Given a matrix $G$, let its singular value decomposition be
$$G = Y \Sigma U^H$$
In the Hermitian matrix $G^H = U \Sigma Y^H$, why doesn't $\Sigma ^H$ appear instead of $\Sigma$?
If $G$ is a complex matrix, its singular values should be complex. Or am I wrong?
Singular values are always real and nonnegative even when the entries are complex. In fact if you already know that they are the square roots of the eigenvalues of $A^H A$, then the result follows from the simple observation $\| Ax \|^2 \geq 0$.
