$p$ can be any polynomial and the matrices are square matrices
Now I know that by the Cayley–Hamilton theorem any matrice A satisfies it's characteristic polynomial and I feel that the key to the proof lies there, but not sure how to capitalize on it.
My guess is since $A$ and $B$ are similar they have the same $p(\lambda)$ so doesn't that imply (by what was stated earlier) that the matrices $p(A)$ and $p(B)$ are similar?
Best Answer
Let $p(x) = \sum_{k=0}^n a_kx^k$.
If $A, B$ are similar, there is an invertible $S$ such that $B = SAS^{-1}$.
Let $p(A) = P$.
What is $p(B)$ ?
$$p(B) = \sum_{k=0}^n a_kB^k = \sum_{k=0}^n a_k(SAS^{-1})^k = \\ \sum_{k=0}^n a_kSA^kS^{-1} = S\left(\sum_{k=0}^n a_kA^k\right)S^{-1} = Sp(A)S^{-1} = SPS^{-1}$$
and the matrices are similar.