Let $A,B$ be two real positive semidefinite matrices of same size. Denote the trace operator by tr. We define the square root of a positive semidefinite matrix $A$ by the unique matrix $\sqrt A$ such that $A= \sqrt A ^\top \sqrt A$.
Why is $\text{tr}(\sqrt{\sqrt A B \sqrt A})= \text{tr}(\sqrt{A B}) $ true ?
Is $\text{tr}(\sqrt{\sqrt A B \sqrt A + \text{Id}})=\text{tr}(\sqrt{A B + \text{Id} }) $ also true, where Id is the identity matrix of same size?
Best Answer
$\sqrt{A} B \sqrt{A}$ and $AB$ are similar matrices, since $AB = \sqrt{A} (\sqrt{A} B \sqrt{A}) \sqrt{A}^{-1}$. So one square root of $AB$ (presumably the one you want) is $\sqrt{A} \sqrt{\sqrt{A} B \sqrt{A}} \sqrt{A}^{-1}$. By similarity it has the same eigenvalues, and thus the same trace, as $\sqrt{\sqrt{A} B \sqrt{A}}$.
In the same way, one square root of $AB + I$ is $ \sqrt{A} \sqrt{\sqrt{A} B \sqrt{A} + I} \sqrt{A}^{-1}$, and this has the same trace as $\sqrt{\sqrt{A} B \sqrt{A}+I}$.