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I'm given a matrix below and been asked to find its eigenvalues and also its eigenvectors.

$$ \left( \begin{matrix} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0\\ \end{matrix} \right) $$

I found the 2 eigenvalues which are $\lambda_1 = 2$ and $\lambda_2 = -1$.

With that I also found the eigenvectors.

Case 1 when $\lambda_1 = 2$, $$ \left( \begin{matrix} 1\\ 1\\ 1\\ \end{matrix} \right) $$

Case 2 when $\lambda_2 = -1$, $$ \left( \begin{matrix} -1\\ 0\\ 1\\ \end{matrix} \right) $$

and also, I found a second eigenvector for $\lambda_2 = -1$, $$ \left( \begin{matrix} -1\\ 1\\ 0\\ \end{matrix} \right) $$

I'm particularly concerned for the $\lambda_2 = -1$, I'm very unsure about whether they are the definitive answers. Can you help me verify that they really are the answer?

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    $\begingroup$ Your work is correct. (To werify you can use WolframAlpha. Here wolframalpha.com/input/…) $\endgroup$ Mar 17 '15 at 20:06
  • $\begingroup$ Appreciate the aid @EmilioNovati $\endgroup$
    – apple
    Mar 17 '15 at 20:09
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You're correct, but I'll add a new method to your arsenal.

This time take the trace, add up the diagonal elements in the matrix starting from the top left to bottom right. You'll get zero. You only listed two Eigenvalues, which are correct, but there should be three. The cool thing about the trace is that its equal to the sum of the eigenvalues. So I don't need to do any more work because $0=2-1+x$ with x=-1 so the last Eigenvalue is also -1. Also the determinant equals the product of the eigenvalues, so as an extra bonus you know the determinant is 2.

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  • $\begingroup$ Are you saying that $\lambda_3 =-1$ as well? @zach466920 $\endgroup$
    – apple
    Mar 17 '15 at 20:24
  • $\begingroup$ Yes, by direct calculation, with determinants, and with the indirect trace method. $\endgroup$
    – Zach466920
    Mar 17 '15 at 20:30

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