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1.8 Introduction to Linear Transformations: Class Notes

Definitions

$R^2$ is the $xy$-plane, and $R^3$ is 3-space.

All vectors in $R^n$ can be expressed as linear combinations of these vectors:

$$\overrightarrow{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right]=x_1{\overrightarrow{e}}_1+x_2{\overrightarrow{e}}_2+\cdots +x_n{\overrightarrow{e}}_n$$

Definitions

$f(a)=b$: $b$ is the image of $a$ under $f$, where $a\in$ domain of $f$ and $b\in$ codomain of $f$.

Consider $f(x)=|x|$. Domain: $ℝ$. Codomain: $ℝ$. Range: $[0,\infty )$. The range is a subset of the codomain.

Transformations

Consider the system

$$\begin{array}{cc}\hfill {\displaystyle 4}& {\displaystyle =2x-3y+5z}\hfill \\ \hfill {\displaystyle 6}& {\displaystyle =7x+2y-z}\hfill \end{array}$$

In matrix form, this is

$$\left[\begin{array}{c}4\\ 6\end{array}\right]=\left[\begin{array}{ccc}2& -3& 5\\ 7& 2& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$$

The $2\times 3$ matrix $A=\left[\begin{array}{ccc}2& -3& 5\\ 7& 2& -1\end{array}\right]$ maps vectors in $R^3$ into vectors in $R^2$. $A$ is called a matrix transformation from $R^3$ to $R^2$.

Definition

#5

Find the domain and codomain of the transformation defined by the matrix product.

a. $\left[\begin{array}{ccc}3& 1& 2\\ 6& 7& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$  b. $\left[\begin{array}{cc}2& -1\\ 4& 3\\ 2& -5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$

Solution

a. A $2\times 3$ matrix times a $3\times 1$ vector gives a $2\times 1$ vector.

Definitions

#8

Find the domain and codomain of the transformation $T$ defined by the formula.

a. $T(x_1,x_2,x_3,x_4)=(x_1,x_2)$
b. $T(x_1,x_2,x_3)=(x_1,x_2-x_3,x_2)$

Solution

For example, $T(3,4,5,6)=(3,4)$ in a, and $T(5,6,7)=(5,-1,6)$ in b.

#12

Find the standard matrix for the transformation defined by the equations.

a.

$$\begin{array}{cc}\hfill {\displaystyle w_1}& {\displaystyle =-x_1+x_2}\hfill \\ \hfill {\displaystyle w_2}& {\displaystyle =3x_1-2x_2}\hfill \\ \hfill {\displaystyle w_3}& {\displaystyle =5x_1-7x_2}\hfill \end{array}$$

b.

$$\begin{array}{cc}\hfill {\displaystyle w_1}& {\displaystyle =x_1}\hfill \\ \hfill {\displaystyle w_2}& {\displaystyle =x_1+x_2}\hfill \\ \hfill {\displaystyle w_3}& {\displaystyle =x_1+x_2+x_3}\hfill \\ \hfill {\displaystyle w_4}& {\displaystyle =x_1+x_2+x_3+x_4}\hfill \end{array}$$

Solution

a. In matrix form, $\overrightarrow{w}=\left[\begin{array}{cc}-1& 1\\ 3& -2\\ 5& -7\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$.

b.

#13

Find the standard matrix for the transformation $T$ defined by the formula.

a. $T(x_1,x_2)=(2x_1-x_2,x_1+x_2)$
b. $T(x_1,x_2,x_3)=(4x_1+x_2,x_1+x_2)$

Solution

a. As a column, $T\left(\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\right)=\left[\begin{array}{c}2x_1-x_2\\ x_1+x_2\end{array}\right]$. The coefficients of the first component form row 1, and the coefficients of the second component form row 2.

b.

Two special transformations:

Definitions

Theorem 1.8.1 Properties of Matrix Transformations

Here $T_A(\overrightarrow{x})=A\overrightarrow{x}$. These properties follow from properties of matrix arithmetic.

Tip: Compare with functions of real numbers: $3(x+y)=3x+3y$ has the additivity property, but $(x+y{)}^2\ne x^2+y^2$.

Theorem 1.8.2

Proof

$(\Rightarrow )$ From properties of matrix arithmetic: Theorem 1.8.1 (b) and (c).

$(\Leftarrow )$ Assume (i) and (ii) hold. The proof will be complete if there is an $m\times n$ matrix $A$ such that $T(\overrightarrow{x})=A\overrightarrow{x}$ for all $\overrightarrow{x}\in R^n$. By (i) and (ii),

$$T(k_1{\overrightarrow{u}}_1+k_2{\overrightarrow{u}}_2+\cdots +k_n{\overrightarrow{u}}_n)=k_{1}T({\overrightarrow{u}}_1)+k_{2}T({\overrightarrow{u}}_2)+\cdots +k_{n}T({\overrightarrow{u}}_n)$$

for all vectors ${\overrightarrow{u}}_i\in R^n$ and all scalars $k_i\in ℝ$. Let

$$A=\left[\begin{array}{cccc}\hfill {\displaystyle T({\overrightarrow{e}}_1)}\hfill & \hfill {\displaystyle T({\overrightarrow{e}}_2)}\hfill & \hfill {\displaystyle \cdots }\hfill & \hfill {\displaystyle T({\overrightarrow{e}}_n)}\hfill \end{array}\right],$$

where ${\overrightarrow{e}}_i$ is a standard basis vector for $R^n$. Then, for $\overrightarrow{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right]$,

$$A\overrightarrow{x}=x_{1}T({\overrightarrow{e}}_1)+x_{2}T({\overrightarrow{e}}_2)+\cdots +x_{n}T({\overrightarrow{e}}_n),$$

a linear combination of the column vectors of $A$ in which the coefficients are the entries of $\overrightarrow{x}$ (Theorem 1.3.1). We then have

$$A\overrightarrow{x}=T(x_1{\overrightarrow{e}}_1+x_2{\overrightarrow{e}}_2+\cdots +x_n{\overrightarrow{e}}_n)=T(\overrightarrow{x})$$

by linearity of $T$.

Definitions

Theorem 1.8.3

This is vocabulary: for transformations from $R^n$ to $R^m$, “linear transformation” and “matrix transformation” mean the same thing.

Theorem 1.8.4

#22

Use Theorem 1.8.2 to show that $T$ is a matrix transformation.

a. $T(x,y,z)=(x+y,y+z,x)$
b. $T(x_1,x_2,x_3)=(x_1,x_3,x_1+x_2)$

Solution

We need to show that (i) $T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$ and (ii) $T(k\overrightarrow{u})=kT(\overrightarrow{u})$.

a. Here $T:R^3\to R^3$. Let $\overrightarrow{u}=(u_1,u_2,u_3)$, $\overrightarrow{v}=(v_1,v_2,v_3)$, and $k\in ℝ$. Then $\overrightarrow{u}+\overrightarrow{v}=(u_1+v_1,u_2+v_2,u_3+v_3)$. The three components of $\overrightarrow{u}+\overrightarrow{v}$ play the roles of $x$, $y$, and $z$:

$$\begin{array}{cc}\hfill {\displaystyle T(\overrightarrow{u}+\overrightarrow{v})}& {\displaystyle =T(u_1+v_1,u_2+v_2,u_3+v_3)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(u_1+v_1+u_2+v_2,u_2+v_2+u_3+v_3,u_1+v_1)}\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle T(\overrightarrow{u})+T(\overrightarrow{v})}& {\displaystyle =(u_1+u_2,u_2+u_3,u_1)+(v_1+v_2,v_2+v_3,v_1)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(u_1+u_2+v_1+v_2,u_2+u_3+v_2+v_3,u_1+v_1)}\hfill \end{array}$$

These are equal, so $T(\overrightarrow{u}+\overrightarrow{v})=T(\overrightarrow{u})+T(\overrightarrow{v})$.

$$\begin{array}{cc}\hfill {\displaystyle T(k\overrightarrow{u})}& {\displaystyle =T(k{u}_1,k{u}_2,k{u}_3)=(k{u}_1+k{u}_2,k{u}_2+k{u}_3,k{u}_1)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =k(u_1+u_2,u_2+u_3,u_1)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =kT(\overrightarrow{u})}\hfill \end{array}$$

#23

Use Theorem 1.8.2 to show that $T$ is not a matrix transformation.

a. $T(x,y)=(x^2,y)$
b. $T(x,y,z)=(x,y,xz)$

Solution

a. Homogeneity fails:

$$\begin{array}{cc}\hfill {\displaystyle T(kx,ky)}& {\displaystyle =(k^2{x}^2,ky)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =k(k{x}^2,y)}\hfill \\ \hfill {\displaystyle }& {\displaystyle \ne kT(x,y)}\hfill \end{array}$$

Or, with numbers: let $k=5$, $x=3$, $y=4$. Then

$$\begin{array}{cc}\hfill {\displaystyle T(5·3,5·4)}& {\displaystyle =(25·9,5·4)=(225,20)}\hfill \\ \hfill {\displaystyle 5T(3,4)}& {\displaystyle =5(9,4)=(45,20)\ne (225,20)}\hfill \end{array}$$

Tip: The $(\Leftarrow )$ part of the proof of Theorem 1.8.2 suggests a technique for finding the standard matrix for a transformation: find what it does to the standard basis vectors. These images are the columns.

#28

The images of the standard basis vectors for $R^3$ are given for a linear transformation $T:R^3\to R^3$. Find the standard matrix for the transformation, and find $T(𝐱)$.

$T({𝐞}_1)=\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right]$, $T({𝐞}_2)=\left[\begin{array}{c}-3\\ 1\\ 0\end{array}\right]$, $T({𝐞}_3)=\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]$; $𝐱=\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]$

Solution

The images of the standard basis vectors are the columns of the standard matrix $A$:

$$T(\overrightarrow{x})=\left[\begin{array}{ccc}2& -3& 1\\ 1& 1& 0\\ 3& 0& 2\end{array}\right]\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}5\\ 2\\ 8\end{array}\right]$$

Example

Find the standard matrix $A$ for the linear transformation $T:R^2\to R^3$ for which

$$T\left(\left[\begin{array}{c}-1\\ 2\end{array}\right]\right)=\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]\text{and}T\left(\left[\begin{array}{c}3\\ -5\end{array}\right]\right)=\left[\begin{array}{c}5\\ -7\\ 1\end{array}\right]$$

and use it to compute $T\left(\left[\begin{array}{c}-4\\ 3\end{array}\right]\right)$.

Solution

Thought process: let $\overrightarrow{u}=\left[\begin{array}{c}-1\\ 2\end{array}\right]$ and $\overrightarrow{v}=\left[\begin{array}{c}3\\ -5\end{array}\right]$. If we can find $a$, $b$ such that ${\overrightarrow{e}}_1=a\overrightarrow{u}+b\overrightarrow{v}$, then we can form $T({\overrightarrow{e}}_1)=aT(\overrightarrow{u})+bT(\overrightarrow{v})$. Likewise for ${\overrightarrow{e}}_2$.

We then want $a$, $b$ such that $a\overrightarrow{u}+b\overrightarrow{v}={\overrightarrow{e}}_1$ (or ${\overrightarrow{e}}_2$):

$$a\left[\begin{array}{c}-1\\ 2\end{array}\right]+b\left[\begin{array}{c}3\\ -5\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]\text{ or }\left[\begin{array}{c}0\\ 1\end{array}\right]\Rightarrow \left[\begin{array}{c}-a+3b\\ 2a-5b\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]\text{ or }\left[\begin{array}{c}0\\ 1\end{array}\right]$$

Solve both systems simultaneously, with ${\overrightarrow{e}}_1$ and ${\overrightarrow{e}}_2$ as two right-hand-side columns:

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

$R_2\to 2R_1+R_2$:

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

$R_1\to -(R_1-3R_2)$:

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

The third column gives the $a$ and $b$ that give ${\overrightarrow{e}}_1$; the fourth column gives the $a$ and $b$ that give ${\overrightarrow{e}}_2$.

$${\overrightarrow{e}}_1=5\overrightarrow{u}+2\overrightarrow{v},{\overrightarrow{e}}_2=3\overrightarrow{u}+\overrightarrow{v}$$

$$\begin{array}{cc}\hfill {\displaystyle T({\overrightarrow{e}}_1)}& {\displaystyle =5T(\overrightarrow{u})+2T(\overrightarrow{v})=5\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]+2\left[\begin{array}{c}5\\ -7\\ 1\end{array}\right]=\left[\begin{array}{c}20\\ -9\\ 2\end{array}\right]}\hfill \\ \hfill {\displaystyle T({\overrightarrow{e}}_2)}& {\displaystyle =3T(\overrightarrow{u})+T(\overrightarrow{v})=3\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]+\left[\begin{array}{c}5\\ -7\\ 1\end{array}\right]=\left[\begin{array}{c}11\\ -4\\ 1\end{array}\right]}\hfill \end{array}$$

$$T\left(\left[\begin{array}{c}-4\\ 3\end{array}\right]\right)=\left[\begin{array}{cc}20& 11\\ -9& -4\\ 2& 1\end{array}\right]\left[\begin{array}{c}-4\\ 3\end{array}\right]=\left[\begin{array}{c}-47\\ 24\\ -5\end{array}\right]$$

Matrix operators on ${𝐑}^2$

Reflection operators

• Reflection about the $x$-axis: $T(x,y)=(x,-y)$, $T_A=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$
• Reflection about the $y$-axis: $T(x,y)=(-x,y)$, $T_A=\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]$
• Reflection about the line $y=x$: $T(x,y)=(y,x)$, $T_A=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

A vector $(x,y)$ above the $x$-axis and its reflection $(x,-y)$, the same distance below the $x$-axis.

Projection operators

• Orthogonal projection onto the $x$-axis: $T(x,y)=(x,0)$, $T_A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$
• Orthogonal projection onto the $y$-axis: $T(x,y)=(0,y)$, $T_A=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$

A vector $(x,y)$ and its orthogonal projection $(x,0)$ (drawn thicker) on the $x$-axis, directly below its tip.

Rotation operator

• Counterclockwise rotation about the origin through an angle $\theta$: $R_{\theta }=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$

The first column is $T({\overrightarrow{e}}_1)$ and the second column is $T({\overrightarrow{e}}_2)$.

The standard basis vectors ${\overrightarrow{e}}_1=(1,0)$ and ${\overrightarrow{e}}_2=(0,1)$ are each rotated counterclockwise through the angle $\theta$. The image of ${\overrightarrow{e}}_1$ is $(\cos \theta ,\sin \theta )$. The image of ${\overrightarrow{e}}_2$ makes the angle $\theta +\pi /2$ with the positive $x$-axis, so it is $(\cos (\theta +\pi /2),\sin (\theta +\pi /2))$.

$$\begin{array}{cc}\hfill {\displaystyle \cos (\theta +\pi /2)}& {\displaystyle =\cos \theta \cos \frac{\pi }{2}-\sin \theta \sin \frac{\pi }{2}=\cos \theta ·0-\sin \theta ·1=-\sin \theta }\hfill \end{array}$$

Similarly, $\sin (\theta +\pi /2)=\sin \theta \cos \frac{\pi }{2}+\cos \theta \sin \frac{\pi }{2}=\cos \theta$, so $T({\overrightarrow{e}}_2)=(-\sin \theta ,\cos \theta )$.

Matrix operators on ${𝐑}^3$

Reflection operators

• Reflection about the $xy$-plane: $T(x,y,z)=(x,y,-z)$, $T_A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right]$
• Reflection about the $xz$-plane: $T(x,y,z)=(x,-y,z)$, $T_A=\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 1\end{array}\right]$
• Reflection about the $yz$-plane: $T(x,y,z)=(-x,y,z)$, $T_A=\left[\begin{array}{ccc}-1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Projection operators

• Orthogonal projection onto the $xy$-plane: $T(x,y,z)=(x,y,0)$, $T_A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$
• Orthogonal projection onto the $xz$-plane: $T(x,y,z)=(x,0,z)$, $T_A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]$
• Orthogonal projection onto the $yz$-plane: $T(x,y,z)=(0,y,z)$, $T_A=\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$