I am interested in finding the gap between the two least eigenvalues of $A(t)$, a Hermitian $N\times N$ sparse matrix whose diagonal elements are $a_it+b_i\,(1\leq i\leq N)$, and all off-diagonal non-zero elements are the same and equal to $ct+d$. Furthermore, $A(t)$ is a banded matrix, that is, $a_{ij}=0$ for $( 1\leq i\leq N) \& (j>i+k)$ and $(1\leq j\leq N) \& (i> j+k)$, with $k<N$ fixed. Since real-scaled $N$ is larger than $10^6$, determining the eigenvalues and then calculating the gap $(g)$ is not practical. Thus, even bounding the gap can be satisfactory.

However, *interlacing theorem* and *Courant-Weyl inequalities* provide some bounds for the eigenvalues, but it is not enough since they involve the most and the least eigenvalues, so the bounds are to some extent wide. Another note I should mention is that while I need to calculate $g$ for a range of values for $t$, so my main (and optimistic) goal is to derive $g$ in terms of $t$. However, calculation of the gap (or its bounds) for some fixed $t_0 \neq 0$ can also do some good.

Now my question is that, according to the structure of $A(t)$, can I calculate $g$ or derive any better bound for it? Is there any sharp inequality that can help?

Any other useful comment would be appreciated.

Edit: To be more illustrative, here is an example of $A(t)$ (with $N=8$): $$\begin{bmatrix} 3-t & t-1 & t-1 & 0 & t-1 & 0 & 0 & 0 \\ t-1 & 3 (t+1) & 0 & t-1 & 0 & t-1 & 0 & 0 \\ t-1 & 0 & 3 (t+1) & t-1 & 0 & 0 & t-1 & 0 \\ 0 & t-1 & t-1 & 3-3 t & 0 & 0 & 0 & t-1 \\ t-1 & 0 & 0 & 0 & 7 t+3 & t-1 & t-1 & 0 \\ 0 & t-1 & 0 & 0 & t-1 & 3-t & 0 & t-1 \\ 0 & 0 & t-1 & 0 & t-1 & 0 & t+3 & t-1 \\ 0 & 0 & 0 & t-1 & 0 & t-1 & t-1 & 5 t+3 \\ \end{bmatrix}$$

and the plot for $g(t)$ for this example with $-1\leq t\leq 1 $ is:

Edit: Here is a long shot of an almost real-scaled example of $A(t)$ (with $N=32768$). While all off-diagonal non-zero elements are the same, Indeed, diagonal elements still vary from others (and may be from each other) -- though this is not visible at this scale.