Can I find the gap between the two least eigenvalues of this special matrix A(t)?‎ 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.‎
           

 A: You say that computing the gap for a fixed $t_0$ could also be good. I think I can do $t_0=0$ in a way that generalizes to examples of larger dimensions. (You haven't defined your matrix exactly for larger dimensions, but that is how I assume it works).
That matrix for $t=0$ is $A(0) = B\otimes I\otimes I + I\otimes B \otimes I + I\otimes I \otimes B$, where $I$ is the ($2\times 2$ in this example) identity matrix and $B=\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix}$. Here $\otimes$ is the Kronecker product. By the properties of the Kronecker product, The eigenvalues of such a matrix are given by $\{\lambda_i+\lambda_j+\lambda_k: i,j,k \in \{1,2\}\}$, where $\lambda_1,\lambda_2$ are the eigenvalues of $B$. So in this case you get 0,2,2,2,4,4,4,6.
I imagine that what happens for larger dimensions is that $B$ is $I+L$, where $L$ is the tridiagonal matrix with (-1,0,-1), for which eigenvalues are known explicitly as $\lambda_j = 2\cos(j\pi/(n+1))+2$, $j=1,2,\dots,n$. So all eigenvalues are known and the gap is $\lambda_n-\lambda_{n-1}$.
This stuff is quite standard in numerical analysis (the key terms are "3d finite difference Laplace matrix").
I suspect that a similar trick may work for a generic $t$, but it looks tricky to make sense of the coefficients of $t$ on the diagonal. Do they come out of a formula?
