find the smallest interval in which the eigen value of the matrix lie

Question

$$ \begin{bmatrix} 3 & 2 & 2 \\ 2 & 5 & 2 \\ 2 & 2 & 3 \\ \end{bmatrix} $$

I was practicing questions on Matrices & Determinants and was going well with the topic until this question! shows up.

I have no idea on how to solve this question.

This problem is taking a lot of my mind and I am not able to go forward while this problem is still stuck in my mind.

Please Help!!

Thank You

Answer 1

Compute $\operatorname{det}(A-X\cdot Id)$ the characteristic polynomial of $A$. One should find

$$-X^3+11X^2-27X+17$$

The roots of this polynomial in ascending order are $1$, $5-2\sqrt{2}$ and $5+2\sqrt{2}$.

Answer 2

Since $A$ is symmetric, the eigenvalues are real. They are in the union of the intervals (why?) $[3-4,3+4],[5-4,5+4],[3-4,3+4]$, that is in the interval $[-1,9]$. If you want exactly the smallest interval, then, of course, you must calculate (all !) the eigenvalues and I do not see any interest to your question.