NoteNextra-origin/content/Math429/Math429_L19.md

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.