# Finding the eigenvalues - missing a trick?

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:

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



• "I'm certain I am missing a trick that will make this computation less tedious" Why? – user587192 Dec 16 '18 at 1:22
• 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. – user3424575 Dec 16 '18 at 1:25
• What would a rank of the resulting matrix be? – user58697 Dec 16 '18 at 1:36
• The first two rows of the product matrix are incorrect. I'm not sure if this helps at all with the question though. – Pseudo Professor Dec 16 '18 at 1:43
• 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. – Matthias Dec 16 '18 at 1:45

The non-zero eigenvalues of $$X_1 X_2$$ are the same as the non-zero eigenvalues of $$X_2 X_1$$.