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1.9 Compositions of Matrix Transformations: Class Notes

Definitions

Theorem 1.9.1

Proof

Let $T_A$ and $T_B$ be as described, and let $\overrightarrow{x}\in R^n$. Then $T_A(\overrightarrow{x})=A\overrightarrow{x}$ and $T_B(\overrightarrow{x})=B\overrightarrow{x}$ by the definition of a matrix transformation. So

$$\begin{array}{cc}\hfill {\displaystyle (T_B\circ T_A)(\overrightarrow{x})}& {\displaystyle =T_B(T_A(\overrightarrow{x}))}\hfill \\ \hfill {\displaystyle }& {\displaystyle =T_B(A\overrightarrow{x})}\hfill \\ \hfill {\displaystyle }& {\displaystyle =B(A\overrightarrow{x})}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(BA)\overrightarrow{x}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =T_{BA}(\overrightarrow{x})}\hfill \end{array}$$

The first line is the definition of composition, and $B(A\overrightarrow{x})=(BA)\overrightarrow{x}$ by associativity of matrix multiplication.

Tip: This is why matrix multiplication is defined the way it is.

#8

Find the standard matrix for the stated composition in $R^2$.

a. A rotation about the origin of 60°, followed by an orthogonal projection onto the $x$-axis, followed by a reflection about the line $y=x$.

b. An orthogonal projection onto the $x$-axis, followed by a rotation about the origin of 45°, followed by a reflection about the $y$-axis.

c. A rotation about the origin of 15°, followed by a rotation about the origin of 105°, followed by a rotation about the origin of 60°.

Solution

a. Recall that

$$R_{\theta }=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$$

so

$${\displaystyle R_{{60}^{\circ }}=\left[\begin{array}{cc}\cos {60}^{\circ }& -\sin {60}^{\circ }\\ \sin {60}^{\circ }& \cos {60}^{\circ }\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right]}$$

For the projection $T_A(x,y)=(x,0)$, $A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$.

For the reflection $T_B(x,y)=(y,x)$, $B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$.

The rotation is applied first, so its matrix goes on the right:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle T_B\circ T_A\circ R_{{60}^{\circ }}}& {\displaystyle =BA{R}_{{60}^{\circ }}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right]}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ 0& 0\end{array}\right]}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left[\begin{array}{cc}0& 0\\ \frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right]}\hfill \end{array}}$$

Note that this means

$${\displaystyle (T_B\circ T_A\circ R_{{60}^{\circ }})\left(\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\right)=\left[\begin{array}{c}0\\ \frac{1}{2}{x}_1-\frac{\sqrt{3}}{2}{x}_2\end{array}\right]}$$

c. Rotating by 15°, then 105°, then 60° rotates by 15° + 105° + 60° = 180° in all:

$$R_{{60}^{\circ }}\circ R_{{105}^{\circ }}\circ R_{{15}^{\circ }}=R_{{180}^{\circ }}=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$$

#12

Let $T_1(x_1,x_2,x_3)=(4x_1,-2x_1+x_2,-x_1-3x_2)$ and $T_2(x_1,x_2,x_3)=(x_1+2x_2,-x_3,4x_1-x_3)$.

a. Find the standard matrices for $T_1$ and $T_2$.

b. Find the standard matrices for $T_2\circ T_1$ and $T_1\circ T_2$.

c. Use the matrices obtained in part (b) to find formulas for $T_1(T_2(x_1,x_2,x_3))$ and $T_2(T_1(x_1,x_2,x_3))$.

Solution

Tip: Parts a and b ask for matrices; part c asks for formulas.

a. Each row of the standard matrix holds the coefficients of $x_1$, $x_2$, $x_3$ in one component of the formula.

$$[T_1]=\left[\begin{array}{ccc}4& 0& 0\\ -2& 1& 0\\ -1& -3& 0\end{array}\right],[T_2]=\left[\begin{array}{ccc}1& 2& 0\\ 0& 0& -1\\ 4& 0& -1\end{array}\right]$$

b.

$$[T_2\circ T_1]=[T_2][T_1]=\left[\begin{array}{ccc}1& 2& 0\\ 0& 0& -1\\ 4& 0& -1\end{array}\right]\left[\begin{array}{ccc}4& 0& 0\\ -2& 1& 0\\ -1& -3& 0\end{array}\right]=\left[\begin{array}{ccc}0& 2& 0\\ 1& 3& 0\\ 17& 3& 0\end{array}\right]$$

$$[T_1\circ T_2]=[T_1][T_2]=\left[\begin{array}{ccc}4& 0& 0\\ -2& 1& 0\\ -1& -3& 0\end{array}\right]\left[\begin{array}{ccc}1& 2& 0\\ 0& 0& -1\\ 4& 0& -1\end{array}\right]=\left[\begin{array}{ccc}4& 8& 0\\ -2& -4& -1\\ -1& -2& 3\end{array}\right]$$

c. Read each formula off the rows of the matrix from part b:

$$T_1(T_2(x_1,x_2,x_3))=(T_1\circ T_2)(x_1,x_2,x_3)=(4x_1+8x_2,-2x_1-4x_2-x_3,-x_1-2x_2+3x_3)$$

$$T_2(T_1(x_1,x_2,x_3))=(T_2\circ T_1)(x_1,x_2,x_3)=(2x_2,x_1+3x_2,17x_1+3x_2)$$

Definition

#20

Determine whether the matrix operator $T:R^3\to R^3$ defined by the equations is invertible; if so, find the standard matrix for the inverse operator, and find $T^{-1}(w_1,w_2,w_3)$.

a.

$$\begin{array}{cc}\hfill {\displaystyle w_1}& {\displaystyle =x_1-2x_2+2x_3}\hfill \\ \hfill {\displaystyle w_2}& {\displaystyle =2x_1+x_2+x_3}\hfill \\ \hfill {\displaystyle w_3}& {\displaystyle =x_1+x_2}\hfill \end{array}$$

b.

$$\begin{array}{cc}\hfill {\displaystyle w_1}& {\displaystyle =x_1-3x_2+4x_3}\hfill \\ \hfill {\displaystyle w_2}& {\displaystyle =-x_1+x_2+x_3}\hfill \\ \hfill {\displaystyle w_3}& {\displaystyle =-2x_2+5x_3}\hfill \end{array}$$

Solution

a. Row reduce $[A\mid I]$, where $A$ is the standard matrix of $T$:

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

Create the zeros in this order: (1) below the first pivot, (2) below the second pivot, (3) above the third pivot, (4) above the second pivot, and (5) make every pivot 1. In the matrices below, the entries made in each step are shown in that step’s color.

Step 1: $R_2\to 2R_1-R_2$ and $R_3\to R_1-R_3$

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

Step 2: $R_3\to 5R_3-3R_2$

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

Step 3: $R_1\to R_1-2R_3$ and ${\displaystyle R_2\to -\frac{1}{5}(R_2-3R_3)}$

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -6}\hfill & \hfill {\displaystyle 10}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -3}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -5}\hfill \end{array}\right]$$

Step 4: $R_1\to R_1+2R_2$

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

Step 5: every pivot is already 1, so the left side is $I$.

The left side reduced to $I$, so $A$ is invertible, and so is $T$. The standard matrix for the inverse operator is the right side:

$$[T^{-1}]=A^{-1}=\left[\begin{array}{ccc}1& -2& 4\\ -1& 2& -3\\ -1& 3& -5\end{array}\right]$$

so

b. Row reduce $[A\mid I]$:

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

$R_2\to R_2+R_1$:

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

Rows 2 and 3 now have the same left side, so $R_3\to R_3-R_2$ would give a row of zeros on the left. The left side can’t reduce to $I$, so $A$ is not invertible.

Answer: $T$ is not invertible.