Commutativity of matrix square root

Question

Let $A, B \in \mathbb{R}^{n \times n}$ and let us assume that $A^{1/2}$ exists. I have often seen people write something like $$ AB = A^{1/2}\, B\; A^{1/2} $$ when both $A$ and $B$ are symmetric, in order to preserve the symmetry of the problem. Nevertheless, I do not really understand how that can be proved and whether it is true in general or just for symmetric matrices. Any clue would be really appreciated.

Answer 1

Note that it is not true (in general) that $AB=A^{1/2}BA^{1/2}$ unless $A^{1/2}$ and $B$ commute.

However, if $A$ is symmetric positive definite (SPD) and thus has a unique SPD square root $A^{1/2}$, we have that $AB$ is similar to $A^{-1/2}(AB)A^{1/2}=A^{1/2}BA^{1/2}$ (this is true also if $A$ is only semidefinite (SPSD), but it requires some more effort to prove since $A^{-1/2}$ is not invertible).

There are many areas where such a transformation is useful. You can use it, e.g., to show that if $A$ is SPSD and $B$ is symmetric, then $AB$ (although generally nonsymmetric) has real eigenvalues (nonnegative if $B$ is SPSD too) and is diagonalizable.

If $A$ is nonsymmetric and has an invertible square root, the same similarity transformation is possible. I'm not sure though it is very useful in this case.