TITLE: Show that $A$ is diagonalisable when $P(A)$ is diagonalisable and $P'(A)$ is invertible
QUESTION [6 upvotes]: Let $n \geq 1$ and $A \in M_n(\mathbb{C})$.
Assume there exists $P \in \mathbb{C}[X]$ such that:

$P(A)$ is diagonalisable
$P'(A)$ is invertible

I have to show that $A$ is diagonalisable.
My try:
I solved the problem when $\textrm{deg}(P) \leq 1$. I also know that $A$ is trigonalisable, then I deduced that: $\forall \lambda \in \textrm{Sp}(A), P'(\lambda) \neq 0$. Also, there exists a diagonal matrix $D$ and $S \in GL_n(\mathbb{C})$ such that $P(A) = SDS^{-1}$. But I can't say anything more. Any help is welcome.

REPLY [6 votes]: Let $\sigma(A) = \{\lambda_1, \ldots, \lambda_k\}$ be the eigenvalues of $A$. We know:

$p(A)$ is diagonalizable so its minimal polynomial $m_{p(A)}$ splits into linear factors. The eigenvalues of $p(A)$ are precisely $\sigma(p(A)) = \{p(\lambda_1), \ldots, p(\lambda_k)\}$ so
$$m_{p(A)}(x) = (x-p(\lambda_1))\cdots(x-p(\lambda_k)).$$

$p'(A)$ is invertible so $0 \notin \sigma(p'(A)) = \{p'(\lambda_1), \ldots, p'(\lambda_k)\}$ or $p'(\lambda_i) \ne 0$ for all $1 \le i\le k$.


Notice that the polynomial $m_{p(A)} \circ p$ annihilates $A$ so the minimal polynomial $m_A$ of $A$ divides it, i.e. $m_A \mid m_{p(A)} \circ p$.
We wish to show that $A$ is diagonalizable, i.e. that $m_A$ splits into linear factors. The zeroes of $m_A$ are among $\lambda_1,\ldots, \lambda_k$ so let's assume that $(x-\lambda_i)^2$ divides $m_A$. Then it also divides $m_{p(A)} \circ p$ so there exists a polynomial $q \in \Bbb{C}[x]$ such that
$$m_{p(A)}(p(x)) = (x-\lambda_i)^2q(x).$$
Taking the derivative gives
$$m'_{p(A)}(p(x))p'(x) = 2(x-\lambda_i)q(x)+(x-\lambda_i)^2q'(x)$$
and plugging in $x = \lambda_i$ yields
$$m'_{p(A)}(p( \lambda_i))\underbrace{p'( \lambda_i)}_{\ne0} = 0$$
so $m'_{p(A)}(p( \lambda_i)) = 0$. This means that $p(\lambda_i)$ is a zero of $m_{p(A)}$ with multiplicity at least $2$ which contradicts the fact that $m_{p(A)}$ splits into linear factors.
Therefore it is not possible that $(x-\lambda_i)^2$ divides $m_A$ so it has to be
$$m_A(x) = (x-\lambda_1)\cdots (x-\lambda_k)$$
so $A$ is diagonalizable.