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Let

$$A=\begin{pmatrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{pmatrix}.$$

I'm trying to find the eigenvalues of $A$, but when I calculate the characteristic polynomial, I get $$p(\lambda)=-\lambda^3+6\lambda^2+15\lambda+2,$$ and I don't know how to solve $p(\lambda)=0$. I'd appreciate any help. Thanks in advance.

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You made an error in your CP, it should be:

$$p(\lambda)=-\lambda^3+6\lambda^2+15\lambda+8 = -(\lambda-8)(\lambda+1)^2 = 0$$

So you have the two eigenvalues $\lambda_1 = 8$ (single root) and $\lambda_{2,3} = -1$ (double root).

Can you continue with the eigenvectors now?

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  • $\begingroup$ Nice catch (the error)! +1 $\endgroup$ – Namaste Oct 5 '13 at 17:44
  • $\begingroup$ I'm glad to hear you're well enough to go to the gym and hike! $\endgroup$ – Namaste Oct 5 '13 at 17:47
  • $\begingroup$ Just take it easy on the hike. Dr. Amy's orders! $\endgroup$ – Namaste Oct 5 '13 at 17:50
  • $\begingroup$ @amWhy: okay Doc! :-) $\endgroup$ – Amzoti Oct 5 '13 at 17:51

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