Question: when is the eigenvalue decomposition of a matrix equal to its singular value decomposition?
Answer: when A is hermitic and has positive eigenvalues.
I don't really understand. If we have A=A* and $A=U\Sigma V^*$ (where * stands for hermitic conjugate) then we have U=V by the uniqueness of SVD. But is that true? In the SVD, U and V are unique up to a constant with modules 1, right? So why is this true?
Suppose we have that, then $A=U\Sigma U^{-1}$. ($U^*=U^{-1}$ because it is unitair matrix). And suppose the eigenvalue decomposition is $TDT^{-1}$, then we have T=U and D=$\Sigma$, again by uniqueness?
But why did we even need that U=V? I thought that if the eigenvalues are positive (and real?) then $TDT^{-1}$ is a valid SVD so it is the SVD by uniqueness?
And we also dont know the eigenvalues in D are in descending order, right? So thats a problem too?