What is the largest eigenvalue of the special matrix? The question is a simplified question to my last question.
Let $I_m$ denote the $m\times m$ identity matrix, and $\alpha\in\mathbb{R}^{n-1}$.
Let
$$
\mathbf{B}=\left[\begin{array}{ccccc}
-\alpha & 0 & &  &\\
\alpha & -\alpha & 0 & \\
0 & \alpha & \ddots & \ddots \\
& & & & &-\alpha\\
\end{array}\right]\in \mathbb{R}^{(mn-m)\times (m-1)}
$$
I want to calculate the largest eigenvalue of
$$
\left[\begin{array}{ccccc}
I_{mn-m} & B \\ 
B^{\top} & I_{m-1}
\end{array}\right].
$$
The answer should be $\sim3- O(\frac{1}{n m^2})$, but I want to know the exact value.
 A: Outline of one approach:

*

*Note/show that in general, the non-zero eigenvalues of a block matrix of the form
$$
A = \pmatrix{I & B\\ B^\top & I} 
$$
are of the form $1 \pm \sigma_i(B)$, where $\sigma_i(B)$ denotes the $i$th eigenvalue of $B$.


*$B$ can be written as the Kronecker product $B = L \otimes \alpha$, where $L$ is the $m \times (m-1)$ matrix
$$
L = \pmatrix{-1 & 0 & &  &\\
1 & -1 & 0 & \\
0 & 1 & \ddots & \ddots \\
& & & & &-1\\}
$$


*$\sigma_i(B)$ is the square root of the $i$th eigenvalue of $B^\top B$. Write
$$
B^\top B = (L \otimes \alpha)^\top(L \otimes \alpha) = 
(L^\top L) \otimes (\alpha^\top \alpha) = \|\alpha\|^2 \cdot L^\top L.
$$


*Verify that $L^\top L$ is tridiagonal and Toeplitz. So, its eigenvalues can be found using the eigenvalue formula given here.


*Deduce that the largest eigenvalue of $A$ is equal to
$$
\lambda_{\max}(A) = 1 + \|\alpha\| \cdot \sqrt{\lambda_{\max}(L^\top L)}.
$$
From the linked formula, should find that
$$
\lambda_{\max}(L^\top L) = 2 + 2\cos(\pi /m) = 4 - \frac {\pi^2}{2m^2} + O(m^{-4}), 
$$
which means that
$$
\sqrt{\lambda_{\max}(L^\top L)} = 2 - \frac {\pi^2}{8m^2} + O(m^{-4}),
$$
which confirms the asymptotics you suggested relative to $m$. The asymptotics relative to $n$ depend on how $\alpha$ changes as $n$ increases, or the distribution from which $\alpha$ is drawn.
