# Lecture 18

## Chapter III Linear maps

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

### Duality 3F

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Review

#### Theorem 3.128, 3.130

Let $V,W$ be a finite dimensional vector space, $T\in \mathscr{L}(V,W)$

a) $null\ T'=(range\ T)^0$, $dim (null\ T')=dim\ null\ T+dim\ W-dim\ V$   
b) $range\ T'=(null\ T)^0$, $dim (range\ T')=dim (range\ T)$
c) dim(range\ T')= dim(range\ T)

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New materials  

#### Theorem 3.129, 3.131

Let $V,W$ be a finite dimensional vector space, $T\in \mathscr{L}(V,W)$

a) $T$ is injective $\iff T'$ is surjective  
b) $T$ is surjective $\iff T'$ is injective  

Proof:

$T$ is injective $\iff null\ T=\{0\}\iff range\ T'=V'\iff T'$ surjective  

$T$ is surjective $\iff range\ T=W\iff null\ T'=0\iff T'$ injective  

#### Theorem 3.132

Let $V,W$ be a finite dimensional vector space, $T\in \mathscr{L}(V,W)$

Then $M(T')=(M(T))^\top$. Where the basis for $M(T)'$ are the dual basis to the ones for $M(T)$

#### Theorem 3.133

$col\ rank\ A=row\ rank\ A$

Proof: $col\ rank\ A=col\ rank\ (M(T))=dim\ range\ T=dim\ range\ T'=dim\ range\ T'=col\ rank\ (M(T'))=col\ rank\ (M(T)^\top)=row\ rank\ (M(T))$

## Chapter V Eigenvalue and Eigenvectors

### Invariant Subspaces 5A

Goal: Study maps in $\mathscr{L}(V)$ (linear operations)

Question: Given $T\in \mathscr{L}(V)$ when can I restrict to $U\subseteq V$ such that $T\vert_U\in \mathscr{L}(U)$

#### Definition 5.2

Suppose $T\in \mathscr{L}(V)$ and $U\subseteq V$ a subspace is said to be invariant under $T$ if $Tu\in U,\forall u\in U$

Example:

For any $T\in \mathscr{L}(V)$, the following are invariance subspaces.

1. $\{0\}$
2. $V$
3. $null\ T$, $v\in null\ T\implies Tv=0\in null\ T$
4. $range\ T$, $v\in range\ T\subseteq V \implies Tv\in range\ T$

#### Definition 5.5

Suppose $T\in\mathscr{L}(V)$, then for $\lambda \in \mathbb{F}$ is an **eigenvalue** of $T$ if $\exists v\in V$ such that $v\neq 0$ and $Tv=\lambda v$.

#### Definition 5.8

Suppose $T\in\mathscr{L}(V)$ and $\lambda \in \mathbb{F}$ is an eigenvalue of $T$. The $v\in V$ is an **eigenvector** of $T$ corresponding to $\lambda$ if $v\neq 0$ and $Tv=\lambda v$

Note: if $\lambda$ is an eigenvalue of $T$ and $v$ an eigenvector corresponding to $\lambda$, then $U=Span(V)$ is an invariant subspace. and $T\vert_U$ is multiplication by $\lambda$

#### Proposition 5.7

$V$ is finite dimensional $T\in \mathscr{L},\lambda\in \mathbb{F}$ then the following are equivalent: (TFAE)

a) $\lambda$ is an eigenvalue  
b) $T-\lambda I$ is not injective
c) $T-\lambda I$ is not surjective
d) $T-\lambda I$ is not invertible

Proof:

(a)$\iff$ (b) $\lambda$ is an eigenvalue $\iff \exists v\in V$ such that $Tv=\lambda v\iff \exists v\in V, v\neq 0, (T-\lambda I)v=0$

Example:

$T(x,y)=(-y,x)$ what are the eigenvalues of $T$.

If $\mathbb{F}=\mathbb{R}$ rotation by $90\degree$, so no eigenvalues.

what if $\mathbb{F}=\mathbb{C}$? we can solve the system $T(x,y)=\lambda (x,y),(-y,x)=\lambda (x,y)$

$$
-y=\lambda x \\
x=\lambda y
$$

So

$$
-1=\lambda ^2,\lambda =\plusmn i
$$

when $\lambda =-i$, $v=(1,i)$, $\lambda=i$, $v=(1,-i)$