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.