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

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.

• 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. Commented Nov 16, 2020 at 8:41

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.

• 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 Commented Nov 16, 2020 at 17:43
• @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 $A$. Commented Nov 16, 2020 at 17:47
• Yeah sorry @mechanodroid I agree with you. Some morning tiredness ;) Thanks a lot! Commented Nov 17, 2020 at 9:05

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.

• I don't master the concept of Jordan normal form. But thank you @daw. Commented Nov 16, 2020 at 10:14
• @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.
– daw
Commented Nov 16, 2020 at 10:35
• What would I get then @daw? The size 1 block argument is no longer usable in this case, is it? Commented Nov 16, 2020 at 14:18
• Why is $p(J)$ in Jordan normal form? Commented Nov 16, 2020 at 17:10
• @mechanodroid ? I did not claim something like this.
– daw
Commented Nov 16, 2020 at 17:18