Chapter 1 extra note 4_5

Linear transformation as a vector space
Composition of linear transformation
Trace of a square matrix

Selected lecture notes:

Linear transformation as a vector space

Definition:

$\mathcal{L}(V,W)$: A set containing all the linear transformation from

$V$ to

$W$.

Proposition:

$\mathcal{L}(V,W)$ is a vector space.

• Proof:

  Given

  $T_1,T_2\in \mathcal{L}$, then

  $T_1:V\to W$ and

  $T_2:V\to W$ are linear transformations.
  Define

  $T:V\to W$ as

  $T(x)=c_1{T}_1(x)+c_2{T}_2(x)$,

  $c_1,c_2\in F$.
  claim:

  $T$ is linear.

  Given

  $x,y\in V$ and

  $\alpha ,\beta \in F$,
  

  $$\begin{array}{rl}T(\alpha x+\beta y)& =c_1{T}_1(\alpha x+\beta y)+c_2{T}_2(\alpha x+\beta y)\\ & =\alpha (c_1{T}_1(x)+c_2{T}_2(x))+\beta (c_1{T}_1(y)+c_2{T}_2(y))\\ & =\alpha T(x)+\beta T(y).\end{array}$$
  So

  $T$ is linear.

Examples

Let

$V={\mathbb{P}}_n$, then

$T_1(p)=p^{\prime }$,

$T_2(p)=p^{″}$,

$T_3(p)=p^{‴}$ are all belonging to

$\mathcal{L}(V,V)$, and so

$$T(p)=p^{\prime }+2p^{″}+3p^{‴}$$
is a linear transformation.

Composition

Definition:
Let

$T_1:V\to W$ and

$T_2:U\to V$, we define

$$(T_1\circ T_2)(x)=T_1(T_2(x)).$$

Proposition:
If

$T_1$ and

$T_2$ are linear transformation,

$T_1\circ T_2$ is a linear transformation.

• Proof:

  $$\begin{array}{rl}{T}_1(T_2(\alpha x+\beta y))& =T_1(\alpha T_2(x)+\beta T_2(y))\\ & =\alpha T_1(T_2(x))+\beta T_1(T_2(y)).\end{array}$$

Matrix presentation of composition

Proposition:
Let

$A=[T_1{]}_{\beta }^{\gamma }$ and

$B=[T_2{]}_{\mu }^{\beta }$, and assume

$T=T_1\circ T_2$, then

$[T{]}_{\mu }^{\gamma }=AB$.

Properties of matrix multiplication:

1. $A(BC)=(AB)C=ABC$
2. $A(B+C)=AB+AC$
3. $A(\alpha B)=(\alpha A)B=\alpha (AB)=\alpha AB$

Properties of composition of linear transformations:

1. $T_1\circ (T_2\circ T_3)=(T_1\circ T_2)\circ T_3=T_1\circ T_2\circ T_3$
2. $T_1\circ (T_2+T_3)=T_1\circ T_2+T_1\circ T_3$
3. $T_1\circ (\alpha T_2)=(\alpha T_1)\circ T_2=\alpha (T_1\circ T_2)=\alpha T_1\circ T_2$

Trace

Definition:

Let

$A$ be a square

$n\times n$ matrix,

$\text{trace}(A)=\sum _{j=1}^n{a}_{jj}$.

Theorem:

Let

$A\in {\mathbb{M}}_{m\times n}$ and

$B\in {\mathbb{M}}_{n\times m}$, then

$$\text{trace}(AB)=\text{trace}(BA).$$

• Proof:

  Lat

  $L=AB$. By direct computation, we found that its diagonal element
  

  $${\ell }_{kk}=\sum _{i=1}^n{a}_{ki}{b}_{ik}.$$
  Therefore
  

  $$\text{trace}(AB)=\sum _{k=1}^m\sum _{i=1}^n{a}_{ki}{b}_{ik}.$$
  Similarly,
  

  $$\text{trace}(BA)=\sum _{k=1}^n\sum _{i=1}^m{b}_{ki}{a}_{ik}.$$
  Since we can switch double finite sums, we have
  

  $$\sum _{k=1}^m\sum _{i=1}^n{a}_{ki}{b}_{ik}=\sum _{i=1}^n\sum _{k=1}^m{b}_{ik}{a}_{ki}=\sum _{k=1}^n\sum _{i=1}^m{b}_{ki}{a}_{ik}.$$
  So

  $\text{trace}(AB)=\text{trace}(BA)$.

• Proof: (version 2)

  Consider the following two linear transformations
  

  $$\begin{array}{rl}{T}_1:{\mathbb{M}}_{n\times m}\to \mathbb{R},& T_1(X)=\text{trace}(AX),\\ T_2:{\mathbb{M}}_{m\times n}\to \mathbb{R},& T_2(X)=\text{trace}(XA).\end{array}$$
  Claim:

  $T_1(X)=T_2(X)$.

  Proof:
  Let

  $E_{ij}\in {\mathbb{M}}_{n\times m}$ be a matrix that has its element being zero except at the

  $j$-th column

  $i$-th row. Then

  $\{E_{ij},1\le i\le n,1\le j\le m\}$ forms a basis for

  ${\mathbb{M}}_{n\times m}$.
  It is then easy to verify that,

  $\mathrm{\forall }k,\ell$ we have
  

  $$T_1(E_{k\ell })=a_{\ell k}=T_2(E_{k\ell }).$$
  Since

  $T_1=T_2$ at all the basis vectors, it is also true for any linear combination of basis vectors. So

  $T_1=T_2$.

  Finally,

  $T_1(B)=T_2(B)$ leads to

  $\text{trace}(AB)=\text{trace}(BA)$.