This is from a practice exam for my quals.
Let $A$ be an $n \times n$ complex matrix.
Suppose $A$ satisfies the following property:
$(AA^\dagger)^2 = (A^\dagger A)^2$
Prove that $A$ is normal, that is, that $AA^\dagger = A^\dagger A$.
My attempt:
Recall that a matrix is normal if and only if it can be diagonalized by a unitary matrix. I will attempt to show this is the case.
$A$ has a singular value decomposition $A = V \Sigma U^\dagger$, where $V$ and $U$ are unitary and $\Sigma$ is diagonal with non-negative real diagonal entries. This factorization is unique up to permutation of diagonal elements of $\Sigma$, since $A$ is square.
$A = V \Sigma U^\dagger, A^\dagger = U \Sigma^\dagger V^\dagger$
and since $A A^\dagger$ and its conjugate transpose are normal,
$AA^\dagger = V \Sigma U^\dagger U \Sigma^\dagger V^\dagger = V \Sigma \Sigma^\dagger V^\dagger$
and similarly for $A^\dagger A$.
I tried to use all of these assumptions to show that $U = V$, but wasn't able to reach a contradiction by assuming otherwise.
Best Answer
There is a simple proof that uses uniqueness of the square root.
Let $R$ and $T$ be two self-adjoint nonnegative operators (matrices). In this case if $R^2 = T^2$ then $R = T$ (that's the uniqueness of square root). Here is some more information about it Proof that Every Positive Operator on V has a Unique Positive Square Root; Show that the square root of a non-negative operator is unique.
Now observe that $A^\dagger A$ and $A A^\dagger$ are both self-adjoint and nonnegative. Therefore if $(A^\dagger A)^2 = (A A^\dagger)^2$ then $A^\dagger A = A A^\dagger$ and hence $A$ is normal.