# Lecture 12

## Chapter III Linear maps

**Assumption: $U,V,W$ are vector spaces (over $\mathbb{F}$)**

### Matrices 3C

#### Proposition 3.51

Let $C$ be an $m\times c$ matrix and $R$ be a $c\times n$ matrix, then 

1. column $k$ of $CR$ is a linear combination of the columns of $C$ with coefficients given by $R_{\cdot,k}$

    *putting the propositions together...*

2. row $j$ of $CR$ is a linear combination of the rows of $R$ with coefficients given by $C_{j,\cdot}$

#### Column-Row Factorization and Rank

#### Definition 3.52

Let $A$ be an $m \times n$ matrix, then

* The column rank of $A$ is the dimension of the span of the columns in $\mathbb{F}^{m,1}$.
* The row range of $A$ is the dimension of the span of the row in $\mathbb{F}^{1,n}$.

> Transpose: $A^t=A^\top$ refers to swapping rows and columns

#### Theorem 3.56 (Column-Row Factorization)

Let $A$ be an $m\times j$ matrix with column rank $c$. Then there exists an $m\times c$ matrix $C$ and $c\times x$ matrix $R$ such that $A=CR$

Proof: 

Let $V=Span\{A_{\cdot,1},...,A_{\cdot,n}\}$, let $C_{\cdot, 1},...,C_{\cdot, c}$ be a basis of $V$. Since these forms a basis, there exists $R_{j,k}$ such that $A_{i,j}=\sum_{j=1}^c C_{i,j}R_{j,k}$, so $A_{\cdot,j}=\sum_{j=1}^c C_{\cdot,j}R_{j,k}$. This implies that $A=CR$ by construction $C$ is $m\times c$, $R$ is $c\times n$.

Example:

$$
A=\begin{pmatrix}
    1&4&2\\
    2&5&8\\
    3&6&4
\end{pmatrix}=\begin{pmatrix}
    1&4\\
    2&5\\
    3&6\\
\end{pmatrix}\begin{pmatrix}
    1&0&-1\\
    0&1&2\\
\end{pmatrix},rank\ A=4
$$

#### Definition 3.58 Rank

The **rank of a matrix** $A$ is the column rank of $A$ denoted $rank\ A$.

#### Theorem 3.57

Given a matrix $A$ the column rank equals the row rank.

Proof:

Note that by **Theorem 3.56**, if $A$ is $m\times n$ and has column rank $c$. $A=CR$ for some $C$ is a $m\times c$ matrix, $R$ is a $c\times n$ matrices, ut the rows of $CR$ are a linear combination of the rows of $R$, and row rank of $R\leq C$. So row rank $A\leq$ column rank of $A$.

Taking a transpose of matrix, then row rank of $A^\top$ (column rank of $A$) $\leq$ column rank of $A^\top$ (row rank $A$).

So column rank is equal to row rank.

### Invertibility and Isomorphisms 3D

Invertible Linear Maps

#### Definition 3.59

A linear map $T\in\mathscr{L}(V,W)$ is **invertible** if there exists $S\in \mathscr{L}(W,V)$ such that $ST=I_V$ and $TS=I_W$. Such a $S$ is called an **inverse** of $T$.

Note: $ST=I_V$ and $TS=I_W$ must **both be true** for inverse map.

#### Lemma 3.60

Every linear map has an unique inverse.

Proof: Exercise and answer in the book.

Notation: $T^{-1}$ is the inverse of $T$

#### Theorem 3.63

A linear map $T:V\to W$ invertible if and only if its injective and surjective.

Proof:

$\Rightarrow$

$null(T)=\{0\}$ since $T(v)=0\implies (T^{-1}))(T(v))=0\implies range (T)=W$ let $w\in W$ then $T(T^{-1}(w))=w,w\in range (T)$

$\Leftarrow$

Find $S:W\to V$ a function such that $T(S(v))=v$ by letting $S(v)$ be the unique vector in $v$ such that $T(S(v))=v$. Goal: Show $S:W\to V$ is linear

$$
ST(S(w_1)+S(w_2))=S(w_1)+S(w_2)\\
S(T(S(w_1)))+T(S(w_2))=S(w_1+w_2)
$$
