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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1Is there something wrong with calculating $\text{det}(A-\lambda_i I)$ for each eigenvalue? – JimmyK4542 Aug 12 '14 at 02:28
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Subtract from the diagonal and check if the new matrix is singular? – Aug 12 '14 at 02:29
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@JimmyK4542 seems like that would take a long time, and the question only asks to verify – Youbloodywombat Aug 12 '14 at 02:30
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@900sit-upsaday I'll give that a shot – Youbloodywombat Aug 12 '14 at 02:31
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Ah ok, that makes sense. Cheers guys – Youbloodywombat Aug 12 '14 at 02:32
2 Answers
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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.
JimmyK4542
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they must make this determinant zero : $$\left | A-\lambda I \right |=\begin{vmatrix} 7- \lambda& 1&1 \\ 3& 1 -\lambda& 2\\ 1 &3 & 2-\lambda \end{vmatrix}$$
Shabbeh
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