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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$\begingroup$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.
$\endgroup$ $\begingroup$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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