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8.1 General Linear Transformations: Class Notes

Definition: (analogous to Theorem 1.8.2)

If $T:V\to W$ is a mapping from a vector space $V$ to a vector space $W$, then $T$ is called a linear transformation from $V$ to $W$ if the following two properties hold for all vectors $𝐮$ and $𝐯$ in $V$ and for all scalars $k$:
a) $T(k𝐮)=kT(𝐮)$ (homogeneity property)
b) $T(𝐮+𝐯)=T(𝐮)+T(𝐯)$ (additivity property)

In the special case where $V=W$, the linear transformation $T$ is called a linear operator on the vector space $V$.

Theorem 8.1.1 (analogous to Theorem 1.8.1)

If $T:V\to W$ is a linear transformation, then:
a) $T(\mathrm{𝟎})=\mathrm{𝟎}$.
b) $T(𝐮-𝐯)=T(𝐮)-T(𝐯)$ for all $𝐮$ and $𝐯$ in $V$.
c) $T(-𝐯)=-T(𝐯)$ for all $𝐯$ in $V$.

Proof of (a)

Since $T$ is linear, $T(k\overrightarrow{x})=kT(\overrightarrow{x})$ for all scalars $k$. Let $k=0$. Then

$$T(\overrightarrow{0})=T(0\overrightarrow{x})=0T(\overrightarrow{x})=\overrightarrow{0}.$$

Examples of Linear Transformations

• $T:V\to W$ defined by $T(𝐯)=\mathrm{𝟎}$ for all $𝐯\in V$ (the zero transformation)
• $I:V\to V$ defined by $I(𝐯)=𝐯$ (the identity operator)
• $T:V\to V$ defined by $T(𝐱)=c𝐱$ (contraction if $0<c<1$ and dilation if $c>1$)
• $T:V\to R$ defined by $T(𝐱)=⟨𝐱,{𝐱}_0⟩$ (inner product of $𝐱$ with ${𝐱}_0$) [We’ll see this in Ch. 6]
• $T:M_{nn}\to M_{nn}$ defined by $T(A)=A^T$
• $T:V\to R^n$ defined by $T(f)=(f(x_1),f(x_2),\dots ,f(x_n))$, where $V$ is a subspace of $F(-\infty ,\infty )$, and $x_1,x_2,\dots ,x_n$ is a sequence of real numbers (evaluation transformation)
• $D:C^1(-\infty ,\infty )\to F(-\infty ,\infty )$ defined by $D(𝐟)=f\prime (x)$ (differentiation)
• $J:C(-\infty ,\infty )\to C^1(-\infty ,\infty )$ defined by ${\displaystyle J(𝐟)={\int }_0^{x}f(t)dt}$ (integration)

Theorem 8.1.2

Let $T:V\to W$ be a linear transformation, for which the vector space $V$ is finite-dimensional. If $S=\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_n\}$ is a basis for $V$, then the image of any vector $𝐯$ in $V$ can be expressed as
$T(𝐯)=c_{1}T({𝐯}_1)+c_{2}T({𝐯}_2)+\dots +c_{n}T({𝐯}_n)$ where $c_1,c_2,\dots ,c_n$ are the coefficients required to express $𝐯$ as a linear combination of the vectors in the basis $S$.

That is, linear transformations preserve linear combinations.

Proof

Let $\overrightarrow{v}\in V$. Then there are $c_i\in R$ such that $\overrightarrow{v}=\sum _{i=1}^n{c}_i{\overrightarrow{v}}_i$. So

$$\begin{array}{cc}\hfill {\displaystyle T(\overrightarrow{v})}& {\displaystyle =T\left(\sum _{i=1}^n{c}_i{\overrightarrow{v}}_i\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\sum _{i=1}^{n}T(c_i{\overrightarrow{v}}_i)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\sum _{i=1}^n{c}_{i}T({\overrightarrow{v}}_i).}\hfill \end{array}$$

The second line uses additivity and the third uses homogeneity. ✓

#20

Consider the basis $S=\{{𝐯}_1,{𝐯}_2\}$ for $R^2$, where ${𝐯}_1=(-2,1)$ and ${𝐯}_2=(1,3)$, and let $T:R^2\to R^3$ be the linear transformation such that $T({𝐯}_1)=(-1,2,0)$ and $T({𝐯}_2)=(0,-3,5)$. Find a formula for $T(x_1,x_2)$ and use that formula to find $T(2,-3)$.

Solution

When we did this in 1.8, we wanted the standard matrix for the transformation. Here, we will find the formula.

We start by expressing $\overrightarrow{x}=(x_1,x_2)$ as a linear combination of ${\overrightarrow{v}}_1$ and ${\overrightarrow{v}}_2$:

$$(x_1,x_2)=c_1(-2,1)+c_2(1,3)$$

This gives the system below; row reduce its augmented matrix:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle -2c_1+c_2}& {\displaystyle =x_1}\hfill \\ \hfill {\displaystyle c_1+3c_2}& {\displaystyle =x_2}\hfill \end{array}⟶\left[\begin{array}{ccc}\hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle x_1}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle x_2}\hfill \end{array}\right]⟶\left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -\frac{3}{7}{x}_1+\frac{1}{7}{x}_2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle \frac{1}{7}{x}_1+\frac{2}{7}{x}_2}\hfill \end{array}\right]}$$

$${\displaystyle c_1=-\frac{3}{7}{x}_1+\frac{1}{7}{x}_2,c_2=\frac{1}{7}{x}_1+\frac{2}{7}{x}_2}$$

$${\displaystyle (x_1,x_2)=(-\frac{3}{7}{x}_1+\frac{1}{7}{x}_2)(-2,1)+(\frac{1}{7}{x}_1+\frac{2}{7}{x}_2)(1,3)}$$

Then, by Theorem 8.1.2, with $T({\overrightarrow{v}}_1)$ and $T({\overrightarrow{v}}_2)$ given,

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle T(x_1,x_2)}& {\displaystyle =(-\frac{3}{7}{x}_1+\frac{1}{7}{x}_2)T({\overrightarrow{v}}_1)+(\frac{1}{7}{x}_1+\frac{2}{7}{x}_2)T({\overrightarrow{v}}_2)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(-\frac{3}{7}{x}_1+\frac{1}{7}{x}_2)(-1,2,0)+(\frac{1}{7}{x}_1+\frac{2}{7}{x}_2)(0,-3,5)}\hfill \end{array}}$$

Collect the first, second, and third components:

$${\displaystyle =(\frac{3}{7}{x}_1-\frac{1}{7}{x}_2+0,-\frac{6}{7}{x}_1+\frac{2}{7}{x}_2-\frac{3}{7}{x}_1-\frac{6}{7}{x}_2,0+\frac{5}{7}{x}_1+\frac{10}{7}{x}_2)}$$

$${\displaystyle T(x_1,x_2)=(\frac{3}{7}{x}_1-\frac{1}{7}{x}_2,-\frac{9}{7}{x}_1-\frac{4}{7}{x}_2,\frac{5}{7}{x}_1+\frac{10}{7}{x}_2)}$$

So

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle T(2,-3)}& {\displaystyle =(\frac{6}{7}+\frac{3}{7},-\frac{18}{7}+\frac{12}{7},\frac{10}{7}-\frac{30}{7})}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(\frac{9}{7},-\frac{6}{7},-\frac{20}{7}).}\hfill \end{array}}$$

Definition: (analogous to definitions seen in Section 4.2 (kernel) and Section 1.8 (range); also related to Definition 2 of Section 4.8 (null space) and, by Theorem 4.8.1, column space)

If $T:V\to W$ is a linear transformation, then the set of vectors in $V$ that $T$ maps into $\mathrm{𝟎}$ is called the kernel of $T$ and is denoted by $\ker (T)$. The set of all vectors in $W$ that are images under $T$ of at least one vector in $V$ is called the range of $T$ and is denoted by $R(T)$.

A linear transformation $T:V\to W$ drawn as two planes, $V$ on the left and $W$ on the right. Inside $V$ is a smaller region, $\ker (T)$; solid lines (in blue) show every vector in $\ker (T)$ going to the single point $\mathbf{0}$ in $W$. Dotted lines (in magenta) from the corners of $V$ show all of $V$ going onto a smaller region of $W$, the range $R(T)$, which contains $\mathbf{0}$.

#7

Determine whether the mapping $T$ is a linear transformation, and if so, find its kernel.
$T:P_2\to P_2$, where
a. $T(a_0+a_{1}x+a_2{x}^2)$ $=a_0+a_1(x+1)+a_2(x+1{)}^2$
b. $T(a_0+a_{1}x+a_2{x}^2)$ $=(a_0+1)+(a_1+1)x+(a_2+1)x^2$

Solution

Let $\overrightarrow{p}=a_0+a_{1}x+a_2{x}^2$ and $\overrightarrow{q}=b_0+b_{1}x+b_2{x}^2$.

a) Checking $T(k\overrightarrow{p}+\overrightarrow{q})$ checks both properties at once:

$$\begin{array}{cc}\hfill {\displaystyle T(k\overrightarrow{p}+\overrightarrow{q})}& {\displaystyle =T(k{a}_0+b_0+(k{a}_1+b_1)x+(k{a}_2+b_2)x^2)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =k{a}_0+b_0+(k{a}_1+b_1)(x+1)+(k{a}_2+b_2)(x+1{)}^2}\hfill \\ \hfill {\displaystyle }& {\displaystyle =k[a_0+a_1(x+1)+a_2(x+1{)}^2]+b_0+b_1(x+1)+b_2(x+1{)}^2}\hfill \\ \hfill {\displaystyle }& {\displaystyle =kT(\overrightarrow{p})+T(\overrightarrow{q}).}\hfill \end{array}$$

$T$ is linear.

$$\ker (T)=\{\overrightarrow{p}\in P_2\mid a_0+a_1(x+1)+a_2(x+1{)}^2=0\text{ for all }x\}$$

Since $1$, $x+1$, and $(x+1{)}^2$ are linearly independent, this forces $a_0=a_1=a_2=0$, so

$$\ker (T)=\{0\}.$$

b) $T(0)=1+x+x^2\ne 0$, so $T$ is not linear (a linear transformation sends $\overrightarrow{0}$ to $\overrightarrow{0}$ by Theorem 8.1.1).

#10

Let $T:P_2\to P_3$ be the linear transformation defined by $T(p(x))=xp(x)$. Which of the following are in $\ker (T)$?
a. $x^2$
b. $0$
c. $1+x$
d. $-x$

Solution

$$\ker (T)=\{p(x)\mid xp(x)=0\}.$$

The only polynomial with $xp(x)=0$ is the zero polynomial. It follows that $\ker (T)=\{0\}$, so only b is in $\ker (T)$.

#11

Let $T:P_2\to P_3$ be the linear transformation in Exercise 10. Which of the following are in $R(T)$?
a. $x+x^2$
b. $1+x$
c. $3-x^2$
d. $-x$

Solution

$$R(T)=\{xp(x)\mid p(x)\in P_2\}.$$

These are polynomials of degree less than or equal to 3 with constant term 0. Of the choices, only $x+x^2$ and $-x$ have constant term 0.

#25

Let $T_A:R^4\to R^3$ be multiplication by $A$. Find a basis for the kernel of $T_A$, and then find a basis for the range of $T_A$ that consists of column vectors of $A$.

Solution

$${\displaystyle \left[\begin{array}{cccc}1& 2& -1& -2\\ -3& 1& 3& 4\\ -3& 8& 4& 2\end{array}\right]⟶\left[\begin{array}{cccc}1& 0& 0& -\frac{10}{7}\\ 0& 1& 0& -\frac{2}{7}\\ 0& 0& 1& 0\end{array}\right]}$$

So ${\displaystyle x_1=\frac{10}{7}{x}_4}$, ${\displaystyle x_2=\frac{2}{7}{x}_4}$, and $x_3=0$. With $x_4=t$,

$${\displaystyle \overrightarrow{x}=\left[\begin{array}{c}\frac{10}{7}\\ \frac{2}{7}\\ 0\\ 1\end{array}\right]t.}$$

Letting $t=7$, a basis for $\ker (T_A)$ is $\left\{\left[\begin{array}{c}10\\ 2\\ 0\\ 7\end{array}\right]\right\}$.

The pivots are in columns 1, 2, and 3, so a basis for $R(T_A)$ is

$$\{\left[\begin{array}{c}1\\ -3\\ -3\end{array}\right],\left[\begin{array}{c}2\\ 1\\ 8\end{array}\right],\left[\begin{array}{c}-1\\ 3\\ 4\end{array}\right]\}.$$

Theorem 8.1.3

If $T:V\to W$ is a linear transformation, then:
a) The kernel of $T$ is a subspace of $V$.
b) The range of $T$ is a subspace of $W$.

Definition: (analogous to Definition 1 of Section 4.9)

Let $T:V\to W$ be a linear transformation. In the case that the range of $T$ is finite-dimensional its dimension is called the rank of T, denoted by $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(T)$; and if the kernel of $T$ is finite-dimensional, then its dimension is called the nullity of T, denoted by $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(T)$.

Theorem 8.1.4 Dimension Theorem for Linear Transformations

(generalization of Theorem 4.9.2)
If $T:V\to W$ is a linear transformation from a finite-dimensional vector space $V$ to a vector space $W$, then the range of $T$ is finite-dimensional, and
$\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(T)+\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(T)=\dim (V)$.

#13

In each part, use the given information to find the nullity of the linear transformation $T$.
a. $T:R^5\to P_5$ has rank 3.
b. $T:P_4\to P_3$ has rank 1.
c. The range of $T:M_{mn}\to R^3$ is $R^3$.
d. $T:M_{22}\to M_{22}$ has rank 3.

Solution

Nullity is $\dim [\ker (T)]$. Recall that the vectors in $\ker (T)$ are domain elements, so $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(T)=\dim (V)-\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(T)$, where $V$ is the domain.

a. $\dim (R^5)=5$: $5-3=2$, so $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(T)=2$.

b. $\dim (P_4)=5$: $5-1=4$, so $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(T)=4$.

c. $\dim (M_{mn})=mn$ and the rank is 3, so $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(T)=mn-3$.

d. $\dim (M_{22})=4$: $4-3=1$, so $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(T)=1$.