Are there sometimes only finitely many square roots of a positive matrix?

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] 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.

• If the matrix has size $1\times 1$, then it has exactly two square roots. – Henning Makholm Nov 11 '13 at 15:37
• Good point! If the eigenvalues are distinct, can we say they are finite? – user88203 Nov 11 '13 at 15:39

Claim: Let $A$ be an $n\times n$ positive semi-definite matrix with eigenvalues $a_1\ge \dots\ge a_n$.
(i) If $a_i=a_j$ for some $i\ne j$, then $A$ has uncountably many square roots.
(ii) If $a_1>\cdots>a_n>0$, then $A$ has $2^n$ square roots.
(iii) If $a_1>\cdots>a_n=0$, then $A$ has $2^{n-1}$ square roots.
Sketch of proof: Case (i) has been shown in the body of your question, so let us assume that $a_1>\cdots>a_n$. Since $A$ is unitarily diagonalizable, without loss of generality, we may assume that $A={\rm diag}\{a_1,\dots,a_n\}$. Let $Q$ be some square root of $A$. From $AQ=QA$ and $a_1>\cdots>a_n$ we can see that $Q$ must be diagonal. Then cases (ii) and (iii) follow.