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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 $L$.

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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 $\lambda_{\min}(L) = 0.3190$.

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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.

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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.

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