If A and B are unitarily equivalent, then they have the same singular values using uniqueness?

Question

I'm trying to prove the following statement: if A and B are unitarily equivalent, then they have the same singular values

so my proof goes like this: unitarily equivalent means $A=QBQ^*$ so if $A=U_A\Sigma_AV_A$ and $B=U_B\Sigma_BV_B$ then $A=U_A\Sigma_AV_A=Q(U_B\Sigma_BV_B)Q^*$ . Multiplying on both sides, we can get $I\Sigma_AI^* = (U_A^*QU_B)\Sigma_A(V_B^*Q^*V_A)$.

From here I am stuck. I want to say that I can use uniqueness of the SVD to help me here but I am not sure how. How do I use the uniqueness of the SVD (and to what extent is it unique?) to help complete this proof?

Answer 1

Suppose $B=U\Sigma V^*$ is a SVD for $B$, with $U$ and $V$ unitary and $\Sigma$ diagonal with decreasing values on the diagonal. If $A=QBQ^*$ with $Q$ unitary, then $$ A=(QU)\Sigma (V^*Q^*) $$ The product of unitary matrices is unitary. Thus this is a SVD for $A$.

The uniqueness for the SVD is only related to the matrix $\Sigma$ (in the canonical diagonal form with decreasing values on the diagonal).

Anyway, this is overkill: the singular values of $A$ are the positive square roots of the nonzero eigenvalues of $A^*A$. If $A$ and $B$ are unitarily similar, then also $A^*A$ and $B^*B$ are.