ 0 \\ -1 \\ 0 \end{array}\right], \left[ \begin{array}{r} 0 \\ 0 \\ 0 \\ 1 \end{array}\right] \right\}$ 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 定理 \PageIndex{3}：复合的矩阵

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 $M_{B_3B_1}(ST)=M_{B_3B_2}(S) M_{B_2B_1}(T)$.

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

Theorem $\PageIndex{4}$: Isomorphism 定理 \PageIndex{4}：同构

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 $\PageIndex{4}$ 示例 \PageIndex{4}

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

1. 求 $M_{B_2B_1}(T)$。
2. 通过证明 $M_{B_2B_1}(T)$ 可逆来验证 $T$ 是同构。
3. 求 $M_{B_1B_2}(T^{-1})$，并验证 $M_{B_1B_2}(T^{-1}) = \left[ M_{B_2B_1}(T)\right]^{-1}$。
4. 利用 $M_{B_1B_2}(T^{-1})$ 求 $T^{-1}$。

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. $T^{-1}\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] = 1$, $T^{-1}\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]= x$, $T^{-1}\left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]= x^2$, $T^{-1}\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]=x^3$, so $T^{-1}\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right] = \frac{1+x^3}{2}$, $T^{-1}\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]= \frac{x-x^2}{2}$, $T^{-1}\left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right] = \frac{x+x^2}{2}$, $T^{-1}\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]= \frac{1-x^3}{2}$. Therefore, $M_{B_1B_2}(T^{-1})=\frac{1}{2}\left[ \begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & -1 \end{array}\right]$ 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. $\begin{aligned} C_{B_1}\left(T^{-1}\left[ \begin{array}{cc} p & q \\ r & s \end{array}\right]\right) & = M_{B_1B_2}(T^{-1}) C_{B_2}\left( \left[ \begin{array}{cc} p & q \\ r & s \end{array}\right]\right) \\ T^{-1}\left[ \begin{array}{cc} p & q \\ r & s \end{array}\right] & = C_{B_1}^{-1}\left(M_{B_1B_2}(T^{-1}) C_{B_2}\left( \left[ \begin{array}{cc} p & q \\ r & s \end{array}\right]\right)\right) \\ & = C_{B_1}^{-1}\left( \frac{1}{2}\left[ \begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & -1 \end{array}\right] \left[ \begin{array}{c} p \\ q\\ r\\ s\end{array}\right]\right) \\ & = C_{B_1}^{-1}\left(\frac{1}{2}\left[ \begin{array}{c} p+s \\ q+r \\ r-q \\ p-s \end{array}\right]\right) \\ & = \frac{1}{2}(p+s)x^3 +\frac{1}{2}(q+r)x^2 +\frac{1}{2}(r-q)x + \frac{1}{2}(p-s).\end{aligned}$

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. $\det(M_{B_2B_1}(T))=4$，故该矩阵可逆，从而 $T$ 是同构。
3. 由 $T^{-1}\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] = 1$、$T^{-1}\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]= x$、$T^{-1}\left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]= x^2$、$T^{-1}\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]=x^3$，可得 $T^{-1}\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right] = \frac{1+x^3}{2}$、$T^{-1}\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]= \frac{x-x^2}{2}$、$T^{-1}\left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right] = \frac{x+x^2}{2}$、$T^{-1}\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]= \frac{1-x^3}{2}$。因此 $M_{B_1B_2}(T^{-1})=\frac{1}{2}\left[ \begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & -1 \end{array}\right]$。应当验证 $M_{B_2B_1}(T) M_{B_1B_2}(T^{-1}) = I_4$，由此可得 $M_{B_2B_1}(T)^{-1}= M_{B_1B_2}(T^{-1})$。
4. $\begin{aligned} C_{B_1}\left(T^{-1}\left[ \begin{array}{cc} p & q \\ r & s \end{array}\right]\right) & = M_{B_1B_2}(T^{-1}) C_{B_2}\left( \left[ \begin{array}{cc} p & q \\ r & s \end{array}\right]\right) \\ T^{-1}\left[ \begin{array}{cc} p & q \\ r & s \end{array}\right] & = C_{B_1}^{-1}\left(M_{B_1B_2}(T^{-1}) C_{B_2}\left( \left[ \begin{array}{cc} p & q \\ r & s \end{array}\right]\right)\right) \\ & = C_{B_1}^{-1}\left( \frac{1}{2}\left[ \begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & -1 \end{array}\right] \left[ \begin{array}{c} p \\ q\\ r\\ s\end{array}\right]\right) \\ & = C_{B_1}^{-1}\left(\frac{1}{2}\left[ \begin{array}{c} p+s \\ q+r \\ r-q \\ p-s \end{array}\right]\right) \\ & = \frac{1}{2}(p+s)x^3 +\frac{1}{2}(q+r)x^2 +\frac{1}{2}(r-q)x + \frac{1}{2}(p-s).\end{aligned}$

Exercises 习题

Exercise 1. Suppose you have $\mathbb{R}^2$ and the $+$ operation is as follows: $(a,b) + (c,d) = (a+d,b+c)$. Scalar multiplication is defined in the usual way. Is this a vector space? Explain why or why not.

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

Exercise 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 4. Suppose you have $\mathbb{R}^2$ and the $+$ operation is defined as follows: $(a,b) + (c,d) = (a-c,b-d)$. Scalar multiplication is same as usual. Is this a vector space? Explain why or why not.

Exercise 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 6. Denote by $\mathbb{R}^{\mathbb{N}}$ the set of real valued sequences. For $\vec{a} \equiv (a_n)_{n=1}^\infty$, $\vec{b} \equiv (b_n)_{n=1}^\infty$ two of these, define their sum to be given by $\vec{a}+\vec{b}=(a_n+b_n)_{n=1}^\infty$ and define scalar multiplication by $c\vec{a}=(ca_n)_{n=1}^\infty\text{ where }\vec{a}=(a_n)_{n=1}^\infty$ Is this a special case of Exercise $\PageIndex{5}$? Is this a vector space?

Exercise 7. Let $\mathbb{C}^2$ be the set of ordered pairs of complex numbers. Define addition and scalar multiplication in the usual way. $(z,w) + (\hat{z},\hat{w}) = (z+\hat{z},w+\hat{w}),\quad u(z,w) \equiv (uz,uw)$ Here the scalars are from $\mathbb{C}$. Show this is a vector space.

Exercise 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 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, $(A+B)_{ij} = A_{ij} + B_{ij},\quad (cA)_{ij} = c(A_{ij})$.

Exercise 10. Consider the set of all vectors in $\mathbb{R}^2 ,(x, y)$ such that $x + y \ge 0$. Let the vector space operations be the usual ones. Is this a vector space? Is it a subspace of $\mathbb{R}^2$?

Exercise 11. 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 12. 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 13. 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 14. 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 15. 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 $(a+bx+cx^2)+(\hat{d}+\hat{b}x+\hat{c}x^2)=(a+\hat{a})+(b+\hat{b})x+(c+\hat{c})x^2$ and scalar multiplication is defined as $d(a+bx+cx^2)=da+dbx+cdx^2$ 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 16. Let $M,N$ be subspaces of a vector space $V$ and consider $M +N$ defined as the set of all $m+n$ where $m \in M$ and $n \in N$. Show that $M +N$ is a subspace of $V$.

Exercise 17. Let $M,N$ be subspaces of a vector space $V$. Then $M \cap N$ consists of all vectors which are in both $M$ and $N$. Show that $M \cap N$ is a subspace of $V$.

Exercise 18. Let $M,N$ be subspaces of a vector space $\mathbb{R}^2$. Then $N \cup M$ consists of all vectors which are in either $M$ or $N$. Show that $N \cup M$ is not necessarily a subspace of $\mathbb{R}^2$ by giving an example where $N \cup M$ fails to be a subspace.

Exercise 19. Let $X$ consist of the real valued functions which are defined on an interval $[a,b]$. For $f ,g \in X$, $f +g$ 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, $-f (x) \equiv (-f (x))$. Then $(f + (-f)) (x) \equiv f (x) + (-f) (x) \equiv f (x) + (-f (x)) = 0$ Hence $f + -f = 0$. For each $x \in [a,b]$, let $f_x (x) = 1$ and $f_x (y) = 0$ if $y\neq x$. Then these vectors are obviously linearly independent.

Exercise 20. 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 $\vec{x} \in \mathbb{R}^n$. Thus a typical element in $\mathbb{R}^n$ is $(f (1),\cdots , f (n))$.

Exercise 21. 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 22. 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 $\operatorname{span}\{\vec{x}_1,\cdots ,\vec{x}_k\}$.

Answer. $$ \sum\limits_{i=1}^k0\vec{x}_k=\vec{0} $$

Exercise 23. Determine if $p(x) = 4x^2 -x$ is in the span given by $\operatorname{span}\{x^2+x, x^2-1, -x+2\}$.

Exercise 24. Determine if $p(x) = -x^2 +x+2$ is in the span given by $\operatorname{span}\{x^2 +x+1, 2x^2 +x\}$.

Exercise 25. Determine if $A=\left[\begin{array}{cc}1&3\\0&0\end{array}\right]$ is in the span given by $\operatorname{span}\left\{\left[\begin{array}{cc}1&0\\0&1\end{array}\right],\left[\begin{array}{cc}0&1\\1&0\end{array}\right],\left[\begin{array}{cc}1&0\\1&1\end{array}\right],\left[\begin{array}{cc}0&1\\1&1\end{array}\right]\right\}$.

Exercise 26. 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 27. Consider the vector space of polynomials of degree at most $2$, $\mathbb{P}_2$. Determine whether the following is a basis for $\mathbb{P}_2$. $\{x^2 +x+1, 2x^2 +2x+1, x+1\}$ Hint: There is a isomorphism from $\mathbb{R}^3$ to $\mathbb{P}_2$. It is defined as follows: $T\vec{e}_1 = 1,\quad T\vec{e}_2 = x,\quad T\vec{e}_3= x^2$ Then extend $T$ linearly. Thus $T\left[\begin{array}{c}1\\1\\1\end{array}\right]=x^2+x+1,\quad T\left[\begin{array}{c}1\\2\\2\end{array}\right]=2x^2+2x+1,\quad T\left[\begin{array}{c}1\\1\\0\end{array}\right]=1+x$ It follows that if $\left\{\left[\begin{array}{c}1\\1\\1\end{array}\right],\left[\begin{array}{c}1\\2\\2\end{array}\right],\left[\begin{array}{c}1\\1\\0\end{array}\right]\right\}$ 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 28. Find a basis in $\mathbb{P}_2$ for the subspace $\operatorname{span}\{1+x+x^2, 1+2x, 1+5x-3x^2\}$ 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 29. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{1+x-x^2 +x^3, 1+2x+3x^3, -1+3x+5x^2 +7x^3, 1+6x+4x^2 +11x^3\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 30. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{1+x-x^2 +x^3, 1+2x+3x^3, -1+3x+5x^2 +7x^3, 1+6x+4x^2 +11x^3\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 31. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -2x^2 +x+2, 3x^3 -x^2 +2x+2, 7x^3 +x^2 +4x+2, 5x^3 +3x+2\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 32. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 +2x^2 +x-2, 3x^3 +3x^2 +2x-2, 3x^3 +x+2, 3x^3 +x+2\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 33. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -5x^2 +x+5, 3x^3 -4x^2 +2x+5, 5x^3 +8x^2 +2x-5, 11x^3 +6x+5\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 34. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -3x^2 +x+3, 3x^3 -2x^2 +2x+3, 7x^3 +7x^2 +3x-3, 7x^3 +4x+3\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 35. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -x^2 +x+1, 3x^3 +2x+1, 4x^3 +x^2 +2x+1, 3x^3 +2x-1\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 36. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -x^2 +x+1, 3x^3 +2x+1, 13x^3 +x^2 +8x+4, 3x^3 +2x-1\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 37. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -3x^2 +x+3, 3x^3 -2x^2 +2x+3, -5x^3 +5x^2 -4x-6, 7x^3 +4x-3\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 38. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -2x^2 +x+2, 3x^3 -x^2 +2x+2, 7x^3 -x^2 +4x+4, 5x^3 +3x-2\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 39. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -2x^2 +x+2, 3x^3 -x^2 +2x+2, 3x^3 +4x^2 +x-2, 7x^3 -x^2 +4x+4\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 40. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -4x^2 +x+4, 3x^3 -3x^2 +2x+4, -3x^3 +3x^2 -2x-4, -2x^3 +4x^2 -2x-4\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 41. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 +2x^2 +x-2, 3x^3 +3x^2 +2x-2, 5x^3 +x^2 +2x+2, 10x^3 +10x^2 +6x-6\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 42. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 +x^2 +x-1, 3x^3 +2x^2 +2x-1, x^3 +1, 4x^3 +3x^2 +2x-1\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 43. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -x^2 +x+1, 3x^3 +2x+1, x^3 +2x^2 -1, 4x^3 +x^2 +2x+1\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 44. Find a basis in $\mathbb{P}_3$ for the subspace $\operatorname{span}\{x^3 -x^2 +x+1, 3x^3 +2x+1, 13x^3 +x^2 +8x+4, 3x^3 +2x-1\}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Exercise 45. Here are some vectors. $\{x^3 +x^2 -x-1, 3x^3 +2x^2 +2x-1\}$ If these are linearly independent, extend to a basis for all of $\mathbb{P}_3$.

Exercise 46. Here are some vectors. $\{x^3 -2x^2 -x+2, 3x^3 -x^2 +2x+2\}$ If these are linearly independent, extend to a basis for all of $\mathbb{P}_3$.

Exercise 47. Here are some vectors. $\{x^3 -3x^2 -x+3, 3x^3 -2x^2 +2x+3\}$ If these are linearly independent, extend to a basis for all of $\mathbb{P}_3$.

Exercise 48. Here are some vectors. $\{x^3 -2x^2 -3x+2, 3x^3 -x^2 -6x+2, -8x^3 +18x+10\}$ If these are linearly independent, extend to a basis for all of $\mathbb{P}_3$.

Exercise 49. Here are some vectors. $\{x^3 -3x^2 -3x+3, 3x^3 -2x^2 -6x+3, -8x^3 +18x+40\}$ If these are linearly independent, extend to a basis for all of $\mathbb{P}_3$.

Exercise 50. Here are some vectors. $\{x^3 -x^2 +x+1, 3x^3 +2x+1, 4x^3 +2x+2\}$ If these are linearly independent, extend to a basis for all of $\mathbb{P}_3$.

Exercise 51. Here are some vectors. $\{x^3 +x^2 +2x-1, 3x^3 +2x^2 +4x-1, 7x^3 +8x+23\}$ If these are linearly independent, extend to a basis for all of $\mathbb{P}_3$.

Exercise 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. $\{x+1, x^2 +2, x^2 -x-3\}$

Exercise 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. $\{x^2 +x, -2x^2 -4x-6, 2x-2\}$

Exercise 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. $\left\{\left[\begin{array}{cc}1&2\\0&1\end{array}\right],\left[\begin{array}{cc}-7&2\\-2&-3\end{array}\right],\left[\begin{array}{cc}4&0\\1&2\end{array}\right]\right\}$

Exercise 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. $\left\{\left[\begin{array}{cc}1&0\\0&1\end{array}\right],\left[\begin{array}{cc}0&1\\0&1\end{array}\right],\left[\begin{array}{cc}1&0\\1&0\end{array}\right],\left[\begin{array}{cc}0&0\\1&1\end{array}\right]\right\}$

Exercise 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 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 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. $\{x^3 +1, x^2 +x, 2x^3 +x^2 , 2x^3 -x^2 -3x+1\}$

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 $c_1(x^3 +1)+c_2 (x^2 +x) +c_3( 2x^3 +x^2) +c_4 (2x^3 -x^2 -3x+1) = 0$ Then combine the terms according to power of $x$. $(c_1 +2c_3 +2c_4) x^3 + (c_2 +c_3 -c_4) x^2 + (c_2 -3c_4) x+ (c_1 +c_4) = 0$ 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: $(c_1 = 0, c_2 = 0, c_3 = 0, c_4 = 0)$ Therefore, these are linearly independent.

Exercise 59. In the context of the above problem, consider polynomials $\{a_ix^3 +b_ix^2 +c_ix+d_i , i = 1, 2, 3, 4\}$ Show that this collection of polynomials is linearly independent on an interval $[s,t]$ if and only if $\left[\begin{array}{cccc}a_1&b_1&c_1&d_1 \\ a_2&b_2&c_2&d_2 \\ a_3&b_3&c_3&d_3 \\ a_4&b_4&c_4&d_4\end{array}\right]$ 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 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 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 $φ : \mathbb{R}^n \mapsto V,$ $ψ :\mathbb{R}^m\mapsto V$ by $φ (\vec{a}) = \sum\limits_{k=1}^n a_k\vec{x}_k ,$ $ψ(\vec{b}) =\sum\limits_{j=1}^m b_j\vec{y}_j$ 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 62. Let $M = \{\vec{u} = (u_1, u_2, u_3, u_4)\in \mathbb{R}^4 : \|u_1\| \le 4\}$. 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 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 64. Let $W$ be a subset of $M_{22}$ given by $W = \{ A \mid A\in M_{22}, A^T = A\}$. In words, $W$ is the set of all symmetric $2\times 2$ matrices. Is $W$ a subspace of $M_{22}$?

Exercise 65. Let $W$ be a subset of $M_{22}$ given by $W=\left\{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] \mid a,b,c,d\in\mathbb{R}, a+b=c+d\right\}$. Is $W$ a subspace of $M_{22}$?

Exercise 66. Let $W$ be a subset of $P_3$ given by $W = \{ ax^3 +bx^2 +cx+d \mid a,b,c,d\in\mathbb{R}, d = 0\}$. Is $W$ a subspace of $P_3$?

Exercise 67. Let $W$ be a subset of $P_3$ given by $W = \{ p(x) = ax^3 +bx^2 +cx+d \mid a,b,c,d\in\mathbb{R}, p(2) = 1\}$. Is $W$ a subspace of $P_3$?

Exercise 68. Let $T$: $\mathbb{P}_2\to\mathbb{R}$ be a linear transformation such that $T(x^2)=1$, $T(x^2+x)=5$, $T(x^2+x+1)=-1$. 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 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. $T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}\sin x+2y+3z \\ 2y+3x+z\end{array}\right]$
4. $T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}x+2y+3z \\ 2y+3z-\ln z\end{array}\right]$

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

Exercise 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.

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

Exercise 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.

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

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

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 73. Suppose $\left[A_1\cdots A_n\right]^{-1}$ 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 74. Let $V$ and $W$ be subspaces of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively and let $T$: $V \to W$ 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 $\sum\limits_i^r a_i\vec{v}_r=0$, then using linearity properties of $T$ we get $0=T(0)=T\left(\sum\limits_i^r a_i\vec{v}_r\right)=\sum\limits_i^r a_iT(\vec{v}_r).$ 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 75. Let $V=span\left\{\left[\begin{array}{c}1\\1\\2\\0\end{array}\right],\left[\begin{array}{c}0\\1\\1\\1\end{array}\right],\left[\begin{array}{c}1\\1\\0\\1\end{array}\right]\right\}$. Let $T\vec{x}=A\vec{x}$ where $A$ is the matrix $\left[\begin{array}{cccc}1&1&1&1\\0&1&1&0\\0&1&2&1\\1&1&1&2\end{array}\right]$. Give a basis for $Im(T)$.

Exercise 76. Let $V=span\left\{\left[\begin{array}{c}1\\0\\0\\1\end{array}\right],\left[\begin{array}{c}1\\1\\1\\1\end{array}\right],\left[\begin{array}{c}1\\4\\4\\1\end{array}\right]\right\}$. Let $T\vec{x}=A\vec{x}$ where $A$ is the matrix $\left[\begin{array}{cccc}1&1&1&1\\0&1&1&0\\0&1&2&1\\1&1&1&2\end{array}\right]$. 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: $V=span\left\{\left[\begin{array}{c}2\\0\\1\\3\end{array}\right],\left[\begin{array}{c}4\\2\\4\\5\end{array}\right]\right\}$

Exercise 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 78. Let $V$ and $W$ be subspaces of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively and let $T$: $V \to W$ 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 $\text{span}\{T\vec{v}_1,\cdots ,T\vec{v}_r\} = W$.

Exercise 79. Define $T$: $\mathbb{R}^4 \to \mathbb{R}^3$ as follows. $T\vec{x}=\left[\begin{array}{rrrr}3&2&1&8\\2&2&-2&6\\1&1&-1&3\end{array}\right]\vec{x}$. Find a basis for $Im(T)$. Also find a basis for $\text{ker}(T)$.

Exercise 80. Define $T$: $\mathbb{R}^4 \to \mathbb{R}^3$ as follows. $T\vec{x}=\left[\begin{array}{rrr}1&2&0\\1&1&1\\0&1&1\end{array}\right]\vec{x}$ 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 81. Suppose $T$: $\mathbb{R}^3 \to \mathbb{R}^3$ is a linear transformation given by $T\vec{x}=A\vec{x}$ where $A$ is a $3\times 3$ matrix. Show that $T$ is an isomorphism if and only if $A$ is invertible.

Exercise 82. Suppose $T$: $\mathbb{R}^3 \to \mathbb{R}^3$ is a linear transformation given by $T\vec{x}=A\vec{x}$ 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