I checked this statement numerically even if I cannot find it anywhere and didn't manage to prove it: $$\text{Tr}(\sqrt{B^TB})=\text{Tr}(\sqrt{BB^T}) \qquad \forall B\in M_{n,r}(\mathbb{R}).$$ The result is well-known without the square root, but here we have one and the matrix $B$ is not necessarily square. Does someone have any idea?
2026-03-26 09:16:41.1774516601
Trace of the square root inside commutation property
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