Let $A \in \mathbb{C}^{n \times n}$ and suppose we have $A = PU = VQ$ are left and right polar factorisations, respectively. That is, $P^*=P$, $Q^*=Q$, and $U^*U = V^*V= I$. I want to prove the following:
- If A is of full rank, then $U=V$.
- $P=Q$ if and only if $A$ is normal.
I am a little confused about how to go about the first problem. I do know that SVD and polar factorisation are closely related, so I think that has some potential, at least for question 1. For 2 I think I have the work done for 2. Consider $$A^*A = QV^*VQ = Q^2 = P^2 = PUU^*P = AA^*.$$ This is one direction. For the other direction if $A^*A = AA^*$, then we would have $$U^*P^2U = P^2 \\ VQ^2V^*=Q^2$$ by substituting $A = PU$ and $A = VQ$ in the first and second line respectively. So we have $A^*A = P^2=Q^2$, and since $A^*A$ is semi-positive it has aa unique square root and so $P=Q$. Is this correct?
I am still unsure about 1, so any help is welcomed.
Krull.
If $A$ is of full rank, then $A=USV^*.$ $A=(UV^*)(VSV^*)=(USU^*)(UV^*)$ is the two polar factorization. As you see, $UV^*=UV^*$ is what you want in (1).
By the way, you can see this method also works when $A$ is not of full rank. When doing SVD, require $U,S,V$ to be $n\times n.$ (If $r(A)=k,$ don't let $S$ be a $k\times k$ matrix). You can do this by Schmidt orthogonalization. So this means that you can ask $U=V$ when doing polar factorizations. But polar factorizations are not unique when $r(A)<n,$ so you don't have $U=V$ for arbitrary polar factorizations.
What's more, if you already have a $A=PU,$ then $A=VDV^*U=U(U^*VD(U^*V)^*).$ So when you find a polar factorization of one side, you can find the version of the other side, requring $U=V$ in $A=PU=VQ.$ When $r(A)=n,$ the factorization is unique, so $U=V$ always happens.