Comparison of the smallest eigenvalues of two tridiagonal matrices Let $n\geq2$ be an integer and $E_{ii}$ for an integer $2\leq i\leq n$ be the $n\times n$-matrix with its $ii$-entry equal to 1 and remaining entries equal zero. Furthermore, let $H_n:=\mathrm{tridiag}(-1,2,-1)$ denote a tridiagonal matrix with all main diagonal entries equal to 2 and off-diagonal entries equal to -1. The spectrum of $B_n:=H_n+E_{11}$ is analytically well-known. If we define $A_n^i:=H_n-E_{11}+2\cdot E_{ii}$, how can we show that for small values of $i$ (I assume $i\leq\mathrm{floor}(n/4)$), it holds $$2-2\cos\left(\frac{2\pi}{2n+1}\right)=\lambda_{\min}(B_n)<\lambda_\min(A_n^i)$$? At least some idea for $i=2$ would be helpful.
 A: I'll assume that $n$ is large and $i\ll n$. However, I believe it should be possible to treat the general situation in the same way; as we'll see, we only need a certain simple property of the recursion (1) below.
By oscillation theory, we can characterize the smallest eigenvalue as follows: Let $u$ be the solution of the difference equation
$$
-u_{j+1}-u_{j-1} + (2+V_j)u_j = \lambda u_j
$$
with the initial values $u_0=0$, $u_1=1$. (Here $V_j$ records the changes made by the matrices $E$, so $V_1=1$ and $V_j=0$ otherwise for $B=B_n$, and the non-zero $V_j$'s for $A=A_n^i$ are $V_1=-1$, $V_i=2$.) Then the smallest eigenvalue is the first (= smallest) $\lambda$ for which $u_{n+1}=0$. Moreover, $\lambda$ is strictly smaller than the smallest eigenvalue precisely if $u$ is positive on $\{ 1,\ldots , n+1\}$.
The quotients $q_j=u_j/u_{j-1}$ solve
$$
q_{j+1} =2-\lambda+V_j - \frac{1}{q_j} \quad\quad\quad (1)
$$
Now consider this equation for $\lambda=\lambda_0\equiv \lambda_{\textrm{min}}(B)$. Since $\lambda_0\approx 0$ (more precisely, $\lambda_0\sim c/n^2$), we have that $q_2(B)\approx 3$, $q_2(A)\approx 1$. Next, approximately solving up to $i$, we find that $q_{i+1}(A)\approx 3$, $q_{i+1}(B)<2$.
From this point on, both $q$'s solve the same equation (1), with $V\equiv 0$. Now the inequality $q_j(A)>q_j(B)$ is preserved by (1), so the claim follows from the oscillation theory facts that I summarized above: $u_{n+1}(A)>0$ and $u_j(A)$ was positive on the whole interval $\{1,\ldots , n\}$, so we are still strictly below the spectrum of $A$.
