Eigenvalues of $A \overline{A}$

Question

Let $A \in \mathbb{C}^{n \times n}$, with $A_{ij} = a_{ij} \in \mathbb{C}$. I am mostly interested in the case $n =3$, but a general pattern is fine too. Define $$(\overline{A})_{ij} := \overline{a_{ij}},$$ i.e., $\overline{A}$ is the matrix obtained by taking the complex conjugate of each entry of $A$. What can we say about the eigenvalues of $$A \overline{A}?$$ It seems that this matrix always has a purely real eigenvalue. This might be related to the fact that $$\text{det}(A \overline{A}) = | \text{det}(A) |^2 \in \mathbb{R},$$ which equals the product of the eigenvalues. Specifically, when $A$ is unitary, matrix $A \overline{A}$ seems to always have one eigenvalue $1$. Can someone prove this?

Answer 1

As @user10354138 pointed out, in general the eigenvalues can be all non-real. Even in the unitary case, for $n$ even, this might happen: for instance, let $\theta \in (0, 2\pi)$ and $R_{\theta/2}\colon \mathbb C^2 \to \mathbb C^2$ the rotation matrix in the (real) plane extended to the complex numbers. Then $R_{\theta/2}\overline{R_{\theta/2}} = (R_{\theta/2})^2 = R_\theta$, that has no real eigenvalues. For the general even case, we may take the operator $\bigoplus\limits_{i=1}^k R_{\theta/2}\colon \mathbb{C}^{2k} \to \mathbb{C}^{2k}$ that has the same (non-real) eigenvalues.

However, we can get our hopes up in the odd case! I'll assume that you already know that every unitary operator is diagonalizable and that all the eigenvalues have norm $1$. Let $A\colon \mathbb C^n \to \mathbb C^n$ be unitary and $p_M$ the characteristic polynomial of a matrix $M$.

Set $U:= A\bar{A}$. So $U U^\dagger = A\bar{A}A^T A^{-1} =A(\overline{A A^\dagger}) A^\dagger = AIA^\dagger = A A^\dagger = I$; i.e. $U$ is also unitary. In this case, $$\begin{align}p_U &:= \det(tI - A\bar A) \\&= \det(tAA^\dagger- A\bar{A}) \\&= \det(tA^\dagger - \bar{A})\det(A) \\&= \det (tA^\dagger A - \bar{A}A) \\&= \det(\overline{tI - A\bar{A}}) \\&= \overline{\det(tI - U)} = \bar{p_U}\end{align}$$

i.e., $p_U$ is a polynomial with real coefficients. In particular, when $n$ is odd, this implies that $p_U$ has a real root $\lambda$. Since $U$ is unitary, $\lambda = \pm 1$.

For the sake of contradiction, suppose that $1$ is not an eigenvalue. So $\lambda = -1$ is the only real eigenvalue. Thus $\dim E_{-1}$ is odd; otherwise, $\dim E_{-1}^\perp = n - \dim E_{-1}$ would be odd (thus $\ne 0$) and, therefore, the same argument can be applied to $U|_{E_{-1}^\perp}$ that yields a real eigenvalue $\ne -1$, which can be only $1$ - a contradiction. Since $\dim E_{-1}$ is odd and complex roots come in conjugate pairs (whose products $= 1$), it follows that $\det U = (-1)^{\dim E_{-1}} = -1 = |\det(A)|^2 > 0$, a contradiction! Therefore $1$ is always an eigenvalue, as we wanted to prove.

Answer 2

Let $$ A=\begin{pmatrix}1&i\\1&1\end{pmatrix}. $$ Then $$ A\bar{A}=\begin{pmatrix}1+i&0\\2&1-i\end{pmatrix} $$ whose eigenvalues $1\pm i$ are obviously nonreal.

What you can say is that the nonreal eigenvalues of $A\bar{A}$ comes in complex conjugate pairs, because the characteristic polynomials of $AB$ is the same as the char poly of $BA$.