If $A$ is diagonizable then $p(A)$ is diagonalizable

Question

Show that if a matrix $A$ of size $n \times n$ is diagonalizable, then $p(A)$ is diagonalizable for each polynomial $p$.

Answer 1

We can write $p$ into this form:$$p(x) = a_mx^m + a_{m−1}x^{m−1} + · · · + a_1x + a_0$$ Now let $\Bbb v$ be an eigenvector of $A$ with eigenvalue $\lambda$.
Since $A^k\Bbb v = \lambda ^k\Bbb v$ for every $k$, we see that$$p(A)v = a_mA^m\Bbb v + a_{m−1}A^{m−1}\Bbb v + · · · + a_1A\Bbb v + a_0I\Bbb v \\= a_mλ^m\Bbb v + a_{m−1}λ^{m−1}\Bbb v + · · · + a_1λ\Bbb v + a_0\Bbb v\\ = (a_mλ^m + a_{m−1}λ^{m−1} + · · · + a_1λ + a_0)v$$ Therefore, $p(A)\Bbb v = p(\lambda)\Bbb v$ which shows that $p(\lambda)$ is an eigenvalue of the matrix $p(A)$, which means $p(A)$ is diagnolizable.

Answer 2

There is no need to consider eigenvalues or eigenvectors. You have: $$ A = B^{-1} D B $$ with $D$ being a diagonal matrix. It follows that for any $n\in\mathbb{N}$ $$ A^n = (B^{-1}DB)\cdot(B^{-1}DB)\cdot\ldots\cdot(B^{-1}DB)= B^{-1} D^n B$$ holds, so for every polynomial $p\in\mathbb{R}[x]$ $$ p(A) = B^{-1} p(D) B$$ holds too. Since $p(D)$ is a diagonal matrix, the claim follows.