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).
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Yes, if $n>m$ whichever one maps $n$-columns into $m$-columns must be degenerate. There are not enough linearly independent columns in the image, so some must be mapped to $0$. – Conifold May 15 '17 at 23:27
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I might be missing a bit of fundamental. Why do you say that we map $n$-columns into $m$-columns? – Morris C. May 15 '17 at 23:34
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Because matrices represent linear transformations, $m\times n$ matrix maps $n$-columns into $m$-columns, or $n$-vectors into $m$-vectors in different terminology. – Conifold May 15 '17 at 23:38
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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 $n-m$.
Robert Israel
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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.
dantopa
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