# Lecture 10

## Chapter III Linear maps

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

### Vector Space of Linear Maps 3A

Review

#### Theorem 3.21 (The Fundamental Theorem of Linear Maps, Rank-nullity Theorem)

Suppose $V$ is finite dimensional, and $T\in \mathscr{L}(V,W)$, then $range(T)$ is finite dimensional ($W$ don't need to be finite dimensional). and 

$$
dim(V)=dim(null (T))+dim(range(T))
$$

Proof:

Let $u_1,...,u_m$ be a basis for $null(T)$, then we extend to a basis of $V$ given by $u_1,...,u_m,v_1,...,v_m$, we have $dim(V)=m+n$. Claim that $Tv_1,...,Tv_n$ forms a basis for $range (T)$. Need to show 

* Linearly independent. (in Homework 3)
* These span $range(T)$.  
  
  Let $w\in range(T)$ the there exists $v\in V$ such that $Tv=W$, $u_1,...,u_m,v_1,...,v_m$ are basis so $\exists a_1,...,a_m,b_1,...,b_n$ such that $v=a_1u_1+...+a_mu_m+b_1v_1+...+b_n v_n$. $Tv=a_1Tu_1+...+a_mTu_m+b_1Tu_1+...+b_nTv_n$. 

  Since $u_k\in null(T)$, So $Tv_1,...,Tv_n$ spans range $T$ and so form a basis. Thus $range(T)$ is finite dimensional and $dim(range(T))=n$. So $dim(V)=dim(null (T))+dim(range(T))$

#### Theorem 3.22

Suppose $V,W$ are finite dimensional with $dim(V)>dim(W)$, then there are no injective maps from $V$ to $W$.

#### Theorem 3.24

Suppose $V,W$ are finite dimensional with $dim(V)<dim(W)$, then there are no surjective maps from $V$ to $W$.

Ideas of Proof: relies on **Theorem 3.21** $dim(null(T))>0$

### Linear Maps and Linear Systems 3EX-1

Suppose we have a homogeneous linear system * with $m$ equation and $n$ variables.

$$
A_{11} x_1+ ... + A_{1n} x_n=0\\
...\\
A_{m1} x_1+ ... + A_{mn} x_n=0
$$

which is equivalent to 

$$
A\begin{bmatrix}
    x_1\\...\\x_n
\end{bmatrix}=\vec{0}
$$

also equivalent to

$$
T(v)=0,\textup{ for some }T
$$

$$
T(x_1,...,x_n)=(A_{11} x_1+ ... + A_{1n},...,A_{m1} x_1+ ... + A_{mn} x_n),T\in \mathscr{L}(\mathbb{R}^n,\mathbb{R}^m)
$$

Solution to * is $null(T)$.

#### Proposition 3.26

A homogeneous linear system with more variables than equations has non-zero solutions.

Proof:

Using $T$ as above, note that since $n>m$, use **Theorem 3.22**, implies that $T$ cannot be injective. So, $null (T)$ contains a non-zero vector.

#### Proposition 3.28

An in-homogenous system with more equations than variables has no solutions for some choices of constants. ($A\vec{x}=\vec{b}$ for some $\vec{b}$ this has no solution)

### Matrices 3A

#### Definition 3.29

For $m,n>0$ and $m\times n$ matrix $A$ is a rectangular array with elements of the $\mathbb{F}$ given by

$$
A=\begin{pmatrix}
    A_{1,1}& ...&A_{1,n}\\
    ... & & ...\\
    A_{n,1}&...&A_{m,n}\\
\end{pmatrix}
$$

### Operations on matrices

Addition:

$$
A+B=\begin{pmatrix}
    A_{1,1}+B_{1,1}& ...&A_{1,n}+B_{1,n}\\
    ... & & ...\\
    A_{n,1}+A_{n,1}&...&A_{m,n}+B_{m,n}\\
\end{pmatrix}
$$

**for $A+B$, $A,B$ need to be the same size**

Scalar multiplication: 

$$
\lambda A=\begin{pmatrix}
    \lambda A_{1,1}& ...& \lambda A_{1,n}\\
    ... & & ...\\
    \lambda A_{n,1}&...& \lambda A_{m,n}\\
\end{pmatrix}
$$

#### Definition 3.39

$\mathbb{F}^{m,n}$ is the set of $m$ by $n$ matrices.

#### Theorem 3.40

$\mathbb{F}^{m,n}$ is a vector space (over $\mathbb{F}$) with $dim(\mathbb{F}^{m,n})=m\times n$

### Matrix multiplication 3EX-2

Let $A$ be a $m\times n$ matrix and $B$ be an $n\times s$ matrix

$$
(A,B)_{i,j}= \sum^n_{r=1} A_{i,r}\cdot B_{r,j}
$$

Claim:

This formula comes from multiplication of linear maps.

#### Definition 3.44

Linear maps to matrices, let $V$, $W$, $Tv_i$ written in terms of $w_i$.

$$
M(T)=\begin{pmatrix}
    Tv_1\vert Tv_2\vert ...\vert Tv_n
\end{pmatrix}
$$