Given the SVD of $A$, what's the SVD of $A^*$?

Question

Given the SVD (ie. the $U, \Sigma$, and $V^*$ of $A$), how can we determine the SVD of $A^*$? Do we have to recompute the SVD for $A^*$ or is there a way we can derive $A^*$ from $A$?

Answer 1

Assuming that $A^*$ is the conjugate transpose of $A$, then $A=U\Sigma V^*$ means that $A^* = V\Sigma^*U^*$. $\Sigma$ is real, though, so $\Sigma^*=\Sigma$ (assuming you're using the form of the SVD where $\Sigma$ is square; if not, that's only true for the diagonal part). This means that the SVD of $A^*$ is just $(V,\Sigma,U^*)$.

Answer 2

If $A=U\Sigma V^*$, then $A^*=V\Sigma ^* U^*=V\Sigma ^T U^*$.

We still need to guarantee that this a SVD decomposition for $A^*$. Obviously $V$ and $U^*$ are unitary matrices. Now for $\Sigma ^T$, does $A^*$ have the same singular values as $A$? Yes, because $AA^*$ and $A^*A$ have the same non-negative eigenvalues.