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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.

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    $\begingroup$ I didn't do the entire thing, but as a first idea, I would consider the minimal polynomial $Q$ of $P(A)$, and try to show that the roots of $Q(P)$ have multiplicity $1$. Then since the minimal polynomial of $A$ divides $Q(P)$ it would also only have roots of multiplicity $1$, which would make $A$ diagonalizable. Though it would also imply that $Q(P)$ is already the minimal polynomial of $A$, which to me seems like a strong result, so I'm not entirely sure that this will work. $\endgroup$ Commented Nov 16, 2020 at 8:41

2 Answers 2

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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.

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  • $\begingroup$ Why do you say that the zeroes of $m_A$ are exactly $\lambda_1, \dots, \lambda_k$? I would say they are among $\lambda_1, \dots, \lambda_k$. I'll edit your wonderful post :) Anyway, the proof is still valid, great job! I'll validate the answer after the edit @mechanodroid $\endgroup$ Commented Nov 16, 2020 at 17:43
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    $\begingroup$ @MrMaths I disagree, the zeroes of $m_A$ are exactly the eigenvalues of $A$ which we denoted by $\lambda_1, \ldots, \lambda_k$. It is not hard to prove: if $Ax = \lambda x$ where $x \ne 0$ then $$0 = m_A(A)x = m_A(\lambda) x \implies m_A(\lambda) = 0.$$ The converse follows from the fact that $m_A$ divides the characteristic polynomial of $A$. $\endgroup$ Commented Nov 16, 2020 at 17:47
  • $\begingroup$ Yeah sorry @mechanodroid I agree with you. Some morning tiredness ;) Thanks a lot! $\endgroup$ Commented Nov 17, 2020 at 9:05
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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.

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  • $\begingroup$ I don't master the concept of Jordan normal form. But thank you @daw. $\endgroup$ Commented Nov 16, 2020 at 10:14
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    $\begingroup$ @MrMaths maybe one can apply a similar argument to the triangularized $A$. One has to compute the entries of $p(A)$ and $p'(A)$ for the diagonal above the main diagonal. $\endgroup$
    – daw
    Commented Nov 16, 2020 at 10:35
  • $\begingroup$ What would I get then @daw? The size 1 block argument is no longer usable in this case, is it? $\endgroup$ Commented Nov 16, 2020 at 14:18
  • $\begingroup$ Why is $p(J)$ in Jordan normal form? $\endgroup$ Commented Nov 16, 2020 at 17:10
  • $\begingroup$ @mechanodroid ? I did not claim something like this. $\endgroup$
    – daw
    Commented Nov 16, 2020 at 17:18

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