# If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable.

I'm attempting to understand a proof presented on this Wikipedia page: https://en.wikipedia.org/wiki/Diagonalizable_matrix

The claim is as follows:

Let $$A$$ be a matrix over $$F$$. If $$A$$ is diagonalizable, then so is any power of it. Conversely, if $$A$$ is invertible, $$F$$ is algebraically closed, and $$A^n$$ is diagonalizable for some $$n$$ that is not an integer multiple of the characteristic of $$F$$, then $$A$$ is diagonalizable.

This is the provided proof:

If $$A^n$$ is diagonalizable, then $$A$$ is annihilated by some polynomial $${\displaystyle \left(x^{n}-\lambda _{1}\right)\cdots \left(x^{n}-\lambda _{k}\right)}$$, which has no multiple root (since $${\displaystyle \lambda _{j}\neq 0})$$ and is divided by the minimal polynomial of $$A$$.

Here are my questions:

(a) Why is it clear that $$A$$ is annihilated by that polynomial? $$A$$ certainly is annihilated by $$(x-\lambda_1)\cdots(x-\lambda_k)$$ and $$A^n$$ is annihilated by $$(x-\lambda_1^n)\cdots(x-\lambda_k^n)$$ (as powers of matrices have powers of eigenvalues as their eigenvalues), but I don't see the connection to Wikipedia's claim.

(b) What tells us that there is no multiple root? It's clear to me that $$\lambda_j\neq0$$ ($$A$$ is invertible implies $$A^n$$ is invertible implies $$A^n$$ does not have $$0$$ as an eigenvalue), but why can't two of the $$\lambda_j$$'s be equal? What says that we have distinct eigenvalues?

(c) Any polynomial that annihilates a matrix certainly is a multiple of the minimal polynomial...but why does this tell us that $$A$$ is diagonalizable?

(d) Earlier in the article, the following is claimed: A matrix or linear map is diagonalizable over the field $$F$$ if and only if its minimal polynomial is a product of distinct linear factors over $$F$$. If questions (a) and (b) are resolved, I can see how this would imply (c), but why is this claim true?

Here is another approach to this problem, but this one seems to be more complicated than what is presented on Wikipedia. Positive power of an invertible matrix with complex entries is diagonalizable only if the matrix itself is diagonalizable.

That passage was written by a user with a Japanese user name "TakuyaMurata". I suppose what he meant was actually this: suppose $$F$$ is an algebraically closed field and $$n>0$$ is not an integer multiple of $$\operatorname{char}(F)$$. If $$A$$ is invertible and $$A^n$$ is diagonalisable over $$F$$, then:

1. $$f(x)=(x^n-\lambda_1)\cdots(x^n-\lambda_k)$$ annihilates $$A$$, where $$\lambda_1,\ldots,\lambda_k$$ are the distinct eigenvalues of $$A^n$$.
2. Since the $$\lambda_i$$s are distinct, $$x^n-\lambda_i$$ and $$x^n-\lambda_j$$ have not any common root when $$i\ne j$$. As $$g(x):=x^n-\lambda_i$$ and $$g'(x)=nx^{n-1}$$ also have no common roots (because $$\lambda_i\ne0$$ and $$n\ne0$$ --- we have used the assumptions that $$A$$ is invertible and $$n$$ is not an integer multiple of $$\operatorname{char} F$$ here), we see that $$f$$ has not any multiple roots.
3. As the minimal polynomial $$m(x)$$ of $$A$$ divides $$f(x)$$, $$m$$ also has not any multiple roots.
4. Therefore, $$m$$ is a product of distinct linear factors (note that $$m$$ splits because $$F$$ by assumption is algebraically closed). Hence $$A$$ is diagonalisable over $$F$$.

https://en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)

quote: An endomorphism φ of a finite dimensional vector space over a field F is diagonalizable if and only if its minimal polynomial factors completely over F into distinct linear factors. The fact that there is only one factor X − λ for every eigenvalue λ means that the generalized eigenspace for λ is the same as the eigenspace for λ: every Jordan block has size 1.

Some quick justifications here: https://www.maths.ed.ac.uk/~tl/minimal.pdf

So since we supposed $$A^n$$ diagonalisable, it has such a minimal polynomial $$\mu(x)=\prod(x-\lambda_i)$$ with distinct $$\lambda_i$$ that satisfies $$\mu(A^n)=0$$

But $$\mu(A^n)=0$$ also means that $$M(A)=0$$ where $$M(x)=\prod(x^n-\lambda_i)$$ or equivalently $$M(x)=\mu(x^n)$$.

Now if the $$\lambda_i=r_ie^{i\theta_i}$$ are distinct then so are their k-roots.

Indeed $$(\lambda_1)^{1/n}=(\lambda_2)^{1/n}\implies\begin{cases} r_1=r_2\\ \frac{\theta_1+2n_1\pi}n=\frac{\theta_2+2n_2\pi}n+2m\pi\end{cases}\implies\begin{cases} r_1=r_2\\ \theta_1=\theta_2\pmod{2\pi}\end{cases}\implies \lambda_1=\lambda_2$$

(We get the result by contraposition).

Considering now the minimal polynomial of $$A$$, it has to divide $$M(x)$$, but we have just seen that because of the hypothesis made on $$F$$, that $$M$$ is completely split, and all the linear factors are distinct.

Coming back to the quote at the beginning, it means that $$A$$ itself should be diagonalisable.

• It seems as though your proof that distinct eigenvalues have distinct roots is the converse of what we need. We know that the $\lambda_i$'s are distinct eigenvalues of $A^k$, and we need to show that $(\lambda_i)^{1/k}$ are distinct eigenvalues of $A$. Also, you changed notation from the original question ($n$ to $k$), but that's not a big deal. Dec 5, 2018 at 22:56
• Nevermind, I see you proved the contrapositive. Dec 5, 2018 at 23:00
• yep, I edited. ;
– zwim
Dec 6, 2018 at 0:07
• I've accepted your answer, but now I'm wondering where the assumption that $A$ is invertible comes into play Dec 6, 2018 at 0:41
• The invertibility is just to avoid these $\lambda_i=0$, since then a divisor of $x^n$ is not necessarily $x$ but any power $x^k$ with $k=1..n$. @Guido, yes I detailled only because OP seemed confused about it.
– zwim
Dec 6, 2018 at 7:37