The polynomial of a square matrix and its eigenvalues

Question

For any polynomial

$$p(x) = a_0 + a_1 x + \cdots + a_k x^k$$

and any square matrix $A$, polynomial $p(A)$ is defined as

$$p(A) = a_0 I + a_1 A + \cdots + a_k A^k$$

Show that if $v$ is any eigenvector of $A$ and $\chi_A(x)$ is the characteristic polynomial of $A$, then $\chi_A(A) v = 0$. Deduce that if $A$ is diagonalizable then $\chi_A(A)$ is the zero matrix.

I don't get "if $v$ is any eigenvector of $A$ and $\chi_A(x)$ is the characteristic polynomial of $A$, then $\chi_A(A)v = 0$". I have gotten $p(\lambda)$ is an eigenvalue of the matrix $p(A)$, but how do I continue?

Answer 1

Sketch: Observe that if $\lambda$ is an eigenvalue of $A$ with corresponding eigenvector $v$, then

$$ \chi_A(A)v=\sum_{i=0}^k a_iA^iv=\left(\sum_{i=0}^k a_i\lambda^i\right)v=\chi_A(\lambda)v=0 $$