12 - Cayley-Hamilton

Cayley-Hamilton

Let $V$ be a finite dimensional vector space

If $T:V\to V$ is a linear transformation then

$${\chi }_T(T)=0$$

In particular

$$m_T(x)\mid {\chi }_T(x)$$

Proof

Let $A$ be the matrix of $T$ with respect to some basis for $V$
Let $\overline{\mathbf{F}}\subseteq \mathbf{F}$ be the algebraic closure of $\mathbf{F}$

So in $\overline{\mathbf{F}}[x]$ then every polynomial factors into linear terms
So by upper triangularisation of a matrix there exists $P\in M_n(\overline{\mathbf{F}})$ such that

$$P^{-1}AP\text{ is upper triangular }\text{ with }\text{ diagonal entries }{\lambda }_1,\cdots ,{\lambda }_n$$

Hence

$${\chi }_{P^{-1}AP}(x)=(-1{)}^{\dim V}\prod _{k=1}^n(x-{\lambda }_k)$$

By Annihilating Polynomial of an Upper Triangular Matrix then

$${\chi }_{P^{-1}AP}(P^{-1}AP)=0$$

As ${\chi }_T(x):={\chi }_{P^{-1}AP}(x)$ then

$${\chi }_T(T)=0$$

As Minimal Polynomial divides Annihiliating Polynomials then

$$m_T(x)\mid {\chi }_T(x)$$