Vtyjkyuk

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
$$A^-1=beginbmatrix
2& -1& 0\
-1& 2& -1\
0 & -1& 1
endbmatrix.$$

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 $det(A)=1$ $operatornametr(A)=6$ and $operatornametr(operatornameadj(A))=5$, I think it is $lambda<5$. But I try with five and $operatornametr(operatornameadj(A))$ little bit bigger then five, so I think there has to be a better way to do this.

A lower bound for eigenvalues of $A$

Let $ e_1, e_2, e_3 $ be the standard basis. Then
$$
A^-1 e_1 = beginbmatrix2 \ -1 \ 0endbmatrix, quad
A^-1 e_2 = beginbmatrix-1 \ 2 \ -1endbmatrix, quad
A^-1 e_3 = beginbmatrix0 \ -1 \ 1endbmatrix
$$
and we have
$$
|A^-1 e_1| = sqrt5, quad
|A^-1 e_2| = sqrt6, quad
|A^-1 e_3| = sqrt2.
$$

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

$$beginalign
| A^-1 u |
& leq | u_1 A^-1 e_1 | + | u_2 A^-1 e_2 | + | u_3 A^-1 e_3 | \
& leq |u_1| | A^-1 e_1 | + |u_2| | A^-1 e_2 | + |u_3| | A^-1 e_3 | \
& leq sqrt^2 + sqrt A^-1 e_3 \
& = |u| sqrt(sqrt5)^2 + (sqrt6)^2 + (sqrt2)^2 \
& = sqrt13 |u|
endalign$$
Thus an upper bound for eigenvalues of $A^-1$ is $sqrt13.$

Now, the eigenvalues of $A^-1$ are the reciprocals of the eigenvalues of $A,$ so $1/sqrt13$ 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 $A.$

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 $frac14$ 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 $operatornametr A$, but I don't see how you calculate the trace of $A$ without first inverting $A^-1$.