Definition of Spectral Radius / Eigenvalues of Product of a Matrix and its Complex Conjugate Transpose For any matrix $\textbf{A}$, I know that in the Euclidean L2 induced/operator norm, $\|\textbf{A}\|=\sqrt{\rho(\textbf{A}^{*}\textbf{A})}$, with * being the complex conjugate transpose and $\rho$ being the spectral radius. I was told that $\sqrt{\rho(\textbf{A}^{*}\textbf{A})}=\sqrt{\rho(\textbf{A}\textbf{A}^{*})}$ because $\textbf{A}\textbf{A}^{*}$ and $\textbf{A}^{*}\textbf{A}$ have the same eigenvalues. Why is this true? I can't seem to prove this. Any help would be appreciated. Thank you!
 A: There may be a difference between $A^{\star}A$ and $AA^{\star}$ when it comes to eigenvalue $0$, but that won't affect the spectral radius. For $\lambda \ne 0$, assume that $A^{\star}Ax=\lambda x$ for some $x \ne 0$. Now apply $A$ to both sides:
$$
         AA^{\star}[Ax] = \lambda [Ax].
$$
So $\lambda$ is an eigenvalue because $Ax \ne 0$; indeed, if $Ax$ were $0$, then $A^{\star}Ax=\lambda x$ would be $0$ and, hence, $x=0$ would hold because $\lambda \ne 0$, which is a contradiction. This works the other way around, too. So the non-zero eigenvalues of $AA^{\star}$ and $A^{\star}A$ are the same.
A: For $\lambda \ne 0$, suppose $\lambda I - AB$ is invertible.  Call its inverse $U$.  Then $V = \lambda^{-1} I + \lambda^{-2} BUA$ is an inverse for $\lambda I - BA$:
$$ (\lambda^{-1} I + \lambda^{-2} BUA)(\lambda I - BA) \\= I - \lambda^{-1} BA + \lambda^{-1} BUA - \lambda^{-2} BUABA \\= I + \lambda^{-1} BA - \lambda^{-1} B U(\lambda I - AB) A \\ = I + \lambda^{-1} BA - \lambda^{-1} BIA \\ = I,$$
and similarly for $(\lambda I - BA)(\lambda^{-1} I + \lambda^{-2} BUA)$.
Hence the non-zero eigenvalues of $AB$ are the same as the non-zero eigenvalues of $BA$.  Hence $\rho(AB) = \rho(BA)$.
This trick (due, I think, to von Neumann) also works for infinite dimensional operators.
