Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $J, L$ be two symmetric positive definite tridiagonal matrices of positive diagonal entries, $\mbox{diag}(J)=(a_1, a_2, \ldots, a_n)$, $\mbox{diag}(L)=(\alpha_1, \alpha_2, \ldots, \alpha_n)$, with $\min_{i=1}^n(a_i)<\min_{i=1}^n(\alpha_i)$ and $\mbox{trace}(L-J)>0$. The two matrices have constant off diagonal entries, $\mbox{offdiag}(J)=\mbox{offdiag}(L)=(b,b,b,\ldots, b)$, $b<0$.

I'm interested whether or not these conditions are sufficient to ensure that the smallest eigenvalue of $J$ is less than the smallest eigenvalue of $L$.

share|improve this question

3 Answers 3

No, your conditions are not sufficient to ensure that the smallest eigenvalue of $J$ is less than the smallest eigenvalue of $L$. Here is a simple counterexample,

$$ \begin{equation*} J = \begin{bmatrix} 1.9 & -1 & 0\\\\ -1 & 4 & -1\\\\ 0 &-1 & 7 \end{bmatrix},\qquad L = \begin{bmatrix} 2 & -1 & 0\\\\ -1 & 3 & -1\\\\ 0 & -1 & 10 \end{bmatrix}. \end{equation*} $$

Here $1.9 = \min(\mbox{diag}(J)) < \min(\mbox{diag}(L))=2$, $\mbox{trace}(L-J) = 2.1$, but $\lambda_{\min}(J) = 1.4736$ while $\lambda_{\min}(L) = 1.3488$.


Older, messier counterexample. $$ \begin{equation*} J = \begin{bmatrix} 0.3309 & -0.0463 & 0\\\\ -0.0463 & 0.5364 & -0.0463\\\\ 0 &-0.0463 & 0.5951 \end{bmatrix},\qquad L = \begin{bmatrix} 0.3432 & -0.0463 & 0\\\\ -0.0463 & 0.4117 & -0.0463\\\\ 0 & -0.0463 & 0.7940 \end{bmatrix}. \end{equation*} $$

Here we see that the $0.3309 = \min(\mbox{diag}(J)) < \min(\mbox{diag}(L))=0.3432$, $\mbox{trace}(L-J) = 0.0865$, but $\lambda_{\min}(J) = 0.3206$ while $\lambda_{\min}(L) = 0.3190$.

share|improve this answer

Well, very roughly speaking, if $b$ is small relative to the diagonals, then the eigenvalues of $J,L$ will be approximately their diagonal values and your conclusion will hold.

share|improve this answer

Felix is correct in the small $b$ limit. In the very large $b$ limit, on the other hand, the spectra of $J$ and $L$ will coincide, so one needs some more finesse. The following is a rough, approximate attempt at making some intuition.

Consider the case where the first diagonal entry of $J$ is $a_1<0$ with $|a_1|\ll |b|^2$, and all other diagonal entries in $J$ and $L$ are zero. Then you can expand $\det(tI-J)$ along the first row to get $$\det(tI-J)=b^{2n} \left(U_n(t/2)-\frac{a_1 }{b^2} U_{n-1}(t/2)\right),$$ where $U_n$ is the $n$th Chebyshev polynomial of the second kind. For small $a_1$, then, the question is which way does the second term "push" the eigenvalue? I don't have a proof but it's graphically clear that the signs of $U_n$ and $U_{n-1}$ coincide just inside of the former's leftmost zero, which means that a small, negative $a_1$ will push $J$'s minimum eigenvalue left, in accord with your conjecture.

This can be extended to $L$ having a nonzero element in its diagonal, which will push its minimum eigenvalue either left or right but no more than $J$'s, at least at first order.

Since your result is (or appears to be) true both for big and small $b$, I should expect it to be true everywhere, or to have some pretty interesting mathematics in the middle.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.