# Lecture 13

## Chapter III Linear maps

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

### Matrices 3C

#### Theorem 3.63

A linear map is invertible if and only if it is injective and surjective.

#### Example

Consider $T:\mathscr{P}(\mathbb{F})\to \mathscr{P}(\mathbb{F})$, $T(f)=xf$

$T$ is injective but not surjective. Since you cannot get constant from multiply $x$. So it is not invertible.

#### Theorem 3.65

Let $V$ and $W$ be finite-dimensional with the same dimension, and $T\in\mathscr{L}(V,W)$, then $T$ is invertible, if and only if $T$ is injective if and only if, $T$ is surjective.

Proof:

Suppose $T$ is injective, then $null\ T={0}$, i.e $dim(null\ T)=0$, since $dim\ V=dim\ null\ T+dim\ range\ T$, we have $dim\ V=dim\ range\ T$ but $dim\ V\dim\ W$, so $dim\ W=dim\ range\ T$. Thus $W=range\ T$. This shows that $T\ injective \implies T\ surjective$.

If $T$ is surjective, then $dim\ range\ T=dim\ W$ but then $dim\ V=dim\ null\ T+dim\ W\implies dim\ null\ T=0$, so $T$ is injective, $T\ surjective\implies T\ injective$.

#### Theorem 3.68

Suppose $V,W$ finite dimensional $dim\ V=dim\ W$, then for $T\in\mathscr{L}(V,W)$ and $S\in \mathscr{L}(W,V)$, then $ST=I\implies TS=I$

#### Example 3.67

Show that for a polynomial $q$ with degree $m$, there exists a unique polynomial $p$ of degree $m$ such that $((x^2+5x+7)p)''=q$

Solution:

Let $T:\mathscr{P}_m(\mathbb{F})\to \mathscr{P}_m(\mathbb{F})$ given by $T(p)=((x^2+5x+7)p)''$ by $T$ is injective since $(x^2+5x+7)$ has degree $\geq 2$ for $p\neq 0$, therefore, $p$ is surjective. (by **Theorem 3.68**)

#### Isomorphisms

#### Definition 3.69

An **isomorphism** of vector spaces is a invertible linear map. Two vector spaces $V,W$ are isomorphic if there exists an isomorphism between them.

Notation: $V\cong W$ means $V$ and $W$ are isomorphic. (Don't use very often, no map is included.)

Example:

$\mathscr{P}_m(\mathbb{F})$ and $\mathbb{F}^{m+1}$ are isomorphic. $T:\mathbb{F}^{m+1}\to \mathscr{P}_m(\mathbb{F}): T((a_0,...,a_m))=a_0+a_1x+...+a_n x^n$

#### Theorem 3.70

Two finite dimensional vector spaces $V,W$ are isomorphic if and only if $dim\ V= dim\ W$

Ideas of Proof:

$\Rightarrow$ use fundamental theorems of linear map

$\Leftarrow$ Let $v_1,...,v_m\in V$ and $w_1,...,w_n\in W$ be bases. Then define $T:V\to W$ by $T(v_k)=w_k$ for $1\leq k\leq n$

Show $T$ is invertible by showing $T$ is injective and surjective.

#### Theorem 3.71

Let $V,W$ be finite dimensional, let $v_1,...,v_n\in V$ and $w_1,...,w_m\in W$ be bases. Then the map 

$$
M(-,(v_1,...,v_n),(w_1,...,w_m)):\mathscr{L}(V,W)\to \mathbb{F}^{m,n}
$$

$T\mapsto M(T)$ or $M(-,(v_1,...,v_n),(w_1,...,w_m))$ is an isomorphism ($M:\mathscr{L}(V,W)\to \mathbb{F}^{m,n}$)

Sketch of Proof:

Need to show $M$ is surjective and injective.

* Injective: i.e need to show if $M(T)=0$, then $T=0$. $M(T)=0\implies Tv_k=0, 1\leq k\leq n$
* Surjective: i.e let $A\in F^{m,n}$ define $T:V\to W$ given by $Tv_k=\sum_{j=1}^m A_{j,k} w_j$ you cna check that $M(T)=A$

#### Corollary 3.72

$dim \mathscr{L}(V,W)=(dim\ V)(dim\ W)$

#### Definition 3.73

$v\in V, v_1,...,v_n$ a basis, then $M(v)=\begin{pmatrix}
    b_1\\
    ...\\
    b_n
\end{pmatrix}, v=a_1v_1,...,a_nv_n$

#### Proposition 3.75, 3.76

$$
M(T)_{\cdot,k}=M(Tv_k)
$$

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