# Lecture 19

## Chapter V Eigenvalue and Eigenvectors

### Invariant Subspaces 5A

#### Proposition 5.11

Suppose $T\in \mathscr{L}(V)$, let $v_1,...,v_n$ be eigenvectors for distinct eigenvalues $\lambda_1,...,\lambda_m$. Then $v_1,...,v_n$ is linearly independent.

Proof: 

Suppose $v_1,...,v_m$ is linearly dependent, we can assume that $v_1,...,v_{m-1}$ is linearly independent. So let $a_1,...,a_{m}$ not all $=0$. such that $a_1v_1+...+a_nv_m=0$, then we apply $(T-\lambda_m I)$ (map $v_n$ to 0)

$$
(T-\lambda_m I)v_k=(\lambda_k-\lambda_m)v_k
$$

so

$$
(T-\lambda_m I)=a_1(\lambda_1-\lambda_m)v_1+...+a_{m-1}(\lambda_{m-1}-\lambda_{m})v_m
$$

but not all of the $a_1,...,a_{m-1}$ are zero and $\lambda_k-\lambda_m\neq 0$ for $1\leq k\leq \lambda$ so they must be linearly independent.

#### Theorem 5.12

Suppose $dim\ V=n$ and $T\in \mathscr{L}(V)$ then $T$ has at most $n$ distinct eigenvalues

Proof:

Since $dim\ V=n$ no linearly independent list has length than $n$ so by **Proposition 5.11**, there are at most $n$ distinct eigenvalues.

#### Polynomials on operators

$p(z)=z+3z+z^3\in \mathscr{P}(\mathbb{R})$

let $T=\begin{pmatrix}
    1&1\\
    0&1
\end{pmatrix}\in \mathscr{L}(\mathbb{R}^2)$

$P(T)=2I+3T+T^3=2I+3T+\begin{pmatrix}
    1&3\\
    0&1
\end{pmatrix}=\begin{pmatrix}
    6&4\\
    0&6
\end{pmatrix}$

#### Notation

$T^m=TT...TT$ (m times) $T$ must be an operator within the same space

$T^0=I$

$T^{-m}=(T^{-1})^m$ (where $T$ is invertible)

if $p\in \mathscr{P}(\mathbb{F})$ with $p(z)=\sum_{i=0}^na_iz^i$ and $T\in \mathscr{L}(V)$ $V$ is a vector space over $\mathbb{F}$

$$
p(T)\sum_{i=0}^na_iT^i
$$

#### Lemma 5.17

Given $p,q\in \mathscr{P}(\mathbb{F})$, $T\in \mathscr{L}(V)$

then 

a) $(pq)T=p(T)q(T)$  
b) $p(T)q(T)=q(T)p(T)$

#### Theorem 5.18

Suppose $T\in \mathscr{L}(V),p\in \mathscr{P}(\mathbb{F})$, then $null\ (P(T))$ and $range\ (P(T))$ are invariant with respect to $T$.

### 5B The Minimal Polynomial

#### Theorem 5.15

Every operator on **finite dimensional complex vector space** has at least on eigenvalues.

Proof: 

Let $dim\ V=n,T\in \mathscr{L}(V), v\in V$ be a nonzero vector.

Now consider $v,Tv,T^2 v,...,T^n v$. Since this list is of length $n+1$, there is a linear dependence. Let $m$ be the smallest integer such that $v,Tv,..T^m v$ is linearly dependent, then 

$$
a_0 v+a_1Tv+...+a_m T^m v=0
$$

Let $p(z)=a_0+a_1 z+...+a_m z^m$, then $p(T)(v)=0,p(z)\neq 0$

$p(z)$ factors as $(z-\lambda) q(z)$ where $degree\ q< degree\ p$

$$
p(T)(v)=((T-\lambda I)q(T))(v)=0
$$

$$
(T-\lambda I)(q(T)(v))=0
$$

but $m$ was minimal so that $p(z)=a_0+a_1 z+...+a_m z^m$ were linearly independent, so $q(T)(v)\neq 0$, so $\lambda$ is an eigenvalue with eigenvector $q(T)(v)$

#### Definition 5.24

Suppose $V$ is finite dimensional $T\in\mathscr{L}(V),p\in \mathscr{P}(\mathbb{F})$, then the **minimal polynomial** is the unique monic (the coefficient of the highest degree is 1) polynomial of minimal degree such that $p(T)=0$

#### Theorem 5.27

Let $V$ be finite dimensional, and $T\in\mathscr{L}(V)$, $p(z)$ the minimal polynomial.

1. The roots of $p(z)$ are exactly the eigenvalues of $T$.
2. If $\mathbb{F}=\mathbb{C}$, $p(z)=(z-\lambda_1)...(z-\lambda_m)$ where $\lambda_1,...,\lambda_m$ are all the eigenvalues.