For arbitary dimensions, showing that the square root of the identity matrix is diagonalizable For arbitrary size, let $A$ be a square matrix such that $A^2=I$ where $I$ is the identity matrix. Then how do I show that $A$ is diagonalizable? Also, is $A$ diagonalizable when $A^2=B$ for some arbitrary diagonalizable matrix $B$?

The polynomial $x^2-1$ annihilates $A$, so its minimal polynomial must be one of $x-1$, $x+1$ or $x^2-1=(x-1)(x+1)$. All of these are products of distinct linear factors, therefore $A$ is diagonalizable.  
One the other hand, consider $B=-I_2$. For $A=\tiny{\begin{bmatrix}0&-1\\1&0\end{bmatrix}}$ we have $A^2=B$, but $A$ is not diagonalizable over the reals.
