# If $A^2$ is diagonalizable, must $A$ be such as well?

Given a diagonalizable matrix $A^2$, must the matrix $A$ be diagonalizable as well?

I can prove that this is true for when $A\in M_{n\times n} (\mathbb{C})$ by using the theorem that the Minimal polynomial for $A^2$ is expressed as a multiplication of linear attributes, and we can simply take $\pm \sqrt{\lambda_i}$ and show that $A$'s minimal polynomial is also a multiplication of linear attributes, thus making $A$ diagonalizable as well.

The problem is that I do now know whether or not this statement is correct for $A\in M_{n\times n} (\mathbb{R})$, I know my proof won't work for when $\lambda_i < 0$, but perhaps there is another proof for this? Or a counterexample?

• Consider the rotation of $\pi/2$ in $\mathbb R^2$ – user99914 Nov 25 '14 at 11:40

$$A:=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\;\;\text{isn't diagonalizable over the reals, but}\;\;A^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$$

obviously is.

$$A = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$$ is not diagonalizable (it has 1 eigenvalue and is not diagonal) but $A^2 = 0$ is.

• This example I also find, but not sure if the zero matrix is considered to be diagonal or not... – Eric_ Nov 25 '14 at 11:46
• Yes it is, @Eric_...by definition, simply. – Timbuc Nov 25 '14 at 11:58
• @Eric_ if you are not convinced with the first one, add ones on the diagonal ;) – mookid Nov 25 '14 at 12:27