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

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.
