Is $A \in \mathcal{M}_n(\mathbb{C})$ diagonalizable?

Question

$A \in \mathcal{M}_n(\mathbb{C})$ 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.

Answer 1

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.

Answer 2

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.