Show that $A$ is diagonalisable when $P(A)$ is diagonalisable and $P'(A)$ is invertible

Question

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.

Answer 1

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.

Answer 2

Since $S^{-1}p(A)S = p(S^{-1}AS)$, we can assume that $A$ is in Jordan normal form. Hence, it is sufficient to consider only one Jordan block $J$ to eigenvalue $\lambda$ of $A$. According to (matrix function of Jordan block), we have $$ p(J) = \pmatrix{ p(\lambda) & p'(\lambda) & \dots\\0 & p(\lambda) & p'(\lambda) & \dots\\&\ddots&\ddots&\ddots}, \quad p'(J) = \pmatrix{ p'(\lambda) & p''(\lambda) & \dots\\0 & p'(\lambda) & p''(\lambda) & \dots\\&\ddots&\ddots&\ddots}. $$ By assumption, $p'(A)$ and hence $p'(J)$ are invertible, so $p'(\lambda)\ne0$. But $p(A)$ and hence $p(J)$ are diagonalizable. This implies that the Jordan block $J$ is of size $1\times1$ (or $p'(\lambda)=0$, which is impossible).

This shows that all Jordan blocks of $A$ are of size $1\times 1$, and $A$ is diagonalizable.