Suppose $A,B\in \mathbb{R}^{m\times n}$, prove that ${\rm tr}(\sqrt{\sqrt{AA^T}BB^T\sqrt{AA^T}})={\rm tr}(\sqrt{B^TAA^TB})$.
The form of LHS comes from fidelity in quantum information (see this Prove $\text{tr}(\sqrt{\sqrt A B \sqrt A})\leq 1$,where both $A$ and $B$ are positive semidefinite, and $\text{tr}(A)=\text{tr}(B)=1$ for details). I try to use the same trick (polar decomposition) in the above link, that is there exists unitary $U$ such that $U\sqrt{\sqrt{AA^T}BB^T\sqrt{AA^T}}=\sqrt{AA^T}B$ but no further ideas.
Best Answer
Let $|A|$ denote $\sqrt{AA^T}$. Then this statement can be written as $$ {\rm tr}(\sqrt{|A|BB^T|A|})={\rm tr}(\sqrt{B^TAA^TB}) $$ Now, we can use polar decomposition to note that $A = |A| U$ for some orthogonal matrix $U$. It follows that the left-hand-side can be written as $$ {\rm tr}(\sqrt{B^T(|A|U)(|A|U)^TB}) = {\rm tr}(\sqrt{B^T|A|UU^T|A|B}) = {\rm tr}(\sqrt{B^T|A||A|B}) $$
Now, writing $P = B^T|A|$, the left-hand side can br written as ${\rm tr}(\sqrt{P^TP})$ and the altered right-hand side can be written as $\operatorname{tr}(\sqrt{PP^T})$. Both sums are simply equal to the sum of the singular values of $P$, so the two sides are equal.