Diagonalizing the matrix $A$, when $A^2$ is diagonalizable If the matrix $A^2$ is diagonalizable and $A$ is invertible, is $A$ diagonalizable? I know it is not true if we leave out the invertibility. For example 
if
$
A=      \begin{pmatrix}
        0 & 1  \\
        0 & 0  \\
        \end{pmatrix}
$ (which is not diagonalizable), then 
$A^2=      
        \begin{pmatrix}
        0 & 0  \\
        0 & 0  \\
        \end{pmatrix}
$ (and that is trivially diagonalizable).
But what if we require $A$ to be invertible? I believe in that case, answer to my original question is yes, but I'm not able to prove that. Can someone please help me?
 A: I'm assuming this is over the complex numbers.
Put $A$ in Jordan canonical form.  The square of a nontrivial Jordan block (with non-zero diagonal) is an upper triangular matrix with nonzero diagonal and above-diagonal elements, and the eigenvalue has geometric multiplicity $1$.  So if $A$ is invertible and not diagonalizable, neither is $A^2$. 
A: Notation: $E_{L}^{B}$ is the set of eigenvectors associated to $L$ for the matrix $B$.
As $A^2$ is diagonalizable, 
$$
E = \bigoplus_{l_i\neq 0} E_{l_i}^{A^2}
$$
On each $E_{l_i}^{A^2}$, $A^2 -l_i I =0 $ and then, as $l_i\neq 0$,
$$E_{l_i}^{A^2} = E_{-\sqrt{l_i}}^{A} \oplus E_{\sqrt{l_i}}^{A}$$
Hence $$
E = \bigoplus_{l_i\neq 0}  \left[E_{-\sqrt{l_i}}^{A} \oplus E_{\sqrt{l_i}}^{A}\right]
$$and $A$ is diagonalizable.
A: Let me formulate what I believe is the precise condition to make this conclusion. No assumptions are made about the base field $K$ over which one wants to diagonalise

Proposition. Let $A$ be a square matrix with entries in a field $K$. Assume that
  
  
*
  
*$A^2$ is diagonalisable over$~K$, with eigenvalues in a set $S\subseteq K$, and
  
*every element of $S$ has two distinct square roots in $K$.
  
  
  Then $A$ is diagonalisable over$~K$, with eigenvalues in $\sqrt S\overset{\rm def}=\{\,r\in K\mid r^2\in S\,\}$.

Note that $S$ need not be just the set of eigenvalues of $A^2$, and certainly $\sqrt S$ can be larger than the set of eigenvalues of$~A$. But one may, and we will, assume $S$ to be finite.
Proof. Let $P_1=\prod_{s\in S}(X-s)$, then by the first assumption $P_1[A^2]=0$. By the second assumption there exists for each $s\in S$ an element $r\in K$ such that $X^2-s=(X-r)(X+r)$, and $r\neq -r$. Choosing one such$~r$ and calling it $\sqrt s$, one has
$$
 P_2\overset{\rm def}=P_1[X^2]=\prod_{s\in S}(X-\sqrt s)(X+\sqrt s)=\prod_{r\in\sqrt S}(X-r).
$$
Then $P_2[A]=P_1[A^2]=0$, and since $P_2$ is split with simple roots (because of $r\neq-r$) in $\sqrt S\subseteq K$, it follows that $A$ is diagonalisable over$~K$, with eigenvalues in $\sqrt S$. $\square$
Note that in fields not of characteristic$~2$ the second condition excludes $0$ and any non-square from the set $S$ of allowed eigenvalues for$~A$, while for fields of characteristic$~2$ it excludes everything (i.e., the proposition will never apply). These are precisely the cases where the conclusion fails miserably.
