Lecture 20

Chapter V Eigenvalue and Eigenvectors

Minimal polynomial 5B

Definition 5.24

Suppose V is finite dimensional, and T\in \mathscr{L}(V) is a linear operator, then the minimal polynomial of T is the unique monic polynomial p of smallest degree satisfying the p(T)=0.

Theorem 5.22

Suppose V is finite dimensional T\in \mathscr{L}(V), then there exists a unique monic polynomial p\in \mathscr{P}(\mathbb{F}) of smallest degree such that p(T)=0. Furthermore deg\ p \leq dim\ V

Proof:

Induct on dim\ V to prove existence.

Base case: dim\ V=0, i.e V={0}. Then any linear operator on V is 0 including the I. So use p(z)=1 then p(T)=I=0.
Inductive step: Suppose the existence holds for all vector spaces with dimension < dim\ V. and dim V\neq 0, Toke v\in V,v\neq 0. Then the list v,Tv,Tv^2,...,T^n v,n= dim\ V is linearly dependent.

then we take the smallest m such that v,Tv,...,T^m v is linearly dependent, then there exists c_0,...,c_{n-1} such that c_0 v+c_1T_v+...+c_{m-1} T^{m-1}+T^mv=0

Now we define p(z)=c_0+c_1z+...+c_{m-1}z^{m-1}+z_m,p(T)v=0, by (c_0 v+c_1T_v+...+c_{m-1} T^{m-1}+T^mv=0)

Moreover, p(T)(T^k v) let q(z)=z^k, then p(T)(T^k)=p(T)q(T)(v)=0, so T^k v\in null(p(T)), thus since v,Tv,..,T^{m-1}v are linearly independent, thus dim\ null\ (p(T))\geq m.

Note that dim\ range\ (p(T))\leq dim\ V-m is invariant with respect to T.

So consider T\vert _{range\ (p(T))}, so by the inductive hypothesis, there exists S\in \mathscr{P}(\mathbb{F}) with deg\ p\leq dim\ range\ (p(T)) such that S(T\vert_{range\ (p(T))}). Now consider (SP)\in \mathscr{P}(\mathbb{F}) to see this let v\in V. then (SP)(T)(v)=(S(T)p(T))(v)=S(T)(p(T)v)=S(T)0=0

deg\ S p=deg\ S+deg\ p\leq dim\ V

uniqueness: Let p be the minimal polynomial, then let q\in \mathscr{L}(\mathbb{F}) monic with q(T)=0 and deg\ q=deg\ p the (p-q)(T)=0 and deg(p-q)\leq deg\ p but then p-q=0 \implies p=q

Finding Minimal polynomials

Idea: Choose v\in V,v\neq 0 find m such that v,Tv,...,T^{dim\ V} v

Find constant (if they exists) such that v_0v+c_1Tv+...+c_{dim\ V-1} T^{dim\ V-1}+ T^{dim\ V}=0

then if the solution is unique (not always true). then p(z)=v_0v+c_1Tv+...+c_{dim\ V-1} T^{dim\ V-1}+ T^{dim\ V} is the minimal polynomial.

Example:

Suppose T\in \mathscr{L}(\mathbb{R}^5) with $M(T)=\begin{pmatrix} 0&0&0&0&-3\ 1&0&0&0&6\ 0&1&0&0&0\ 0&0&1&0&0\ 0&0&0&1&0\ \end{pmatrix}$

let v=e_1,Tv=e_2,T^2v=e_3,T^3 v=e_4, T^4v=e_5, T^5v=-3e_1+6e_2

now T^5v-6Tv+3v=0 this is unique so p(z)=z^5-6z+3 is the minimal polynomial.

Theorem 5.27

If V is finite dimensional and T\in\mathscr{L}(V), with minimal polynomial p, then the zeros of p are (exactly) their eigenvalues.

Theorem 5.29

T\in \mathscr{L}(V), p the minimal polynomial and q\in\mathscr{P}(\mathbb{F}), such that q(T)=0, the p divides q.

Corollary 5.31

If T\in \mathscr{L}(V) with minimal polynomial p U\subseteq V (invariant subspace), then p is a multiple of T\vert_U divides p.

Theorem 5.32

T is not invertible \iff The minimal polynomial has 0 as a constant term.

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