Determine upper and lower limit of eigenvalues

Question

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.

Answer 1

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 $A.$

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

Answer 2

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 $A^{-1}$.