How can we evaluate the characteristic polynomial with a matrix as the parameter? 
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 diagonalisable
then $χ_A(A)$ is the zero matrix

I don't get what it means here to apply the characteristic polynomial with the matrix as the parameter. Does it subtract from each term in p(A)?
 A: The definition given for $p$ is just showing you how to treat $\chi_A(A)$. The characteristic polynomial typically takes scalar values and has roots at the eigenvalues of $A$, but now instead of a function from $\mathbb{R}\to\mathbb{R}$ we treat it as a function $M_{n\times n}(\mathbb{R})\to M_{n\times n}(\mathbb{R})$ where we plug in a matrix and obtain a matrix as a result. The function $p$ is just being used as an example.
A: As mentioned in the other post, if $p(x)$ is a polynomial then $p(A)$ is simply plugging in A. It works analogously (but remember to multiply the constant term by the identity matrix).
Now to help you with the second part, you need to know about the spectral mapping theorem. It says that $$spec(p(A))=p(spec(A))$$
That is, the eigenvalues of p(A) are the eigenvalues of A, evaluated in the polynomial, i.e. $p(\lambda_1),...,p(\lambda_n)$, where $\lambda$ are eigenvalues of p(A).
Here $p=\chi$, the characteristic polynomial. Then by the spectral mapping theorem you know that the eigenvalues of $\chi(A)=\chi(\lambda_1),...,\chi(\lambda_n)$. But by definition of the characteristic polynomial, these are all 0. So the only eigenvalue of $\chi(A)$ is 0. If A is diagonalizable, it can be written in the form $A=VDV^{-1}$ where $D=diag(\lambda)$ and V are the eigenvectors. Suppose the characteric polynomial of A has the form $a_0+a_1x+...+a_nx^n$. Then
$$\begin{split}\chi(A)&=a_0I+a_1A+...+a_nA^n\\
&=a_0I+a_1VDV^{-1}+...+a_nVD^nV^{-1}\text{ a nice property of diagonalization}\\
&=V(a_0I+a_1D+...+a_nD^n)V^{-1}\\
&=Vdiag(a_0+a_1\lambda_i+...+a_n\lambda_i^n)V^{-1}\text{ by nice property of diagonal matrices}\end{split}$$
But the characteristic polynomial of A evaluated at each of its eigenvectors is 0. Thus this becomes
$$\begin{split}\chi(A)&=V0V^{-1}\\&=0\end{split}$$ the nxn zero matrix.
This proves the second part.
I leave the first part to you.
