"Square-normal" matrices are normal

Question

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.

Answer 1

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.

Answer 2

Edit: I now see the mistake in your attempt: You are trying to prove that $U = V$, which would imply $A = A^{\dagger}$ i.e. $A$ being hermitian, which is stronger than normality. If $A$ is unitarily similar to a diagonal matrix (which is equivalent to being normal as you mentioned), the diagonal matrix could also have negative entries, but you eliminate that possibility by trying to show that the two unitary matrices in the SVD of $A$ are the same.

This post (failed answer) now exactly shows how far this approach gets you and where it breaks down.

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As you point out we have (I write $A^H$ instead of $A^{\dagger}$) $$ A A^{H} = V \Sigma \Sigma^H V^H = V \Sigma^2 V^H \quad \text{and} \quad A^H A = U \Sigma^2 U^H, $$ as $\Sigma$ is real valued and diagonal. Thus, your equality reads $$ (A^H A)^2 = U \Sigma^4 U^H = V \Sigma^4 V^H = (A A^H)^2 \implies \Sigma^4 = U^{-1} V \Sigma^4 V^{H} U^{-H} $$ as $U$ and $V$ are unitary. With $B := U^{-1} V$ (which is unitary) we have $$B \Sigma^4 = \Sigma^4 B.$$ Since $\Sigma^4$ is diagonal we have $a_{i,j} = 0 \iff \sigma_{i} = \sigma_j$, where $\Sigma^4 = (\sigma_k)_{k}$. (proposition 1.12 here). If all $\sigma_k$'s are different, this implies that $B$ is diagonal and because its diagonal it must be the unit matrix, i.e $U^{-1} V = I$, implying $U = V$.

As far as I am concerned, there's no reason to assume that the $\sigma_k$'s are different, so we are at a dead end. Also notice that if $A$ were normal and $A^2 = I$ we have $A = A^{\dagger}$, which is what you would prove by $U = V$. Notice that $A^2 = U \Sigma^2 V^{\dagger} = I$ implies $B = U^{-1} V = \Sigma^2$, meaning that $B$ is diagonal.