Suppose I have a Hermitian, positive semidefinite matrix $A$ such that
$$A = U \Sigma U^H$$
The trace is given by
$$\operatorname{tr}(A)=\operatorname{tr}(U \Sigma U^H)=\operatorname{tr}(U ^HU\Sigma)=\operatorname{tr}(\Sigma)=\sum\sigma_i$$
Am I correct to say that the trace of a Hermitian matrix is the sum of its singular values? I have never seen this property written down anywhere though. It seems like there is something I did wrong but didn't notice here.
Best Answer
In general, for a hermitian matrix $A$, a SVD is not the same as a unitary diagonalisation.
As a concrete example, take the $1\times 1$ matrix $A = [-1]$. Then $U = [1]$ gives a diagonalisation, but the singular value of $A$ is $1$, meaning the SVD is $$ A = [-1]\cdot [1]\cdot [1]^H $$ The singular values of a Hermitian matrix $A$ are the absolute values of the eigenvalues of $A$. Thus the trace of $A$ and the trace of $\Sigma$, while related, are not necessarily the same. They are the same in the case where $A$ is positive semidefinite.