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8.4 Matrices for General Linear Transformations: Class Notes

Recall this example from section 1.8

Find the standard matrix A for the linear transformation $T:R^2\to R^3$ for which $T\left(\left[\begin{array}{c}-1\\ 2\end{array}\right]\right)=\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]$ and $T\left(\left[\begin{array}{c}3\\ -5\end{array}\right]\right)=\left[\begin{array}{c}5\\ -7\\ 1\end{array}\right]$ and use it to compute $T\left(\left[\begin{array}{c}-4\\ 3\end{array}\right]\right)$.

Tip: Write each standard basis vector in terms of ${\overrightarrow{v}}_1$ and ${\overrightarrow{v}}_2$: if ${\overrightarrow{e}}_1=a{\overrightarrow{v}}_1+b{\overrightarrow{v}}_2$, then $T({\overrightarrow{e}}_1)=aT({\overrightarrow{v}}_1)+bT({\overrightarrow{v}}_2)$.

And #3 from 4.7

Consider the bases $B=\{{𝐮}_1,{𝐮}_2,{𝐮}_3\}$ and $B\prime =\{{𝐮\prime }_1,{𝐮\prime }_2,{𝐮\prime }_3\}$ for $R^3$, where

$${𝐮}_1=\left[\begin{array}{c}2\\ 1\\ 1\end{array}\right],{𝐮}_2=\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right],{𝐮}_3=\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right]$$

$${𝐮\prime }_1=\left[\begin{array}{c}3\\ 1\\ -5\end{array}\right],{𝐮\prime }_2=\left[\begin{array}{c}1\\ 1\\ -3\end{array}\right],{𝐮\prime }_3=\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]$$

a. Find the transition matrix $B$ to $B\prime$.
b. Compute the coordinate vector $[𝐰{]}_B$, where $𝐰=\left[\begin{array}{c}-5\\ 8\\ -5\end{array}\right]$ and use the transition matrix in part (a) to compute $[𝐰{]}_{B\prime }$.

Now we combine these two concepts.

#5

Let $T:R^2\to R^3$ be defined by $T\left(\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\right)=\left[\begin{array}{c}{x}_1+2x_2\\ -x_1\\ 0\end{array}\right]$.

a. Find the matrix $[T{]}_{B\prime ,B}$ relative to the bases $B=\{{𝐮}_1,{𝐮}_2\}$ and $B\prime =\{{𝐯}_1,{𝐯}_2,{𝐯}_3\}$, where

$${𝐮}_1=\left[\begin{array}{c}1\\ 3\end{array}\right],{𝐮}_2=\left[\begin{array}{c}-2\\ 4\end{array}\right],{𝐯}_1=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],{𝐯}_2=\left[\begin{array}{c}2\\ 2\\ 0\end{array}\right],{𝐯}_3=\left[\begin{array}{c}3\\ 0\\ 0\end{array}\right]$$

Solution

In $[T{]}_{B\prime ,B}$, $B\prime$ is the new basis and $B$ is the old basis. We find $[T{]}_{B\prime ,B}$ using an augmented matrix that has the new basis vectors as columns on the left and the images of the old basis vectors as columns on the right.

a. First find $T({\overrightarrow{u}}_1)$ and $T({\overrightarrow{u}}_2)$.

$$T({\overrightarrow{u}}_1)=T\left(\left[\begin{array}{c}1\\ 3\end{array}\right]\right)=\left[\begin{array}{c}1+2·3\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}7\\ -1\\ 0\end{array}\right]$$

$$T({\overrightarrow{u}}_2)=T\left(\left[\begin{array}{c}-2\\ 4\end{array}\right]\right)=\left[\begin{array}{c}-2+2·4\\ 2\\ 0\end{array}\right]=\left[\begin{array}{c}6\\ 2\\ 0\end{array}\right]$$

$$\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 7}\hfill & \hfill {\displaystyle 6}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

The left side holds the new basis vectors ${𝐯}_1,{𝐯}_2,{𝐯}_3$; the right side holds the images of the old basis vectors. Row reducing this gives us representations of $T({\overrightarrow{u}}_1)$ and $T({\overrightarrow{u}}_2)$ in terms of the $B\prime$ vectors. That is, we’ll have $[T({\overrightarrow{u}}_1){]}_{B\prime }$ and $[T({\overrightarrow{u}}_2){]}_{B\prime }$.

$$\to \left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -\frac{1}{2}}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle \frac{8}{3}}\hfill & \hfill {\displaystyle \frac{4}{3}}\hfill \end{array}\right]$$

$$[T{]}_{B\prime ,B}=[[T({\overrightarrow{u}}_1){]}_{B\prime }|[T({\overrightarrow{u}}_2){]}_{B\prime }]$$

This is the matrix for the transformation $T$ relative to the bases $B$ and $B\prime$. It acts on coordinate vectors relative to $B$ and returns coordinate vectors relative to $B\prime$.

b. Verify that the formula $[T{]}_{B\prime ,B}[𝐱{]}_B=[T(𝐱){]}_{B\prime }$ holds for every vector in $R^2$.

b. Let $\overrightarrow{x}=(x_1,x_2)$.

Left side: find $[\overrightarrow{x}{]}_B$, the coordinate vector relative to $B$ for an arbitrary vector in $R^2$. Solve $c_1{\overrightarrow{u}}_1+c_2{\overrightarrow{u}}_2=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$:

$${\displaystyle \left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle x_1}\hfill \\ \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle x_2}\hfill \end{array}\right]\to \left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle \frac{2x_1+x_2}{5}}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle \frac{-3x_1+x_2}{10}}\hfill \end{array}\right]\Rightarrow [\overrightarrow{x}{]}_B=\left[\begin{array}{c}\frac{2x_1+x_2}{5}\\ \frac{-3x_1+x_2}{10}\end{array}\right]}$$

Now multiply:

$${\displaystyle [T{]}_{B\prime ,B}[\overrightarrow{x}{]}_B=\left[\begin{array}{cc}0& 0\\ -\frac{1}{2}& 1\\ \frac{8}{3}& \frac{4}{3}\end{array}\right]\left[\begin{array}{c}\frac{2x_1+x_2}{5}\\ \frac{-3x_1+x_2}{10}\end{array}\right]=\left[\begin{array}{c}0\\ -\frac{1}{2}{x}_1\\ \frac{2}{3}{x}_1+\frac{2}{3}{x}_2\end{array}\right]}$$

Right side: find $[T(\overrightarrow{x}){]}_{B\prime }$, the coordinate vector relative to $B\prime$ of the image of an arbitrary vector. Since $T\left(\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\right)=\left[\begin{array}{c}{x}_1+2x_2\\ -x_1\\ 0\end{array}\right]$, we use

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle x_1+2x_2}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -x_1}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]\to \left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -\frac{1}{2}{x}_1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle \frac{2}{3}{x}_1+\frac{2}{3}{x}_2}\hfill \end{array}\right]$$

So ${\displaystyle [T(\overrightarrow{x}){]}_{B\prime }=\left[\begin{array}{c}0\\ -\frac{1}{2}{x}_1\\ \frac{2}{3}{x}_1+\frac{2}{3}{x}_2\end{array}\right]=[T{]}_{B\prime ,B}[\overrightarrow{x}{]}_B}$. The two sides are the same.

A specific example

Suppose we want to use $[T{]}_{B\prime ,B}$ to find $T\left(\left[\begin{array}{c}11\\ 3\end{array}\right]\right)$, where $T\left(\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\right)=\left[\begin{array}{c}{x}_1+2x_2\\ -x_1\\ 0\end{array}\right]$.

Step 1: Express $\overrightarrow{x}=(11,3)$ in terms of $B$.

$$\left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 11}\hfill \\ \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 3}\hfill \end{array}\right]\to \left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 11}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -3}\hfill \end{array}\right]\to \left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 5}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -3}\hfill \end{array}\right]$$

So $(11,3)=5{\overrightarrow{u}}_1-3{\overrightarrow{u}}_2$. That is, $[\overrightarrow{x}{]}_B=\left[\begin{array}{c}5\\ -3\end{array}\right]$.

Step 2: Multiply.

$${\displaystyle [T{]}_{B\prime ,B}[\overrightarrow{x}{]}_B=\left[\begin{array}{cc}0& 0\\ -\frac{1}{2}& 1\\ \frac{8}{3}& \frac{4}{3}\end{array}\right]\left[\begin{array}{c}5\\ -3\end{array}\right]=\left[\begin{array}{c}0\\ -\frac{11}{2}\\ \frac{28}{3}\end{array}\right]=[T(\overrightarrow{x}){]}_{B\prime }}$$

This is the same as $5[T({\overrightarrow{u}}_1){]}_{B\prime }-3[T({\overrightarrow{u}}_2){]}_{B\prime }$.

Step 3: Use the new coordinate vector with $B\prime$.

$${\displaystyle 0\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]-\frac{11}{2}\left[\begin{array}{c}2\\ 2\\ 0\end{array}\right]+\frac{28}{3}\left[\begin{array}{c}3\\ 0\\ 0\end{array}\right]=\left[\begin{array}{c}17\\ -11\\ 0\end{array}\right]=T(\overrightarrow{x})}$$

We can visualize the process as follows.

A diagram with four corners. Bottom left: $\overrightarrow{x}=(11,3)$. Arrow (1) goes up to top left: $[\overrightarrow{x}{]}_B=(5,-3)$. Arrow (2), labeled $[T{]}_{B\prime ,B}[\overrightarrow{x}{]}_B$, goes right to top right: $[T(\overrightarrow{x}){]}_{B\prime }=(0,-\frac{11}{2},\frac{28}{3})$. Arrow (3) goes down to bottom right: $T(\overrightarrow{x})=(17,-11,0)$. A direct approach arrow goes straight from $\overrightarrow{x}$ to $T(\overrightarrow{x})$ using the formula $T\left(\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\right)=\left[\begin{array}{c}{x}_1+2x_2\\ -x_1\\ 0\end{array}\right]$.

Definition

Let $V$ be an $n$-dimensional vector space with basis $B=\{{𝐮}_1,{𝐮}_2,\dots ,{𝐮}_n\}$ and let $W$ be an $m$-dimensional vector space with basis $B\prime$. Suppose that $T:V\to W$ is a linear transformation, and that for each vector $𝐱$ in $V$, the coordinate vectors for $𝐱$ and $T(𝐱)$ are $[𝐱{]}_B$ and $[T(𝐱){]}_{B\prime }$, respectively. Then the matrix for $𝐓$ relative to the bases $𝐁$ and $𝐁\prime$ is written [by this author] as

$$[T{]}_{B\prime ,B}=[[T({𝐮}_1){]}_{B\prime }|[T({𝐮}_2){]}_{B\prime }\left|\dots \right|[T({𝐮}_n){]}_{B\prime }].$$

This matrix has the property that $[T{]}_{B\prime ,B}[𝐱{]}_B=[T(𝐱){]}_{B\prime }$.

#8

Let $T:P_2\to P_3$ be the linear transformation defined by $T(p(x))=xp(x-3)$, that is, $T(c_0+c_{1}x+c_2{x}^2)$ $=x(c_0+c_1(x-3)+c_2(x-3{)}^2)$

a. Find $[T{]}_{B\prime ,B}$ relative to the bases $B=\{1,x,x^2\}$ and $B\prime =\{1,x,x^2,x^3\}$.
b. Use the three-step procedure illustrated in Example 2 to compute $T(1+x-x^2)$.
c. Check the result obtained in part (b) by computing $T(1+x-x^2)$ directly.

Solution

a. Apply $T$ to each vector in $B$:

$$T(1)=x,T(x)=x(x-3)=x^2-3x,T(x^2)=x(x-3{)}^2=x^3-6x^2+9x$$

Since $B\prime$ is the standard basis for $P_3$, the coordinate vectors are the coefficients (constant term first):

$$[T(1){]}_{B\prime }=\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],[T(x){]}_{B\prime }=\left[\begin{array}{c}0\\ -3\\ 1\\ 0\end{array}\right],[T(x^2){]}_{B\prime }=\left[\begin{array}{c}0\\ 9\\ -6\\ 1\end{array}\right]$$

b. Step 1: $1+x-x^2=1(1)+1(x)-1(x^2)$, so $[\overrightarrow{x}{]}_B=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$.

Step 2:

$$[T{]}_{B\prime ,B}[\overrightarrow{x}{]}_B=\left[\begin{array}{ccc}0& 0& 0\\ 1& -3& 9\\ 0& 1& -6\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]=\left[\begin{array}{c}0\\ -11\\ 7\\ -1\end{array}\right]=[T(\overrightarrow{x}){]}_{B\prime }$$

Step 3: $T(\overrightarrow{x})=0(1)-11(x)+7(x^2)-1(x^3)$.

c. Directly:

$$T(1+x-x^2)=x[1+(x-3)-(x-3{)}^2]=x[-x^2+7x-11]=-x^3+7x^2-11x$$

This matches part (b).

Definition

In the special case where $T:V\to V$ is a linear operator with $B=B\prime =\{{𝐮}_1,{𝐮}_2,\dots ,{𝐮}_n\}$, the matrix for $T$ is called the matrix for $𝐓$ relative to the basis $𝐁$ and is written [by this author] as

$$[T{]}_B=[[T({𝐮}_1){]}_B|[T({𝐮}_2){]}_B\left|\dots \right|[T({𝐮}_n){]}_B].$$

#18

Let $D:P_2\to P_2$ be the differentiation operator $D(𝐩)=p\prime (x)$.

a. Find the matrix for $D$ relative to the basis $B=\{{𝐩}_1,{𝐩}_2,{𝐩}_3\}$ for $P_2$ in which ${𝐩}_1=2,{𝐩}_2=2-3x,{𝐩}_3=2-3x+8x^2$.
b. Use the matrix in part (a) to compute $D(6-6x+24x^2)$.

Solution

a. Differentiate each basis vector:

$$D({\overrightarrow{p}}_1)=0,D({\overrightarrow{p}}_2)=-3,D({\overrightarrow{p}}_3)=-3+16x$$

That is,

$$D({\overrightarrow{p}}_1)=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right],D({\overrightarrow{p}}_2)=\left[\begin{array}{c}-3\\ 0\\ 0\end{array}\right],D({\overrightarrow{p}}_3)=\left[\begin{array}{c}-3\\ 16\\ 0\end{array}\right]$$

These are relative to the standard basis for $P_2$. To write them relative to $B$, put the $B$ vectors on the left and the images on the right:

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle -3}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 16}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 8}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]\to \left[\begin{array}{cccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -\frac{3}{2}}\hfill & \hfill {\displaystyle \frac{23}{6}}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -\frac{16}{3}}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

b. To find $D(6-6x+24x^2)$, express $\overrightarrow{p}=6-6x+24x^2$ in terms of $B$.

$$\left[\begin{array}{cccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 6}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle -6}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 8}\hfill & \hfill {\displaystyle 24}\hfill \end{array}\right]\to \left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill \end{array}\right]\Rightarrow [\overrightarrow{p}{]}_B=\left[\begin{array}{c}1\\ -1\\ 3\end{array}\right]$$

$${\displaystyle [D{]}_B[\overrightarrow{p}{]}_B=\left[\begin{array}{ccc}0& -\frac{3}{2}& \frac{23}{6}\\ 0& 0& -\frac{16}{3}\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}1\\ -1\\ 3\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}+\frac{23}{2}\\ -16\\ 0\end{array}\right]=\left[\begin{array}{c}13\\ -16\\ 0\end{array}\right]=[D(\overrightarrow{p}){]}_B}$$

So $D(\overrightarrow{p})=13{\overrightarrow{p}}_1-16{\overrightarrow{p}}_2+0{\overrightarrow{p}}_3$ $=13(2)-16(2-3x)$.

Check: differentiating directly, $D(6-6x+24x^2)=-6+48x$.

Theorem 8.4.1

If $T_1:U\to V$ and $T_2:V\to W$ are linear transformations, and if $B,B″,$and $B\prime$ are bases for $U$, $V$, and $W$, respectively, then

$$[T_2\circ T_1{]}_{B\prime ,B}=[T_2{]}_{B\prime ,B″}[T_1{]}_{B″,B}.$$

The matching highlights show that each basis appears on both sides: $B$ goes with $U$, $B″$ with $V$, and $B\prime$ with $W$.

#13

Let $T_1:P_1\to P_2$ be the linear transformation defined by $T_1(c_0+c_{1}x)=2c_0-3c_{1}x$ and let $T_2:P_2\to P_3$ be the linear transformation defined by $T_2(c_0+c_{1}x+c_2{x}^2)=3c_{0}x+3c_1{x}^2+3c_2{x}^3$. Let $B=\{1,x\}$, $B″=\{1,x,x^2\}$, and $B\prime =\{1,x,x^2,x^3\}$.

a. Find $[T_2\circ T_1{]}_{B\prime ,B}$, $[T_2{]}_{B\prime ,B″}$, and $[T_1{]}_{B″,B}$.
b. State a formula relating the matrices in part (a).
c. Verify that the matrices in part (a) satisfy the formula you stated in part (b).

Solution

a. Find $[T_2\circ T_1{]}_{B\prime ,B}$ from $(T_2\circ T_1)(c_0+c_{1}x)$: apply $T_2\circ T_1$ to the vectors in $B$.

$$(T_2\circ T_1)(1)=T_2(T_1(1))=T_2(2)=6x\to \left[\begin{array}{c}0\\ 6\\ 0\\ 0\end{array}\right]$$

$$(T_2\circ T_1)(x)=T_2(T_1(x))=T_2(-3x)=-9x^2\to \left[\begin{array}{c}0\\ 0\\ -9\\ 0\end{array}\right]$$

$$[T_2\circ T_1{]}_{B\prime ,B}=\left[\begin{array}{cc}0& 0\\ 6& 0\\ 0& -9\\ 0& 0\end{array}\right]$$

For $[T_2{]}_{B\prime ,B″}$: $T_2(1)=3x,T_2(x)=3x^2,T_2(x^2)=3x^3$, so

$$[T_2{]}_{B\prime ,B″}=\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]$$

For $[T_1{]}_{B″,B}$: $T_1(1)=2,T_1(x)=-3x$, so

$$[T_1{]}_{B″,B}=\left[\begin{array}{cc}2& 0\\ 0& -3\\ 0& 0\end{array}\right]$$

b. The formula (Theorem 8.4.1):

c. Multiply:

$$[T_2{]}_{B\prime ,B″}[T_1{]}_{B″,B}=\left[\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& -3\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 6& 0\\ 0& -9\\ 0& 0\end{array}\right]=[T_2\circ T_1{]}_{B\prime ,B}$$

Theorem 8.4.2

If $T:V\to V$ is a linear operator, and if $B$ is a basis for $V$, then $T$ is one-to-one if and only if $[T{]}_B$ is invertible. Moreover, when these conditions hold, ${\left[T^{-1}\right]}_B=[T{]}_B^{-1}$.