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I need to find the eigenvalues/eigenvectors of the following matrix:

$A= \begin{bmatrix}-26 & 20\\-30 & 24\end{bmatrix}$

I first used the equation for eigenvectors, and sorted out the determinant:

$P_A(x) = \det (xI - A)$ $= \det\begin{bmatrix}x & 0\\0 & x\end{bmatrix} - \begin{bmatrix}-26 & 20\\-30 & 24\end{bmatrix}$ $ = \det\begin{bmatrix}x + 20 & -20\\ 30 & x - 24\end{bmatrix}$ $= (x + 20)(x - 24) - (-600)$ $= x^2 - 4x + 120$

The problem at this point, is that the trinomial $ x^2 - 4x + 120$ can't be factored into $(x + a)(x + b)$. At least I don't think. The computer on Wolfram hasn't been able to do it, so I don't think it's factorable, which would mean I can't find the two x values that would give me the eigenvalues.

So how can I get the x values from a trinomial that isn't factoable? How else could I find the matrix's eigenvalues?

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    It should be \det\begin{bmatrix}x + 26 & -20\\ 30 & x - 24\end{bmatrix}2012-11-13

2 Answers 2

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You didn't compute the polynomial correctly, it should be $x^2+2x-24 = (x+6)(x-4)$.

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Completing the square, we see that $x^2-4x+120=x^2-4x+4+116=(x-2)^2+116,$ so $x^2-4x+4=0$ if and only if $(x-2)^2=-116$ if and only if $x-2=\pm i\sqrt{116}=\pm 2i\sqrt{29}$ if and only if $x=2\pm 2i\sqrt{29}.$ Thus, we can factor it as $\left(x-2-2i\sqrt{29}\right)\left(x-2+2i\sqrt{29}\right).$ However, you've miscalculated the characteristic polynomial. It is indeed factorable.