# Lecture 14

## Chapter III Linear maps

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

### Matrices 3C

Review

#### Proposition 3.76

$$
M(Tv)=M(T)M(v)
$$

#### Theorem 3.78

Let $V,W$ be finite dimensional vector space, and $T\in \mathscr{L}(V,W)$ then $dim\ range\ T=column\ rank (M(T))=rank(M(T))$

Proof:

$range=Span\{Tv_1,...,Tv_n\}$ compare to $Span\{M(T)_{\cdot,1},...,M(T)_{\cdot, n}\}=Span\{M(T)M(v_1),...,M(T)M(v_n)\}=Span\{M(Tv_1),...,M(Tv_n)\}$

Since $M$ is a isomorphism, then the theorem makes sense.

#### Change of Basis

#### Definition 3.79, 3.80

The identity matrix

$$
I=\begin{pmatrix}
    1.& 0\\
    0& '1\\
\end{pmatrix}
$$

The inverse matrix of an invertible matrix $A$ denoted $A^{-1}$ is the matrix such that

$$
AA^{-1}=I=A^{-1}A
$$

Question: Let $u_1,...,u_n$ and $v_1,...,v_n$ be two bases for $V$. What is $M(I,(u_1,...,u_n),(v_1,...,v_n)),I\in \mathscr{L}(V)$

#### Proposition 3.82

Let $u_1,...,u_n$ and $v_1,...,v_n$ be bases of $V$, then $M(I,(u_1,...,u_n),(v_1,...,v_n)),I\in \mathscr{L}(V)$ and $M(I,(v_1,...,v_n),(u_1,...,u_n)),I\in \mathscr{L}(V)$ are inverse to each other.

Proof:

$$
M(I,(u_1,...,u_n),(v-1,...,v_n)),I\in \mathscr{L}(V) M(I,(v-1,...,v_n),(u_1,...,u_n))=M(I,(u_1,...,u_n),(u_1,...,u_n))
$$

#### Theorem 3.84 Change of Basis

Let $u_1,...,u_n$ and $v_1,...,v_n$ be two bases for $V$ and $T\in \mathscr{L}(v), A=M(T,(u_1,...,u_n)), B=M(T,(v_1,...,v_n)), C=M(I,(u_1,...,u_n),(v_1,...,v_n))$, then $A=C^{-1}BC$

#### Theorem 3.86

Let $T\in \mathscr{L}(v)$ be an invertible linear map, then $M(T^{-1})=M(T)^{-1}$

### Products and Quotients of Vector Spaces 3E

Goals: To construct vectors spaces from other vector spaces.

#### Definition 3.87

Suppose $V_1,...,V_m$ vectors spaces over some field $\mathbb{F}$, then the product is given by 

$$
V_1\times ...\times V_n=\{(v_1,v_2,...,v_n)\vert v_1\in V_1, v_2\in V_2,...,v_n\in V_n\}
$$

with addition given by 

$$
(v_1,...,v_n)+(u_1,...,u_n)=(v_1+u_1,...,v_n+u_n)
$$

and scalar multiplication 

$$
\lambda (v_1,...,v_n)=(\lambda v_1,...,\lambda v_n),\lambda \in \mathbb{F}
$$

#### Theorem 3.89

If $v_1,...,v_n$ are vectors paces over $\mathbb{F}$ then $V_1\times ...\times V_n$ is a vector space over $\mathbb{F}$

Example:

$V=\mathscr{P}_2(\mathbb{R})\times \mathbb{R}^2=\{(p,v)\vert p\in \mathscr{P}_2(\mathbb{R}), v\in \mathbb{R}^2\}=\{(a_0+a_1x+a_2x,(b,c))\vert a_0,a_1,a_2,b,c\in \mathbb{R}\}$

A basis for $V$ would be $(1,(0,0)),(x,(0,0)),(x^2,(0,0)),(0,(1,0)),(0,(0,1))$

#### Theorem 3.92

$$
dim(V_1\times ...\times V_n)=dim(V_1)+...+dim(V_n)
$$

Sketch of proof:

take a basis for each $V_k$, make them vectors in the product then combine the entire list of vector to be basis.

Example:

$\mathbb{R}^2\times \mathbb{R}^3=\{((a,b),(c,d,e))\vert a,b,c,d,e\in \R\}$

$\mathbb{R}^2\times \mathbb{R}^3\cong \mathbb{R}^5,((a,b),(c,d,e))\mapsto(a,b,c,d,e)$

#### Theorem 3.93

Let $V_1,...,V_m\subseteq V$, define $\Gamma: V_1\times...\times V_m\to V_1+...+V_m$. $\Gamma(v_1,...,v_n)=v_1+...+v_n$ then $\Gamma$ is always surjective. And it is injective if and only if $V_1+...+V_m$ is a direct sum.

Sketch of the proof:

injective $\iff null\ T\{ (0,...,0) \} \iff$ the only way to write $0=v_1,...,v_m$ is $v_1=...=v_n=0 \iff$ then $V_1+...+V_m$ is a direct sum

#### Theorem 3.94

$V_1+...+V_m$ is a direct sum if and only if $dim(V_1+...+V_m)=dim(V_1)+...+dim(V_m)$

Proof:

Use $\Gamma$ above is an isomorphism $\iff$ $V_1+...+V_m$ is a direct sum

Use $\Gamma$ above is an isomorphism $\implies dim(V_1+...+V_m)=dim(V_1)+...+dim(V_m)$