The Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem 

Theorem 1. (The Cayley-Hamilton Theorem) Let $T \in \mathcal{L}(V)$, and let $f(t)$ be the characteristic polynomial of $T$. (V is finite-dimensional) Then $f(T) = T_0$, the zero transformation.

Proof. We need to show that $f(T)(v) = \mathbf{0}, \forall v \in V$. If $v = \mathbf{0}$, it is clear. 
Suppose that $v \neq \mathbf{0}$. Let $W$ be the $T$-cyclic subspace of $V$ generated by $v$, and let $k = \dim(W)$. Then $! \exists a_0, ..., a_{k-1} \in F$ such that $T^k(v) = a_0v + \cdots + a_{k-1}T^{k-1}(v)$ By Theorem 2. By denoting $b_i = -a_i$ for each $i$ ($0 \leq ki \leq k-1$), $b_0v + \cdots + b_{k-1}T^{k-1}(v) + T^k(v) = \mathbf{0}$. Then by Theorem 2, the characteristic polynomial $g(t)$ of $T_W$ is $g(t) = (-1)^k(b_0 + \cdots + b_{k-1}t^{k-1} + t^k)$.
By Theorem 1, $f(t) = g(t) \cdot h(t)$ where $h(t) \in P(F)$. Then $f(T)(v) = h(T)g(T)(v) = h(T)(-1)^k(b_0v + \cdots + b_{k-1}T^{k-1}(v) + T^k(v)) = \mathbf{0}$. Thus $f(T) = T_0$. $\blacksquare$

Corollary

Corollary. Let $A \in M_{n \times n}(F)$, and $f(t)$ be the characteristic polynomial of $A$. Then $f(A) = O$.

Proof. Let denote $f(t) = a_0 + a_1t + \cdots + a_nt^n$. Then $f(L_A)(v) = a_0v + a_1Av + \cdots + a_nA^nv = f(A)v = \mathbf{0}$. Thus $f(L_A)(v) = f(A)v = L_{f(A)}(v) = \mathbf{0} = Ov, \forall v \in F^n$. Since $L_{f(A)}(v) = Ov = L_O(v), \forall v \in F^n$, $L_{f(A)} = L_O \Longrightarrow f(A) = O$. $\blacksquare$