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$A \in \mathcal{M}_n(\mathbb{C})$ such that $A^2$ has got $n$ distinct non zero eigenvalues. Show that A is diagonalizable.

Attempt :

As $A^2$ has got $n$ distinct non zero eigenvalues. The characteristic polynomial of $A^2$ :$\mathcal{X}_{A^2}$ is :

$\mathcal{X}_{A^2}(X)=(-1)^n\prod\limits_{k=1}^n(X-\lambda_k)$ since $\mathcal{X}_{A^2}(A^2)=0$, I deduce that

$P(A)=0$ with $P(X)=\mathcal{X}_{A^2}(X^2)=\prod\limits_{k=1}^n(X-\sqrt{\lambda_k})(X+\sqrt{\lambda_k})$

As $P$ is composed with linear factors, we deduce the same for its minimal polynomial. Then A is diagonalizable.

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  • $\begingroup$ I'm sure you want roots in $P$. Further you need to treat the case where one of the $\lambda_k$ is zero separately. $\endgroup$ – s.harp Feb 21 at 22:19
  • $\begingroup$ I don't understand, we have $\lambda_k\ne 0$ $\endgroup$ – Stu Feb 21 at 22:23
  • $\begingroup$ I see, I didn't read that part of your formulation. What you wrote then gives the implication, but you may want to carry out more steps. $\endgroup$ – s.harp Feb 21 at 22:40
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The matrix $A$ has $n$ eigenvalues $\mu_1\ldots,\mu_n$, not necessarily distinct. Of course, if you prove that they are distinct, then your problem is solved. But they are distinct, since the eigenvalues of $A^2$ are ${\mu_1}^2,\ldots,{\mu_n}^2$, and you know that these are distinct.

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  • $\begingroup$ I will add a correction right now $\endgroup$ – Stu Feb 21 at 22:29
  • $\begingroup$ $\lambda_k$ are the eigenvalues of $A^2$ $\endgroup$ – Stu Feb 21 at 22:35
  • $\begingroup$ Right. I've called $\mu_k$ to the eigenvalues of $A$ then. $\endgroup$ – José Carlos Santos Feb 21 at 22:37
  • $\begingroup$ This answer has a deceptive simplicity due to sloppy formulation. By "$n$ eigenvalues not necessarily distinct" you must mean taken with their multiplicities as roots of the characteristic polynomial, since other choices of possible repetition won't make the argument work. Then to claim that their squares give the eigenvalues of $A^2$, presumably with the same kind of multiplicities, is not so obvious, though it is true. One way to prove it is to use a Jordan normal form, but if your proof is using such an argument then it should be more explicit about that. $\endgroup$ – Marc van Leeuwen Feb 25 at 15:04
  • $\begingroup$ See for instance this question. $\endgroup$ – Marc van Leeuwen Feb 25 at 15:10
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The argument in the question can be made complete by adding that your polynomial $P$ has only simple roots; it is that property rather than the number of roots that allows the conclusion of diagonalisability. Here is how I would formulate it.

The matrix $A^2$ has a characteristic polynomial $\chi$ of degree$~n$ with $n$ (distinct) roots, which means it factors $\chi=(X-a_1)\ldots(X-a_n)$ with $a_1,\ldots,a_n\in\Bbb C$ all distinct, and nonzero by hypothesis. Since $\chi$ is an annihilating polynomial of $A^2$, the result $P=\chi[X:=X^2]$ of substituting $X^2$ for $X$ is an annihilating polynomial of$~A$. Moreover rewriting each factor $X^2-a_i=(X-b_i)(X+b_i)$ where $b_i$ is one of the two complex square roots of $a_i$, one has $P=(X-b_1)(X+b_1)\ldots(X-b_n)(X+b_n)$, and this split polynomial has simple roots since $b_i\neq-b_i$ for any $i$, and $b_i\neq\pm b_j$ for $i\neq j$ since $b_i^2=a_i\neq a_j=b_j^2$. Being annihilated by a split polynomial with simple roots is a (necessary and) sufficient condition for being diagonalisable.

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