---
title: Ch1-4_5
tags: Linear algebra
GA: G-77TT93X4N1
---

# 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 $\alpha, \beta\in F$,
> > $$
> > \begin{aligned}
> > T(\alpha x + \beta y) &= c_1 T_1(\alpha x + \beta y) + c_2 T_2(\alpha x + \beta y) \\
> > &= \alpha\left(c_1T_1(x) + c_2T_2(x)\right) + \beta\left(c_1T_1(y) + c_2T_2(y)\right) \\
> > &= \alpha T(x) + \beta T(y).
> > \end{aligned}
> > $$
> > So $T$ is linear.

#### Examples

Let $V = \mathbb{P}_n$, then $T_1(p) = p'$, $T_2(p)=p''$, $T_3(p)=p'''$ are all belonging to $\mathcal{L}(V,V)$, and so
$$
T(p) = p' + 2p'' + 3p'''
$$
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:
> $$
> \begin{aligned}
> T_1(T_2(\alpha x+ \beta y)) &= T_1(\alpha T_2(x)+ \beta T_2(y))\\
> &= \alpha T_1(T_2(x)) + \beta T_1(T_2(y)).
> \end{aligned}
> $$

### Matrix presentation of composition

**Proposition:**
Let $A = [T_1]^{\gamma}_{\beta}$ and $B = [T_2]^{\beta}_{\mu}$, and assume $T=T_1\circ T_2$, then $[T]^{\gamma}_{\mu} = AB$.

**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, $\text{trace}(A) = \sum^n_{j=1} a_{jj}$.

**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
> $$
> \ell_{kk} = \sum^n_{i=1} a_{ki}b_{ik}.
> $$
> Therefore
> $$
> \text{trace}(AB) = \sum^m_{k=1}\sum^n_{i=1} a_{ki}b_{ik}.
> $$
> Similarly,
> $$
> \text{trace}(BA) = \sum^n_{k=1}\sum^m_{i=1} b_{ki}a_{ik}.
> $$
> Since we can switch double finite sums, we have
> $$
> \sum^m_{k=1}\sum^n_{i=1} a_{ki}b_{ik} = \sum^n_{i=1}\sum^m_{k=1} b_{ik}a_{ki}=\sum^n_{k=1}\sum^m_{i=1} b_{ki}a_{ik}.
> $$
> So $\text{trace}(AB) = \text{trace}(BA)$.
* Proof: (version 2)
> Consider the following two linear transformations
> $$
> \begin{aligned}
> T_1:\mathbb{M}_{n\times m}\to \mathbb{R},& \quad T_1(X) = \text{trace}(AX),\\
> T_2:\mathbb{M}_{m\times n}\to \mathbb{R},& \quad T_2(X) = \text{trace}(XA).
> \end{aligned}
> $$
> 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, $\forall k, \ell$ 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 $\text{trace}(AB) = \text{trace}(BA)$.

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### Adjoint linear transformation

**Definition:**
Let $T:V\to W$ be linear, $\beta$ and $\gamma$ be the bases of $V$ and $W$, respectively. We define $T^*:W\to V$, the adjoint transformation of $T$, by setting
$$
[T^*]^{\beta}_{\gamma} = ([T]^{\gamma}_{\beta})^T,
$$
where the superscript denotes matrix transpose.

**Proposition:**
$T^*$ is a linear transformation.