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tags: Giang's linear algebra notes
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# Chapter 4: From linear transformations to matrices

[toc]

## Coordinate matrices and composition

**Composition of linear transformations and matrix multiplication**: Suppose that $X$ is a $l$-dimensional and has an ordered basis $\alpha=\{u_{1},\dots,u_{l}\}$, $Y$ is $m$-dimensional and has an ordered basis $\beta=\{v_{1},\dots,v_{m}\}$, and $Z$ is $n$-dimensional and has a basis $\gamma$ of $n$ elements. Let $T:X\to Y$ and $S:Y\to Z$ be linear transformations. Then
$$[ST]_{\alpha}^{\gamma}=[S]_{\beta}^{\gamma}\cdot[T]_{\alpha}^{\beta}$$

**Intuition of matrix multiplication**: Matrices are originally used to express transformation. Let $u\in U$ be a vector and $[u]^{\alpha}$ be its coordinate vector w.r.t basis $\alpha=\{u_{1},\dots,u_{n}\}$. Now we transform $u$ into $v\in V$ by $T_{1}:U\to V$. Let $\beta=\{v_{1},\dots,v_{p}\}$ be a basis of $V$, we have that
$$[v]^{\beta}=\begin{bmatrix}\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}\\{} [T_{1}(u_{1})]^{\beta} & \cdots & [T_{1}(u_{n})]^{\beta}\\ \rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex} \end{bmatrix}\cdot[u]^{\alpha}$$
Then we transform $v$ to $w\in W$ by $T_{2}:V\to W$. Let $\gamma=\{w_{1},\dots,w_{m}\}$ be a basis of $W$, we have that
$$\begin{aligned}
[w]^{\gamma} & =\begin{bmatrix}\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}\\{}
[T_{2}(u_{1})]^{\gamma} & \cdots & [T_{2}(u_{p})]^{\gamma}\\
\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}
\end{bmatrix}\cdot[v]^{\beta}\\
& =\begin{bmatrix}\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}\\{}
[T_{2}(u_{1})]^{\gamma} & \cdots & [T_{2}(u_{p})]^{\gamma}\\
\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}
\end{bmatrix}\cdot\begin{bmatrix}\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}\\{}
[T_{1}(u_{1})]^{\beta} & \cdots & [T_{1}(u_{n})]^{\beta}\\
\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}
\end{bmatrix}\cdot[v]^{\beta}\end{aligned}$$
The intuition is
$$\begin{aligned}
[w]^{\gamma} & =\begin{bmatrix}\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}\\{}
[T_{2}(u_{1})]^{\gamma} & \cdots & [T_{2}(u_{p})]^{\gamma}\\
\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}
\end{bmatrix}\cdot\begin{bmatrix}\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}\\{}
[T_{1}(u_{1})]^{\beta}\cdot[v]_{1}^{\beta} & \cdots & [T_{1}(u_{n})]^{\beta}\cdot[v]_{n}^{\beta}\\
\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}
\end{bmatrix}\\
& =\sum_{i=1}^{n}\begin{bmatrix}\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}\\{}
[T_{2}(u_{1})]^{\gamma} & \cdots & [T_{2}(u_{p})]^{\gamma}\\
\rule[-1ex]{0.5pt}{2.5ex} & & \rule[-1ex]{0.5pt}{2.5ex}
\end{bmatrix}\cdot[T_{1}(u_{i})]^{\beta}\cdot[v]_{i}^{\beta}\end{aligned}$$

>**NOTE**: The theorem explains why matrix multiplication is defined in a strange
way, i.e. multiplying rows against columnes, etc.

>**NOTE**: The theorem also explains why we need the number of columns of the left
matrix to equal the number of rows of the right matrix, i.e. to compose
$T:X\to Y$ and $S:Y\to Z$ into $ST:X\to Z$, we need the target space of
$T$ to equal the initial space of $S$

## Comparison between linear transformations and matrices

By using bases, we can understand the behavior of linear transformations in terms of matrix multiplication. However, it is not quite saying that linear transformations are the same as matrices, since

1. The correspondence only works for finite dimensional spaces $X,Y,Z$

2. The matrix we get depends on the basis we choose, thus a single linear transformation can correspond to many different matrices, depending on what bases we pick

*Example*: Consider the identity transformation $I_{V}:V\to V$ on a vector space $V$, defined by $I_{V}(v)=v$, If we pick any basis $\beta$ of $V,$we have
$$[I_{V}]_{\beta}^{\beta}=\begin{bmatrix}1 & 0 & \cdots & 0\\
0 & 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1
\end{bmatrix}$$
However, if we consider another basis $\alpha$ of $V$. Then $[I_{V}]_{\beta}^{\alpha}\neq[I_{V}]_{\beta}^{\beta}$

## Matrices as linear transformations

**Linear transformation via matrix**: Let $A$ be a $m\times n$ matrix. Then we define the linear transformation $L_{A}:\textbf{R}^{n}\to\textbf{R}^{m}$ by the rule
$$\forall x\in\textbf{R}^{n},L_{A}(x)=Ax$$
where we think of vectors in $\textbf{R}^{n}$ and $\textbf{R}^{m}$ as column vectors

**Theorem**: Let $L_{A}:\textbf{R}^{n}\to\textbf{R}^{m}$ be a linear transformation, $\alpha$ be the standard basis for $\textbf{R}^{n}$, and $\beta$ be the standard basis for $\textbf{R}^{m}$. If $A$ is an $m\times n$ matrix, then $[L_{A}]_{\alpha}^{\beta}=A$. If $T:\textbf{R}^{n}\to\textbf{R}^{m}$ is a linear transformation, then $L_{[T]_{\alpha}^{\beta}}=T$

**Theorem**: Let $T:X\to Y$, $S:Y\to Z$, and $R:Z\to W$ be linear transformations. Then we have $$R(ST)=(RS)T$$

**Matrix multiplication is associative**. Let $A$ be an $m\times n$ matrix, $B$ be a $l\times m$ matrix, and $C$ be a $k\times l$ matrix. Then $C(BA)=(CB)A$

>**NOTE**: In fact, all familiar rules of algebra apply to matrices, provided that
all the matrix operations make sense. However, matrix multiplication is
not commutative, i.e. $AB\neq BA$

**Properties of matrices and corresponding linear transformations**. If $A$ is an $m\times n$ matrix and $B$ is an $l\times m$ matrix, then

1. $L_{BA}=L_{B}L_{A}$, and

2. $L_{A+B}=L_{A}+L_{B}$

## Invertible linear transformations

**Definition**: Let $T:V\to W$ be a linear transformation. We say that a linear transformation $S:W\to V$ is *the inverse of $T$* if $TS=I_{W}$ and $ST=I_{V}$. We say that $T$ is *invertible* if it has an inverse, and call the inverse $T^{-1}$; thus $TT^{-1}=I_{W}$ and $T^{-1}T=I_{V}$

>**NOTE**: The definition of inverse matrices is symmetric, i.e. if $S$ is the
inverse of $T$, then $T$ is the inverse of $S$

>**NOTE**: Every transformation can have at most one inverse

**Lemma**: Let $T:V\to W$ be a linear transformation, and let $S:W\to V$ and $S':W\to V$ both be inverse of $T$. Then $S=S'$

>**NOTE**: Not every linear transformation has an inverse

**Lemma**: If $T:V\to W$ has an inverse $S:W\to V$, then $T$ must be one-to-one and onto

**Lemma**: If $T:V\to W$ is one-to-one and onto linear transformation, then it has an inverse $S:W\to V$, which is also a linear transformation

**Isomorphisms**: Two vector spaces $V$ and $W$ are said to be *isomorphic* if there is an invertible linear transformation $T:V\to W$ from one space to another

>**NOTE**: Isomorphic spaces tend to have almost identical properties

**Isomorphic vector spaces**: Two finite-dimensional spaces $V$ and $W$ are *isomorphic* if and only if $\dim(V)=\dim(W)$

**From basis to isomorphism**. If $V$ has a finite basis $\beta=\{v_{1},\dots,v_{n}\}$, then the coordinate map $\phi_{\beta}:V\to\textbf{R}^{n}$ defined by
$$\phi_{\beta}(x)=[x]^{\beta}$$
is a linear transformation, and is invertible. Thus $\phi_{\beta}$ is an isomorphism between $V$ to $\textbf{R}^{n}$. In textbook, $\phi_{\beta}$ is called the standard representation of $V$ w.r.t $\beta$

## Invertible linear transformations and invertible matrices

**Inverse matrix**: An $m\times n$ matrix $A$ has to inverse $B$, if $B$ is an $n\times m$ matrix such that $BA=I_{n}$, and $AB=I_{m}$.. In this case, we call $A$
an *invertible matrix,* and denote $B$ by $A^{-1}$

**Invertible linear transformation and invertible matrices**. Let $V$ be a vector space with finite ordered basis $\alpha$ and let $W$ be a vector space with finite ordered basis $\beta$. Then a linear transformation $T:V\to W$ is invertible if and only if the matrix $[T]_{\alpha}^{\beta}$ is invertible. Further more $([T]_{\alpha}^{\beta})^{-1}=[T^{-1}]_{\beta}^{\alpha}$

**Corollary**: An $m\times n$ matrix $A$ is invertible if and only if the linear transformation $L_{A}:\textbf{R}^{n}\to\textbf{R}^{m}$ is invertible. Furthuermore, the inverse of $L_{A}$ is $L_{A^{-1}}$

**Corollary**: In order for a matrix $A$ to be invertible, it must be square, i.e. $m=n$

## Expansions

**Identity matrix**: $I_{n}$ is defined by
$$I_{n}=\begin{bmatrix}1 & 0 & \cdots & 0\\
0 & 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1
\end{bmatrix}$$
or equivalently
$$\forall i,j\in\{1,\dots,n\},(I_{n})_{ij}=\begin{cases}
1 & i=j\\
0 & i\neq j
\end{cases}$$

>**NOTE**: If $A$ and $B$ are matrice with different shapes, then $A+B$ is
undefined