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I can compute the eigenvalues of the product of these matrices $X_1 X_2$ by multiplying out the matrices, but I'm certain I am missing a trick that will make this computation less tedious. Any hints?

$X_1$: $$ \begin{matrix} 1 & 2 \\ 0 & 2 \\ 1 & 0 \\ 2 & 0 \\ \end{matrix} $$

$X_2$:

$$ \begin{matrix} 0 & 2 & 3 & 1\\ 2 & 1 & 3 & -1 \\ \end{matrix} $$

The product of these matrices being:

\begin{matrix} 4 & 6 & 8 & -1\\ 4 & 6 & 12 & 0 \\ 0 & 2 & 3 & 1 \\ 0 & 4 & 6 & 2 \\ \end{matrix}

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  • $\begingroup$ "I'm certain I am missing a trick that will make this computation less tedious" Why? $\endgroup$ – user587192 Dec 16 '18 at 1:22
  • $\begingroup$ Mainly because this was a question on a previous exam - I doubt the instructor would expect us to find the eigenvalues for a 4x4 matrix. $\endgroup$ – user3424575 Dec 16 '18 at 1:25
  • $\begingroup$ What would a rank of the resulting matrix be? $\endgroup$ – user58697 Dec 16 '18 at 1:36
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    $\begingroup$ The first two rows of the product matrix are incorrect. I'm not sure if this helps at all with the question though. $\endgroup$ – Pseudo Professor Dec 16 '18 at 1:43
  • $\begingroup$ The only thing special I can see is that row 4 is twice row 3, so the matrix is not invertible, making $\lambda = 0$ one of the eigenvalues. Also, if you calculate $\det(A-\lambda I) = 0$, using the zeros in the first column to your advantage should make the determinant a lot less tedious. $\endgroup$ – Matthias Dec 16 '18 at 1:45
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The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.

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