Solution
= Solution
For $B=A^{-1}A^\dagger$,
$$
B^\dagger B=A(A^\dagger)^{-1}A^{-1}A^\dagger.
$$
Multiplying $B^\dagger B=I$ on the left by $A^{-1}$ and on the right by $(A^\dagger)^{-1}$ shows that it is equivalent to $(A^\dagger)^{-1}A^{-1}=A^{-1}(A^\dagger)^{-1}$. Taking inverses gives $A^\dagger A=AA^\dagger$. Thus $B$ is unitary exactly when $A$ is a <normal matrix>.