There are matrices for which the spectrum of \(\mat{A}\) and \(\mat{A}\mat{A}^T\) are identical. As an example, take
\[
\mat{A} = \begin{pmatrix}
\frac{1}{2} & 0 & \frac{1}{2} & 0 \\
0 & \frac{1}{2} & 0 & \frac{1}{2} \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}.
\]
Freaky, right? I wouldn’t have believed it. I know one class of matrices for which this is true, but it’s a bit funky: if \(\mat{C}\) and \(\mat{B}\) are rectangular matrices related in a nice way (essentially, \(\mat{B}\) is \(\mat{C}\) plus a perturbation that is orthogonal to the range space of \(\mat{C}\)), then \(\mat{A} = \mat{C} \mat{B}^\dagger\) has this property. Here \(\dagger\) denotes pseudoinversion. The proof of this special case is complicated, and not particularly intuitive. So I wonder if there’s a simpler proof for the general case.