Assuming that $A$ is a diagonalizable matrix, does the singular value decomposition of $A$ coincide with its spectral decomposition?
I think no, because in the spectral decomposition $A = Q^{-1} \Lambda Q$, $Q$ doesn't have to be unitary but the SVD of a matrix $A = UDV^t$ gives unitary matrices $U$ and $V$. However, this could be remedied by diving the columns of $Q$ by their lengths. Right? So, now I think that up to swapping columns and multiplying them by scalars, the SVD and the spectral decomposition of a matrix are the same. Is it correct?