For any polynomial $p(x) = a_0 + a_1x + · · · + a^kx^k$ and any square matrix A, $p(A)$ is defined as $p(A) = a_0I + a_1A + · · · + a_kA^k$. Show that if v is any eigenvector of A and $χ_A(x)$ is the characteristic polynomial of A, then $χ_A(A)v = 0$, Deduce that if A is diagonalizable then $χ_A(A)$ is the zero matrix.
I got that $p(\lambda)$ is the eigenvalue of $p(A)$ and if A is diagonalizable then $χ_A(x)$ is the zero matrix since p(A) is similar to its diagonal matrix p(D)
$$p(D) = a_nD_n + · · · + a_1D + a_0I$$
= $\begin{matrix} p(\lambda_1) &0&...&0\\0& p(\lambda_2)&...& 0 \\ ...&...&...&...\\0&0&... & p(\lambda_n) \end{matrix}$
First if $v$ is an eigenvector of $A$, there is some eigenvalue of $A$, $\lambda$ such that $Av=\lambda v$. then for all positive integer $k$, $A^kv=\lambda^kv$ so for any polynomial $P$ you have $P(A)v=P(\lambda)v$ Then since $\lambda$ is an eigenvalue it is a root of the characteristic polynomial : $\chi_A(\lambda)=0$ so $\chi_A(A)v=0$