Chapter 1 extra note 6

Invertible transformation
one-to-one
onto

Selected lecture notes:

Invertible transformation

Definition:
Identity transformation:

$I_v:V\to V$,

$I_v(\mathbf{v})=\mathbf{v}$,

$\mathrm{\forall }\mathbf{v}\in V$.

Proposition:
For any linear transformation,

$T:V\to W$, we have

$T\circ I_v=T$ and

$I_w\circ T=T$.

• Proof:

  Given

  $\mathbf{v}\in V$,
  

  $$(T\circ I_v)(\mathbf{v})=T(I_v(\mathbf{v}))=T(\mathbf{v}).$$
  Similarly,
  

  $$(I_w\circ T)(\mathbf{v})=I_w(T(\mathbf{v}))=T(\mathbf{v}).$$

Definition:
Let

$T:V\to W$ be a linear transformation,

$T$ is left invertible if

$\mathrm{\exists }B:W\to V$ such that

$B\circ T=I_v$.

Definition:
Let

$T:V\to W$ be a linear transformation,

$T$ is right invertible if

$\mathrm{\exists }C:W\to V$ such that

$T\circ C=I_w$.

Definition:
Let

$T:V\to W$ be a linear transformation,

$T$ is invertible if it is both left and right invertible.

Theorem:
Let

$T:V\to W$ be a linear transformation,

$T$ is invertible if and only if there exists an unique left inverse

$B$ and an unique right inverse

$C$. Moreover,

$B=C$.

• Proof:

  (

  $\Leftarrow$) True by definition.

  (

  $\Rightarrow$)
  Let

  $T:V\to W$ be an invertible linear transformation, then

  $\mathrm{\exists }B$ and

  $C$ such that

  $B\circ T=I_v$ and

  $T\circ C=I_w$.
  Given

  $\mathbf{w}\in W$,
  

  $$B(\mathbf{w})=B(T(C(\mathbf{w})))=(B\circ T)(C(\mathbf{w}))=C(\mathbf{w}).$$
  Therefore

  $B=C$.

  Suppose

  $\mathrm{\exists }{B}_2:W\to V$ such that

  $B_2\circ T=I_v$, then, given

  $\mathbf{w}\in W$,
  

  $$B_2(\mathbf{w})=B_2(T(C(\mathbf{w})))=(B_2\circ T)(C(\mathbf{w}))=C(\mathbf{w}).$$
  Therefore

  $B_2=C=B$. So the left inverse is unique.

  Similarly, the right inverse is unique.

Corollary:

$T:V\to W$ is an invertible linear transformation if and only if

$\mathrm{\exists }{T}^{-1}:W\to V$ such that

$T^{-1}\circ T=I_v$ and

$T\circ T^{-1}=I_w$.

Corollary:
Let

$A:V\to W$ and

$B:U\to V$ be invertible linear transformations, then

$A\circ B:U\to W$ is an invertible linear transformation and

$(A\circ B{)}^{-1}=B^{-1}\circ A^{-1}$.

• Proof:

  $$(A\circ B)\circ (B^{-1}\circ A^{-1})=A\circ (B\circ B^{-1})\circ A^{-1}=A\circ I_v\circ A^{-1}=I_w.$$
  and
  

  $$(B^{-1}\circ A^{-1})\circ (A\circ B)=B^{-1}\circ (A^{-1}\circ A)\circ B=B^{-1}\circ I_v\circ B=I_u.$$

---

Invertibility and solvability

Theorem:
Let

$T:V\to W$ be a linear transformation,

$T$ is invertible if and only if

$\mathrm{\forall }b\in W$, the equation

$T(x)=b$ has an unique solution.

• Proof:

  (

  $\Rightarrow$)
  

  $T$ is invertible, so

  $\mathrm{\exists }{T}^{-1}:W\to V$ that is linear.
  Given

  $b\in W$, define

  $x=T^{-1}(b)$, then
  

  $$T(x)=T(T^{-1}(b))=(T\circ T^{-1})(b)=b.$$

  To check uniqueness, suppose

  $\mathrm{\exists }{x}_2\in V$ such that

  $T(x_2)=b$, then
  

  $$x_2=(T^{-1}\circ T)(x_2)=T^{-1}(T(x_2))=T^{-1}(b)=x.$$
  Therefore, the solution is unique.

  (

  $\Leftarrow$)
  We define a function

  $B:W\to V$ such that
  

  $$B(b)=x,\text{if}T(x)=b.$$
  This function

  $B$ is well-defined since for any

  $b\in W$,

  $\mathrm{\exists }!x\in V$ such that

  $T(x)=b.$

  claim 1:

  $B\circ T=I_v$

  pf:
  Let

  $x\in V$ and

  $T(x)=b\in W$.
  Since

  $T(x)=b$ has only one solution, therefore

  $B(b)=x$.
  Then we have
  

  $$B(T(x))=B(b)=x.$$

  claim 2:

  $T\circ B=I_w$

  pf:
  Let

  $b\in W$,

  $\mathrm{\exists }!x\in V$ such that

  $T(x)=b$, hence

  $B(b)=x$, and
  

  $$T(B(b))=T(x)=b.$$

  claim 3:

  $B$ is linear

  pf:
  Given

  $b_1,b_2\in W$,

  $\mathrm{\exists }!x_1,x_2$ such that

  $T(x_1)=b_1$ and

  $T(x_2)=b_2$.
  Since

  $T$ is linear and

  $T\circ B=I_w$, we have
  

  $$T({\alpha }_{1}B(b_1)+{\alpha }_{2}B(b_2))={\alpha }_{1}T(B(b_1))+{\alpha }_{2}T(B(b_2))={\alpha }_1{b}_1+{\alpha }_2{b}_2.$$
  Therefore,

  $B({\alpha }_1{b}_1+{\alpha }_2{b}_2)={\alpha }_{1}B(b_1)+{\alpha }_{2}B(b_2)$.

One-to-one and onto

Definition:

$T:V\to W$ is one-to-one (1-1, injective) if

$T(\mathbf{u})=T(\mathbf{v})$ implies

$\mathbf{u}=\mathbf{v}$.

Proposition:

$T:V\to W$ is one-to-one if and only if

$\text{Ker}(T)=\{\mathbf{0}\}$.

• Proof:

  (

  $\Rightarrow$)
  

  $T$ is linear so

  $T(\mathbf{0})=\mathbf{0}$.
  Suppose

  $\mathrm{\exists }\mathbf{v}\in V$ such that

  $T(\mathbf{v})=\mathbf{0}$,
  since

  $T$ is one-to-one, we must have

  $\mathbf{v}=\mathbf{0}$.

  (

  $\Leftarrow$)
  Let

  $\mathbf{u}\mathbf{,}\mathbf{v}\in V$ such that

  $T(\mathbf{u})=T(\mathbf{v})$.
  Since

  $T$ is linear
  

  $$\mathbf{0}=T(\mathbf{u})-T(\mathbf{v})=T(\mathbf{u}-\mathbf{v}).$$
  So

  $(\mathbf{u}-\mathbf{v})\in \text{Ker}(T)$ and we must have

  $\mathbf{u}-\mathbf{v}=\mathbf{0}$, that is,

  $\mathbf{u}=\mathbf{v}$.

Proposition:
If

$T:V\to W$ is left invertible, then

$T$ is one-to-one.

• Proof:

  Given

  $\mathbf{u}\mathbf{,}\mathbf{v}\in V$ such that

  $T(\mathbf{u})=T(\mathbf{v})$.
  

  $T$ is left invertible, so

  $\mathrm{\exists }B:W\to V$ such that

  $B\circ T=I_v$.
  Then
  

  $$\mathbf{u}=B(T(\mathbf{u}))=B(T(\mathbf{v}))=\mathbf{v}.$$

Proposition:
If

$T:V\to W$ is not one-to-one,

$T$ is not left invertible.

Definition:

$T:V\to W$ is onto (surjective) if for any

$\mathbf{w}\in W$, there exists a

$\mathbf{v}\in V$ such that

$T(\mathbf{v})=\mathbf{w}$.

Proposition:
If

$T:V\to W$ is right invertible, then

$T$ is onto.

• Proof:

  Since

  $T$ is right invertible,

  $\mathrm{\exists }C:W\to V$ such that

  $T\circ C=I_w$.
  Given

  $\mathbf{w}\in W$, we define

  $\mathbf{v}=C(\mathbf{w})$,
  Therefore

  $T(\mathbf{v})=T(C(\mathbf{w}))=\mathbf{w}$.

Proposition:
If

$T:V\to W$ is not onto,

$T$ is not right invertible.

Theorem:
Let

$T:V\to W$ be a linear transformation,

$T$ is invertible if and only if

$T$ is one-to-one and onto.

• Proof:

  (

  $\Rightarrow$)
  Given

  $T\in \mathcal{L}(V,W)$ that is invertible, then

  $\mathrm{\exists }{T}^{-1}\in \mathcal{L}(W,V)$.

  claim:

  $T$ is one-to-one

  pf:
  Suppose there exists

  $\mathbf{v}\in V$ such that

  $T(\mathbf{v})=\mathbf{0}$.
  Since

  $T^{-1}$ is linear that maps

  $\mathbf{0}$ to

  $\mathbf{0}$, we have
  

  $$\mathbf{v}=T^{-1}(T(\mathbf{v}))=T^{-1}(\mathbf{0})=\mathbf{0}.$$
  Therefore,

  $\text{Ker}(T)=\{\mathbf{0}\}$ and

  $T$ is one-to-one.

  claim:

  $T$ is onto

  pf:
  Given

  $\mathbf{w}\in W$, we have

  $\mathbf{w}=T(T^{-1}(\mathbf{w}))$, that is,

  $\mathbf{w}\in \text{Ran}(T)$.
  It is then clear that

  $\text{Ran}(T)=W$ and

  $T$ is onto.

  (

  $\Leftarrow$)
  

  $T$ is onto, so, for each

  $\mathbf{w}\in W$,

  $\mathrm{\exists }\mathbf{v}\in V$ such that

  $T(\mathbf{v})=\mathbf{w}$.
  We then define a function

  $S:W\to V$ by setting

  $S(\mathbf{w})=\mathbf{v}$.

  claim:

  $S$ is a right inverse of

  $T$

  pf:
  Given

  $\mathbf{w}\in W$, let

  $S(\mathbf{w})=\mathbf{u}$, then

  $T(\mathbf{u})=\mathbf{w}$ and
  

  $$(T\circ S)(\mathbf{w})=T(S(\mathbf{w}))=T(\mathbf{u})=\mathbf{w}.$$

  claim:

  $S$ is a left inverse of

  $T$

  pf:
  Given

  $\mathbf{v}\in V$, let

  $T(\mathbf{v})=\mathbf{w}$, and assume

  $S(\mathbf{w})=\mathbf{u}$, i.e.,

  $T(\mathbf{u})=\mathbf{w}$.
  Since

  $T$ is one-to-one, we must have

  $\mathbf{v}=\mathbf{u}$ and therefore
  

  $$(S\circ T)(\mathbf{v})=S(T(\mathbf{v}))=S(\mathbf{w})=\mathbf{u}=\mathbf{v}.$$

  claim:

  $S$ is a linear transformation

  pf:
  Given

  ${\mathbf{w}}_1,{\mathbf{w}}_2\in W$ and

  ${\alpha }_1,{\alpha }_2\in F$, since

  $T$ is linear and

  $T\circ S=I_w$, we have
  

  $$T({\alpha }_{1}S({\mathbf{w}}_1)+{\alpha }_{2}S({\mathbf{w}}_2))={\alpha }_{1}T(S({\mathbf{w}}_1))+{\alpha }_{2}T(S({\mathbf{w}}_2))]={\alpha }_1{\mathbf{w}}_1+{\alpha }_2{\mathbf{w}}_2.$$
  Since

  $T$ is one-to-one, we must have
  

  $$S({\alpha }_1{\mathbf{w}}_1+{\alpha }_2{\mathbf{w}}_2)={\alpha }_{1}S({\mathbf{w}}_1)+{\alpha }_{2}S({\mathbf{w}}_2).$$
  So

  $S$ is a linear transformation.

---

Examples

Example 1

Let

$T\in \mathcal{L}({\mathbb{P}}_2,{\mathbb{P}}_2)$ with

$T(p)=p^{\prime }+p$.

Choose

$\beta =\{1,x,x^2\}$ be a basis for

${\mathbb{P}}_2$. We have

$$\begin{array}{rl}T(1)=1& =\left[\begin{array}{ccc}1& x& x^2\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\\ T(x)=1+x& =\left[\begin{array}{ccc}1& x& x^2\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right],\\ T(x^2)=2x+x^2& =\left[\begin{array}{ccc}1& x& x^2\end{array}\right]\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right].\end{array}$$
So, the matrix representation of

$T$ is

$$[T{]}_{\beta }=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 2\\ 0& 0& 1\end{array}\right].$$
It is then easy to find that

$$[T^{-1}{]}_{\beta }=\left[\begin{array}{ccc}1& -1& 2\\ 0& 1& -2\\ 0& 0& 1\end{array}\right].$$
Therefore,

$$\begin{array}{rl}{T}^{-1}(a+bx+c{x}^2)& =T^{-1}\left(\left[\begin{array}{ccc}1& x& x^2\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\right)\\ & =\left[\begin{array}{ccc}1& x& x^2\end{array}\right]\left[\begin{array}{c}a-b+2c\\ b-2c\\ c\end{array}\right]\\ & =(a-b+2c)+(b-2c)x+c{x}^2.\end{array}$$

We can also determine the adjoint transformation of

$T$, we have

$$[T^{\ast }{]}_{\beta }=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 2& 1\end{array}\right].$$
Hence,

$$\begin{array}{rl}{T}^{\ast }(a+bx+c{x}^2)& =T^{\ast }\left(\left[\begin{array}{ccc}1& x& x^2\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\right)\\ & =\left[\begin{array}{ccc}1& x& x^2\end{array}\right]\left[\begin{array}{c}a\\ a+b\\ 2b+cc\end{array}\right]\\ & =a+(a+b)x+(2b+c)x^2.\end{array}$$

We can also use a different basis for

${\mathbb{P}}_2$. For example, choose

$\gamma =\{1+x+x^2,x+x^2,x^2\}$. The matrix representation of

$T$ is then

$$[T{]}_{\beta }^{\gamma }=\left[\begin{array}{ccc}1& 1& 0\\ -1& 0& 2\\ 0& -1& -1\end{array}\right],$$
and

$$[T^{-1}{]}_{\gamma }^{\beta }=\left[\begin{array}{ccc}2& 1& 2\\ -1& -1& -2\\ 1& 1& 1\end{array}\right].$$
We again have

$$\begin{array}{rl}{T}^{-1}(a+bx+c{x}^2)& =T^{-1}\left(\left[\begin{array}{ccc}1+x+x^2& x+x^2& x^2\end{array}\right]\left[\begin{array}{c}a\\ b-a\\ c-b\end{array}\right]\right)\\ & =\left[\begin{array}{ccc}1& x& x^2\end{array}\right]\left[\begin{array}{c}a-b+2c\\ b-2c\\ c\end{array}\right]\\ & =(a-b+2c)+(b-2c)x+c{x}^2.\end{array}$$

Example 2

Let

$T\in \mathcal{L}({\mathbb{P}}_2,{\mathbb{P}}_2)$ with

$T(p)=p^{\prime }$.

It is easy to check that

$1\in \text{Ker}(T)$, so

$T$ is not one-to-one, thus is not left invertible and is not invertible.

Also,

$x^2\notin \text{Ran}(T)$, so

$T$ is not onto and is thus not right invertible.

Example 3

Let

$T\in \mathcal{L}({\mathbb{P}}_2,{\mathbb{P}}_1)$ with

$T(p)=p^{\prime }$.

Again,

$1\in \text{Ker}(T)$, so

$T$ is not one-to-one, thus is not left invertible and is not invertible.

But

$T$ is onto, and we can define

$C\in \mathcal{L}({\mathbb{P}}_1,{\mathbb{P}}_2)$ with

$C(p)={\int }_0^{x}p(s)dx$.

One can check easily that

$T\circ C(p)=p$,

$\mathrm{\forall }p\in {\mathbb{P}}_1$.

Example 4

Let

$T\in \mathcal{L}(V,V)$ with

$T(p)=p^{\prime }$, where

$\beta =\{\sin (x),\cos (x)\}$ is a basis of

$V$.

Then

$T$ is on-to-one and onto, and invertible.