Determine upper and lower limit of eigenvalues Without computing the eigenvalues of a matrix $A$, I need to determine an upper and lower bound on them. I don't know $A$ but I know that
$$A^{-1}=\begin{bmatrix}
    2& -1& 0\\
    -1& 2& -1\\
    0 & -1& 1
   \end{bmatrix}.$$
I know that this matrices is positive definite, because it is symmetric and has positive pivots, so zero is a lower bound. But I do not know how to find an upper bound. Because we have that $\det(A)=1$ $\operatorname{tr}(A)=6$ and $\operatorname{tr}(\operatorname{adj}(A))=5$, I think it is $\lambda<5$. But I try with five and $\operatorname{tr}(\operatorname{adj}(A))$ little bit bigger then five, so I think there has to be a better way to do this.
 A: This question is not really precise, since the tightest bound of course is to just compute the eigenvalues. 
Here is one approach: the Gershgorin circle theorem will give you an upper bound of $4$ on the eigenvalues of $A$, and so a lower bound of $\frac{1}{4}$ on the eigenvalues of $A^{-1}$. Assuming that $\det A^{-1}$ is indeed one (I haven't checked your calculation), this gives an upper bound of $16$ on the largest eigenvalue of $A$.
Of course, this is worse than the bound of 6 from $\operatorname{tr} A$, but I don't see how you calculate the trace of $A$ without first inverting $A^{-1}$.
A: A lower bound for eigenvalues of $A$
Let $\{ e_1, e_2, e_3 \}$ be the standard basis. Then
$$
A^{-1} e_1 = \begin{bmatrix}2 \\ -1 \\ 0\end{bmatrix}, \quad
A^{-1} e_2 = \begin{bmatrix}-1 \\ 2 \\ -1\end{bmatrix}, \quad
A^{-1} e_3 = \begin{bmatrix}0 \\ -1 \\ 1\end{bmatrix}
$$
and we have 
$$
\|A^{-1} e_1\| = \sqrt{5}, \quad
\|A^{-1} e_2\| = \sqrt{6}, \quad
\|A^{-1} e_3\| = \sqrt{2}.
$$
For an arbitrary vector $u = u_1 e_1 + u_2 e_2 + u_3 e_3$ we have
$A^{-1} u = u_1 A^{-1} e_1 + u_2 A^{-1} e_2 + u_3 A^{-1} e_3$ so
$$\begin{align}
\| A^{-1} u \| 
& \leq \| u_1 A^{-1} e_1 \| + \| u_2 A^{-1} e_2 \| + \| u_3 A^{-1} e_3 \| \\
& \leq |u_1| \|  A^{-1} e_1 \| + |u_2| \| A^{-1} e_2 \| + |u_3| \|  A^{-1} e_3 \| \\
& \leq \sqrt{|u_1|^2 + |u_2|^2 + |u_3|^2} \sqrt{\|  A^{-1} e_1 \|^2 + \| A^{-1} e_2 \|^2 + \|  A^{-1} e_3 \|^2} \\
& = \|u\| \sqrt{(\sqrt{5})^2 + (\sqrt{6})^2 + (\sqrt{2})^2} \\
& = \sqrt{13} \|u\|
\end{align}$$
Thus an upper bound for eigenvalues of $A^{-1}$ is $\sqrt{13}.$
Now, the eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A,$ so $1/\sqrt{13}$ is a lower bound for eigenvalues of $A.$

Probably a similar reasoning can be used for the lower bound for $A^{-1}$ and thus for the upper bound for $A.$
