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8.2 Compositions and Inverse Transformations: Class Notes

Definition

If $T:V\to W$ is a linear transformation from a vector space $V$ to a vector space $W$, then $T$ is said to be one-to-one if $T$ maps distinct vectors in $V$ into distinct vectors in $W$.
OR $T:V\to W$ is one-to-one if and only if for each vector $𝐰$ in the range of $T$, there is exactly one vector $𝐯$ in $V$ such that $T(𝐯)=𝐰$.
OR $T:V\to W$ is one-to-one if and only if $T(𝐮)=T(𝐯)$ implies $𝐮=𝐯$.

Definition

If $T:V\to W$ is a linear transformation from a vector space $V$ to a vector space $W$, then $T$ is said to be onto (or onto W) if every vector in $W$ is the image of at least one vector in $V$.

Examples

One-to-one and onto:

• Rotation operators on $R^2$ ($0\le \theta <2\pi$)
• $T:P_3\to R^4$ where $T(a+bx+c{x}^2+d{x}^3)=(a,b,c,d)$
• $T:M_{22}\to R^4$ where $T\left(\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\right)=(a,b,c,d)$

One-to-one but not onto:

• $T:P_n\to P_{n+1}$, where $T(\overrightarrow{p})=T(p(x))=xp(x)$

Not one-to-one:

• $D:C^1(-\infty ,\infty )\to F(-\infty ,\infty )$, where $D(\overrightarrow{f})=f\prime (x)$. Example: $D(x^2)=D(x^2+1)=2x$.

Theorem 8.2.1

If $T:V\to W$ is a linear transformation, then $T$ is one-to-one if and only if $\ker (T)=\{\mathrm{𝟎}\}$.

Proof

($\Rightarrow$) By Theorem 8.1.1, $T(\overrightarrow{0})=\overrightarrow{0}$ because $T$ is linear. If $T$ is one-to-one and if $T(\overrightarrow{v})=\overrightarrow{0}$, then $\overrightarrow{v}=\overrightarrow{0}$ by definition. Thus $\ker (T)=\{\overrightarrow{0}\}$.

($\Leftarrow$) Let $\overrightarrow{u},\overrightarrow{v}\in V$ such that $\overrightarrow{u}\ne \overrightarrow{v}$. Then $\overrightarrow{u}-\overrightarrow{v}\ne \overrightarrow{0}$, so, since $\ker (T)=\{\overrightarrow{0}\}$ and $T$ is linear,

$$T(\overrightarrow{u})-T(\overrightarrow{v})=T(\overrightarrow{u}-\overrightarrow{v})\ne \overrightarrow{0}\Rightarrow T(\overrightarrow{u})\ne T(\overrightarrow{v})\Rightarrow T\text{ is one-to-one.}$$

#4

Determine whether the linear transformation is one-to-one by finding its kernel and then applying Theorem 8.2.1.
a. $T:R^2\to R^3$, where $T(x,y)=(x-y,y-x,2x-2y)$
b. $T:R^2\to R^2$, where $T(x,y)=(0,2x+3y)$
c. $T:R^2\to R^2$, where $T(x,y)=(x+y,x-y)$

Solution

a. $\ker (T)=\{(x,y)\mid x-y=0,$ $y-x=0,$ $2x-2y=0\}$

These are all equivalent to $y=x$, a line through the origin, so $\ker (T)\ne \{\overrightarrow{0}\}$. The kernel contains nonzero vectors such as $(1,1)$.

b. $\ker (T)=\{(x,y)\mid 2x+3y=0\}$. Since this means ${\displaystyle y=-\frac{2}{3}x}$, this is again a line through the origin, so $\ker (T)\ne \{\overrightarrow{0}\}$.

c. $\ker (T)=\{(x,y)\mid x+y=0,x-y=0\}$. If

$$\begin{array}{cc}\hfill {\displaystyle x+y}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle x-y}& {\displaystyle =0}\hfill \end{array}$$

then, adding, $2x=0\Rightarrow x=0\Rightarrow y=0$. So $\ker (T)=\{(0,0)\}=\{\overrightarrow{0}\}$.

#5

Determine whether multiplication by $A$ is one-to-one by computing the nullity of $A$ and then applying Theorem 8.2.1.

Solution

a.

$$A=\left[\begin{array}{cc}1& -2\\ 2& -4\\ -3& 6\end{array}\right]⟶\left[\begin{array}{cc}1& -2\\ 0& 0\\ 0& 0\end{array}\right]=\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{f}(A)$$

$\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)=1$. Since $A$ has 2 columns and $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)+\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(A)=n=2$, $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(A)=1$.

Since $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(A)\ne 0$, $\ker (T_A)\ne \{\overrightarrow{0}\}$.

b. $A$ has 4 columns but $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)\le 3$ (it has only 3 rows). Since $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)+\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(A)=4$, $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(A)\ge 1$.

Theorem 8.2.X

If $T:V\to W$ is a one-to-one linear transformation and $\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_k\}$ is a linearly independent subset of $V$, then $\{T({𝐯}_1),T({𝐯}_2),\dots ,T({𝐯}_k)\}$ is linearly independent.

That is, a one-to-one transformation preserves linear independence.

Proof

Suppose

$$c_{1}T({\overrightarrow{v}}_1)+c_{2}T({\overrightarrow{v}}_2)+\cdots +c_{k}T({\overrightarrow{v}}_k)=\overrightarrow{0}.$$

Then, because $T$ is linear,

$$T(c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+\cdots +c_k{\overrightarrow{v}}_k)=\overrightarrow{0}.$$

Since $T$ is one-to-one (so $\ker (T)=\{\overrightarrow{0}\}$),

$$c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+\cdots +c_k{\overrightarrow{v}}_k=\overrightarrow{0}.$$

Then $c_1=c_2=\cdots =c_k=0$ because $\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\dots ,{\overrightarrow{v}}_k\}$ is linearly independent. It follows that $\{T({\overrightarrow{v}}_1),T({\overrightarrow{v}}_2),\dots ,T({\overrightarrow{v}}_k)\}$ is linearly independent.

Theorem 8.2.2

If $V$ and $W$ are finite-dimensional vector spaces with the same dimension, and if $T:V\to W$ is a linear transformation, then the following statements are equivalent.
a) $T$ is one-to-one.
b) $\ker (T)=\{\mathrm{𝟎}\}$.
c) $T$ is onto [i.e., $R(T)=W$].

$R(T)$ is the range of $T$.

Proof

a $\iff$ b is established by Theorem 8.2.1.

b $\iff$ c: Assume $\dim (V)=n=\dim (W)$. We know that $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(T)+\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(T)=n$ (Theorem 8.1.4).

$$\begin{array}{cc}\hfill {\displaystyle \ker (T)=\{\overrightarrow{0}\}}& {\displaystyle \iff \mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(T)=0}\hfill \\ \hfill {\displaystyle }& {\displaystyle \iff \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(T)=n}\hfill \\ \hfill {\displaystyle }& {\displaystyle \iff R(T)=W}\hfill \\ \hfill {\displaystyle }& {\displaystyle (R(T)\text{ is a subspace of }W}\hfill \\ \hfill {\displaystyle }& {\displaystyle \text{with the same dimension})}\hfill \\ \hfill {\displaystyle }& {\displaystyle \iff T\text{ is onto.}}\hfill \end{array}$$

Theorem 8.2.3

If $T_A:R^n\to R^m$ is a matrix transformation, then
a) $T_A$ is one-to-one if and only if the columns of $A$ are linearly independent.
b) $T_A$ is onto if and only if the columns of $A$ span $R^m$.

Proof

a) $T_A$ is one-to-one if and only if $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(A)=0$, which holds if and only if $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)=n$ (since $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)+\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(A)=n$). Thus the $n$ columns of $A$ are linearly independent (a basis for $\mathrm{c}\mathrm{o}\mathrm{l}(A)$ is the $n$ pivot columns).

b) $T_A$ is onto $\iff A\overrightarrow{x}=\overrightarrow{b}$ has a solution for all $\overrightarrow{b}\in R^m$ $\iff$ every $\overrightarrow{b}\in R^m$ is in $\mathrm{c}\mathrm{o}\mathrm{l}(A)$ $\iff$ the columns of $A$ span $R^m$.

#14

Use Theorem 8.2.3 to determine whether multiplication by $A$ is one-to-one, onto, both, or neither. Justify your answer.

Solution

a) Not onto: 2 columns can’t span $R^3$. One-to-one: ${\overrightarrow{c}}_1\ne k{\overrightarrow{c}}_2$ for any $k$, so the two columns are linearly independent.

b) Not one-to-one: ${\overrightarrow{c}}_2=-{\overrightarrow{c}}_1$ and ${\overrightarrow{c}}_1=3{\overrightarrow{c}}_4$, so the columns are not linearly independent.

$$\left[\begin{array}{cccc}3& -3& 1& 1\\ 6& -6& 0& 2\\ 9& -9& 1& 3\end{array}\right]⟶\left[\begin{array}{cccc}3& -3& 1& 1\\ 0& 0& 2& 0\\ 0& 0& 2& 0\end{array}\right]⟶\left[\begin{array}{cccc}3& -3& 1& 1\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]$$

2 pivots, so $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)=2\ne 3$ and the columns do not span $R^3$. Not onto.

c) ${\overrightarrow{c}}_2=-3{\overrightarrow{c}}_1$, so $T_A$ is not one-to-one (the columns are linearly dependent). It is also not onto (the columns don’t span $R^2$).

d) $det(A)=2·1·1=2\ne 0$, so $A$ is invertible and its columns are linearly independent and span $R^3$: one-to-one and onto.

If $T_A(\overrightarrow{x})=A\overrightarrow{x}$ with $A$ shaped like a wide matrix (more columns than rows), then $T_A$ is not one-to-one. If $A$ is shaped like a tall matrix (more rows than columns), $T_A$ can’t be onto.

A wide bracketed matrix, labeled not one-to-one, next to a tall bracketed matrix, labeled can’t be onto.

In the same way, for $T:V\to W$:

Top: $V$ is drawn smaller than $W$, and $T$ can’t be onto. Bottom: $V$ is drawn larger than $W$, and $T$ can’t be one-to-one.

Theorem 8.2.4 Equivalent Statements (extends Theorems 4.9.8, 5.1.5 and 6.4.5, the last two of which we have not seen)

If $A$ is an $n\times n$ matrix, then the following statements are equivalent.
a) $A$ is invertible.
b) $A𝐱=\mathrm{𝟎}$ has only the trivial solution.
c) The reduced row echelon form of $A$ is $I_n$.
d) $A$ is expressible as a product of elementary matrices.
e) $A𝐱=𝐛$ is consistent for every $n\times 1$ matrix $𝐛$.
f) $A𝐱=𝐛$ has exactly one solution for every $n\times 1$ matrix $𝐛$.
g) $det(A)\ne 0$.
h) The column vectors of $A$ are distinct and linearly independent.
i) The row vectors of $A$ are distinct and linearly independent.
j) The column vectors of $A$ span $R^n$.
k) The row vectors of $A$ span $R^n$.
l) The column vectors of $A$ form a basis for $R^n$.
m) The row vectors of $A$ form a basis for $R^n$.
n) $A$ has rank $n$.
o) $A$ has nullity 0.
p) The orthogonal complement of the null space of $A$ is $R^n$.
q) The orthogonal complement of the row space of $A$ is $\{\mathrm{𝟎}\}$.
r) $\lambda =0$ is not an eigenvalue of $A$. [This is from Theorem 5.1.5.]
s) $A^{T}A$ is invertible. [This is from Theorem 6.4.5.]
t) The kernel of $T_A$ is $\{\mathrm{𝟎}\}$.
u) The range of $T_A$ is $R^n$.
v) $T_A$ is one-to-one.

Statements t), u), and v) are the new additions.

Perspectives

If $A$ is an $m\times n$ matrix, a subspace of $R^n$ (input side) can be considered as:

• The null space of $A$ (matrix view)
• The solution space of $A\overrightarrow{x}=\overrightarrow{0}$ (system view)
• The kernel of $T_A$ (transformation view)

A subspace of $R^m$ can be considered as:

• The column space of $A$ (matrix view)
• All $\overrightarrow{b}\in R^m$ such that $A\overrightarrow{x}=\overrightarrow{b}$ is consistent (system view)
• The range of $T_A$ (transformation view)

Definition

If $T:V\to W$ is a one-to-one linear transformation with range $R(T)$, and if $𝐰$ is any vector in $R(T)$, then there is exactly one vector $𝐯$ in $V$ for which $T(𝐯)=𝐰$. The inverse of T maps $𝐰$ back into $𝐯$ and is denoted by $T^{-1}$. (This is analogous to inverse in Section 1.9.)

Definition: (analogous to composition in Section 1.9)

If $T_1:U\to V$ and $T_2:V\to W$ are linear transformations, then the composition of $T_2$ with $T_1$, denoted by $T_2\circ T_1$ (which is read “$T_2$ circle $T_1$”) is the function defined by the formula $(T_2\circ T_1)(𝐮)=T_2(T_1(𝐮))$ where $𝐮$ is a vector in $U$.

Note that the order of composition matters.

Theorem 8.2.5

If $T_1:U\to V$ and $T_2:V\to W$ are linear transformations, then $(T_2\circ T_1):U\to W$ is also a linear transformation.

Tip: This is a good proof to practice.

Proof

Let $\overrightarrow{u},\overrightarrow{v}\in U$ and $k\in R$.

$$\begin{array}{ccc}\hfill {\displaystyle (T_2\circ T_1)(k\overrightarrow{u}+\overrightarrow{v})}& {\displaystyle =T_2(T_1(k\overrightarrow{u}+\overrightarrow{v}))}\hfill & {\displaystyle \text{Definition of composition}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =T_2(k{T}_1(\overrightarrow{u})+T_1(\overrightarrow{v}))}\hfill & {\displaystyle T_1\text{ is linear}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =k{T}_2(T_1(\overrightarrow{u}))+T_2(T_1(\overrightarrow{v}))}\hfill & {\displaystyle T_2\text{ is linear}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =k(T_2\circ T_1)(\overrightarrow{u})+(T_2\circ T_1)(\overrightarrow{v})}\hfill & {\displaystyle \text{Definition of composition}}\hfill \end{array}$$

Therefore $T_2\circ T_1$ is linear.

Theorem 8.2.6

If $T_1:U\to V$ and $T_2:V\to W$ are one-to-one linear transformations, then:
a) $T_2\circ T_1$ is one-to-one.
b) $(T_2\circ T_1{)}^{-1}=T_1^{-1}\circ T_2^{-1}$.

#38

Let $D(𝐟)=f\prime (x)$ and ${\displaystyle J(𝐟)={\int }_0^{x}f(t)dt}$ be the derivative and integral linear transformations. Find $(J\circ D)(𝐟)$ for
a. $𝐟(x)=x^2+3x+2$
b. $𝐟(x)=\sin x$

Solution

a) First apply $D$, then $J$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle (J\circ D)(\overrightarrow{f})}& {\displaystyle =J(f\prime (x))=J(\frac{d}{dx}(x^2+3x+2))}\hfill \\ \hfill {\displaystyle }& {\displaystyle =J(2x+3)={\int }_0^x(2t+3)dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle ={(t^2+3t)|}_0^x=x^2+3x}\hfill \end{array}}$$

b) $D(\overrightarrow{f})=\cos x$, so

$${\displaystyle (J\circ D)(\overrightarrow{f})=J(\cos x)={\int }_0^x\frac{d}{dt}(\sin t)dt={\int }_0^x\cos tdt=\sin x}$$

Key concepts from 8.3

Definition

A linear transformation $T:V\to W$ that is both one-to-one and onto is said to be an isomorphism, and $W$ is said to be isomorphic to $V$.

Theorem 8.3.1

Every real $n$-dimensional vector space is isomorphic to $R^n$. [In this sense, every $n$-dimensional vector space is algebraically equivalent to $R^n$.]

Theorem 8.3.2

If $S$ is an ordered basis for a vector space $V$, then the coordinate map $𝐮\stackrel{T}{⟶}(𝐮{)}_S$ is an isomorphism between $V$ and $R^n$.

Theorem 8.3.X

A linear transformation $T:V\to W$ is an isomorphism if and only if whenever $\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_k\}$ is a basis for $V$, $\{T({𝐯}_1),T({𝐯}_2),\dots ,T({𝐯}_k)\}$ is a basis for $W$.

Examples of Natural isomorphisms

Between $P_{n-1}$ and $R^n$: $a_0+a_{1}x+\dots +a_{n-1}{x}^{n-1}$ $\stackrel{T}{⟶}(a_0,a_1,\dots ,a_{n-1})$
Between $M_{22}$ and $R^4$: $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\stackrel{T}{⟶}(a,b,c,d)$