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I have a question in mind which is a slight generalization of a past question(famous). See here if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ if you want the basic version. Now my question : If $A^2$ is diagonalizable , then is $A$ diagonalizable when $A$ is non-singular? Obviously, the previous counterexample of taking $A$ to be a nilpotent matrix of index will not work here as then $|A|=0$. Any proofs?

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People are showing that the answer is yes. Someone should point out that the answer is yes if we're talking about an algebraically closed field (for example if $A$ is real and we're allowing a complex matrix to do the diaogonalizing.)

For example if $$A=\begin{bmatrix}0&1 \\-1&0\end{bmatrix}$$then $A^2=I$ is certainly diagonalizable, but $A$ has no real eigenvalues, so $A$ is not diagonalizable over $\Bbb R$.

David C. Ullrich
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We use:

$A$ is diagonalizable if and only if the minimal polynomial for $A$ has no repeated roots.

Also:

Given $A$ and a polynomial $p$ such that $p(A)=0,$ then $p(x)$ is divisible by the minimal polynomial of $A$.

Now, let $m(x)$ be the minimal polynomial for $A^2.$

If $m(x)$ has no repeated roots, then $p(x)=m(x^2)$ can only have $x=0$ as a repeated root. (I'll leave that to you to prove.)

But $p(A)=0$, so $p(x)$ is divisible by the minimal polynomial of $A.$ If $A$ is non-singular, then $0$ is not a root of the minimal polynomial for $A.$ So the minimal polynomial for $A$ has no repeated roots.

Thomas Andrews
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    Perhaps it's worth mentioning that you use this fact: if $p(A) = 0$, then the minimal polynomial of $A$ divides $p(x)$. Depending on one's definition of the minimal polynomial, this may or may not be a trivial observation. – Ben Grossmann Jun 23 '18 at 14:33
  • @Omnomnomnom Thanks. Edited. – Thomas Andrews Jun 23 '18 at 17:18