Show that for any invertible $A$, if $P\equiv\sqrt{A^\dagger A}$ then $U\equiv A P^{-1}$ is unitary

Question

Please bare in mind, that I am self-teaching this area at the moment and so additional explanation where possible would be greatly appreciated.

Let $A:V \to V$ be an arbitrary invertible operator. Show that there exists a complex unitary operator U and a hermitian positive definite operator P such that $A =UP$.

I have tried:

Let $P$ be a hermitian positive definite operator such that $P^2 =A^\dagger A$. Then I let $U=AP^{-1}$.

I am having trouble with showing that $U$ is unitary.

Answer 1

This is the so-called polar decomposition of the matrix $A$. If $U$ is defined as in

$$U=AP^{-1}=A(A^\dagger A)^{-1/2},$$

then you have $$U^\dagger U=\underbrace{(A^\dagger A)^{-1/2}A^\dagger}_{U^\dagger} \underbrace{A (A^\dagger A)^{-1/2}}_U =(A^\dagger A)^{-1/2}(A^\dagger A)(A^\dagger A)^{-1/2}=I.$$ Moreover, $$ U U^\dagger=\underbrace{A (A^\dagger A)^{-1/2}}_U \underbrace{(A^\dagger A)^{-1/2}A^\dagger}_{U^\dagger} =A(A^\dagger A)^{-1} A^\dagger\\ = A(A^{-1}(A^\dagger)^{-1})A^\dagger= I. $$ Thus $U$ is unitary.

Answer 2

HINT: Write down $U^*U = (AP^{-1})^*(AP^{-1})$ and use what you know.

Answer 3

As $P=P^*$,$P^{-1}=(P^*)^{-1}$ and $U^*U=P^{-1}A^*A P^{-1}=P^{-1}P^2 P^{-1}=Id$, yherefore $U$is unitary