Does $A^{\star} A$ and $A A^{\star}$ always have same eigenvalues except one has zero as an eigenvalue? I wonder this because for a $m \times n$ matrix $A$, we have that $A^{\star}A$ and $A A^{\star}$ have the same $\textbf{non-zero}$ eigenvalues. Would this mean that one of them (the one with a higher dimension) ALWAYS has zero as an eigenvalue (assuming $m \neq n$ of course).
 A: More generally, $\text{rank}(AB) \le \min(\text{rank}(A), \text{rank}(B)) $.  In particular, an $n \times n$ matrix that factors as the product of an $n \times m$ and an $m \times n$ matrix with $m < n$ will have rank at most $m$, and therefore it will have $0$ as an eigenvalue with geometric multiplicity at least $n-m$.
A: Yes, the product share the same number of nonzero eigenvalues.
Take a matrix $\mathbf{A}\in\mathbb{C}^{\color{blue}{m}\times \color{red}{n}}_{\rho}$. The product matrices $\mathbf{A}^{*}\mathbf{A}$ and $\mathbf{A}\mathbf{A}^{*}$ share the same $\rho$ nonzero eigenvalues.
For $\mathbf{A}^{*}\mathbf{A}$, the $0$ eigenvalues are (after ordering)
$$
 \underbrace{\lambda_{\rho+1}, \lambda_{\rho+2}, \dots, \lambda_{\color{red}{n}}}_{n-\rho} = 0
$$
For $\mathbf{A}\mathbf{A}^{*}$, the $0$ eigenvalues are (after ordering)
$$
 \underbrace{\lambda_{\rho+1}, \lambda_{\rho+2}, \dots, \lambda_{\color{blue}{m}}}_{m-\rho} = 0
$$
Why is this so? See the superb explanation by @Robert Israel.
