Question: Let \(A\in {{\mathbb{R}}^{n\times m}}\), \(B\in {{\mathbb{R}}^{m\times n}}\) and let \(m\ge n\). Observe that \(AB\) has n eigenvalues, while \(BA\) has m eigenvalues. Prove that m of the eigenvalues of \(AB\) are precisely those of \(BA\), while the remaining \(n-m\) eigenvalues of \(AB\) are zero.

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