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I know the general process for finding eigenvalues, but I'm stuck on this particular matrix. The matrix is:
$\begin{bmatrix}-13 & -8 &-4 \\ 12 & 7&4 \\ 24 &16 &7\end{bmatrix}$

When I try to compute the $\text{det}(A-\lambda I)$, I get an intractable algebra problem for the characteristic equation. What's the secret here?

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    $\begingroup$ Welcome to the Math Stack Exchange. What did you get for the characteristic equation? Did you try the rational zeros theorem or an online facility such as MathWay? $\endgroup$ – Paul Sundheim Oct 16 '15 at 16:51
  • $\begingroup$ Use proper LaTeX please. $\endgroup$ – Kushal Bhuyan Oct 16 '15 at 16:54
  • $\begingroup$ The secret is to do the algebra to obtain the char equation real SLOW.....because mistakes are made easily. $\endgroup$ – imranfat Oct 16 '15 at 17:09
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Why is it intractable? I get as the determinant $3 + 5 \lambda + \lambda^2 - \lambda^3$. There is a cubic formula for solving this equation:

http://mathworld.wolfram.com/CubicFormula.html

it yields: $3 + 5 \lambda + \lambda^2 - \lambda^3 = -(\lambda - 3) (\lambda + 1)^2 = 0$, so $\lambda_1=-1$ and $\lambda_2=3$. The first eigenvalue obviously has algebraic multiplicity two. So this matrix has a form $\begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3\end{pmatrix}$.

EDIT: It turns out that the geometric multiplicity of $\lambda = -1$ is two as well. So the matrix is diagonalizable.

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  • $\begingroup$ Hahaha, thanks for the welcoming words :)) We are a pretty big family :D $\endgroup$ – Mischa Oct 16 '15 at 17:11
  • $\begingroup$ I got it now. My algebra was sloppy. The hint to add row 2 to row 1 made for an easier factorization of the characteristic equation though. Thanks all. $\endgroup$ – Tim Raymond Oct 16 '15 at 22:22
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Hint: To calculate the determinant

$\begin{vmatrix} -13 - \lambda & -8 & -4\\ 12 & 7 -\lambda & 4\\ 24 & 16 & 7-\lambda \end{vmatrix}$

do the elementary row operation that replaces row 1 by row 1 plus row 2, and note that this does not change the determinant.

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