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)

so

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 $\mathbb{F}$

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

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 $degree\ q< degree\ p$

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.