Comparing SVD of orthogonal matrix $(\sqrt{W}A)$ and matrix $A$ in order to find the singular values of A

Question

I have a diagonal matrix W and another matrix A. I also have that $(\sqrt{W}A)$ is an orthogonal matrix.

I have that the SVD of $$\sqrt{W}A=U\Sigma V^T=(\sqrt{W}A)II$$ where I is the identity matrix. So, the singular values of $(\sqrt{W}A)$ are all $1$.

I am trying to figure out what the SVD of just $A$ would be so I can find the singular values of A. I tried to say that it is $$A=\sqrt{W}^{-1}\sqrt{W}AII=AII$$ but this isn't an SVD since A is not orthogonal.

I was also thinking that maybe it would be $$A=(\sqrt{W}A)\sqrt{W}^{-1}II$$ but I don't think that $\sqrt{W}^{-1}$ will commute in that way.

Any help would be much appreciated. Thank you.

Answer 1

Since $\sqrt{W}A$ is unitary, $$ I = (\sqrt{W}A)(\sqrt{W}A)^* = \sqrt{W} AA^* \sqrt{W}^*$$

First note that $\sqrt{W}$ has to be full rank since $\sqrt{W}A$ is full rank. Therefore $\sqrt{W}^{-1}$ exists and, assuming $W$ is real, $$ AA^* = \sqrt{W}^{-1}\sqrt{W}^{-1} = W^{-1} $$

Now think about how the singular values of $AA^*$ are related to $A$, and how $W^{-1}$ is related to $W$.