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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.

wwliu
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1 Answers1

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Outline of one approach:

  1. 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 $B$.

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

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

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

  5. 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.

Ben Grossmann
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