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How would you check whether eigenvalues $\lambda_1=8$, $\lambda_2=3$, $\lambda_3=-1$ belong to a matrix? $$ \begin{matrix} 7 & 1 & 1\\ 3 & 1 & 2 \\ 1 & 3 & 2 \\ \end{matrix} $$

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    $\begingroup$ Is there something wrong with calculating $\text{det}(A-\lambda_i I)$ for each eigenvalue? $\endgroup$ – JimmyK4542 Aug 12 '14 at 2:28
  • $\begingroup$ Subtract from the diagonal and check if the new matrix is singular? $\endgroup$ – user147263 Aug 12 '14 at 2:29
  • $\begingroup$ @JimmyK4542 seems like that would take a long time, and the question only asks to verify $\endgroup$ – Youbloodywombat Aug 12 '14 at 2:30
  • $\begingroup$ @900sit-upsaday I'll give that a shot $\endgroup$ – Youbloodywombat Aug 12 '14 at 2:31
  • $\begingroup$ Ah ok, that makes sense. Cheers guys $\endgroup$ – Youbloodywombat Aug 12 '14 at 2:32
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One method would be to simply check that $\text{det}(A-\lambda_i I) = 0$ for each of the three eigenvalues.

Another method would be to check the following conditions:

$\text{tr}(A) = \lambda_1+\lambda_2+\lambda_3$

$\text{det}(A) = \lambda_1\lambda_2\lambda_3$

$\text{tr}(A^2) = \lambda_1^2+\lambda_2^2+\lambda_3^2$

If you know the values of $\lambda_1+\lambda_2+\lambda_3$, and $\lambda_1\lambda_2\lambda_3$, and $\lambda_1^2+\lambda_2^2+\lambda_3^2$, then this uniquely determines the values of $\lambda_1$, $\lambda_2$, $\lambda_3$ up to a permutation.

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they must make this determinant zero : $$\left | A-\lambda I \right |=\begin{vmatrix} 7- \lambda& 1&1 \\ 3& 1 -\lambda& 2\\ 1 &3 & 2-\lambda \end{vmatrix}$$

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