Question: A positive (semi-) definite matrix has a unique positive (semi-) definite square root. What are its other square roots? In some cases there are infinitely many (such as for $aI$). Are there cases where there are finitely many?
In [this][1] question it is shown that the identity matrix has infinitely many square roots (although of course it has a unique positive square root). But it doesn't seem to address the general question of the square roots of a positive operator. The behavior of the class of matrices
$$\left( \begin{matrix} \sqrt{a} & t \\ 0 & -\sqrt{a} \end{matrix} \right)$$
as square roots of $aI$ seems to be due to the "merging" of the $-\sqrt{a}$ and $\sqrt{a}$ eigenspaces when you square. This might suggest that for a positive matrix with distinct eigenvalues, there are only $2^r$ square roots, with $r$ the rank of the matrix?
Ideas: One thing that might narrow it down: if a matrix is normal, will all its square roots be normal, or not necessarily? If we know this, we can go straight to working with eigenvalues.
[1]: In a finite dimensional vector space a positive operator can have infinitely many square roots [2]: https://math.stackexchange.com/questions/558072/how-many-square-roots-not-necessarily-positive-does-a-positive-matrix-have