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It's well-known that the eigenvalues of the Clement-Kac-Sylvester tridiagonal matrix

$$\begin{pmatrix} 0 & n-1 & 0 & \dots & 0 \\\ 1 & 0 & n-2 & \dots & 0\\\ 0 & 2 & 0 & \dots & 0 \\\ \vdots & \vdots & \vdots & \ddots & 1 \\\ 0 & 0 & 0 & n-1 & 0 \end{pmatrix}$$

are $\pm (n), \pm (n-2), ..., (\pm1 \; or \; 0)$. Please refer to Regarding a Paper by Paul.A Clement on Tridiagonal Matrices and https://math.stackexchange.com/questions/405670/regarding-a-paper-by-paul-a-clement-on-tridiagonal-matrices for details.

My question is do we know the eigenvalues of the leading principal submatrix of the Clement-Kac-Sylvester tridiagonal matrix $(1<r\le n/2)$:

$$\begin{pmatrix} 0 & n-1 & 0 & ... & 0 \\\ 1 & 0 & n-2 & & ... \\\ 0 & 2 & 0 & ... & 0 \\\ ... & & ... & & n-r+1 \\\ 0 & ... & 0 & r-1 & 0 \end{pmatrix}$$

or do we have any lower/upper bound on the eigenvalues of this submatrix.

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  • $\begingroup$ What order of $r$ as compared to $n$ are you looking for? After playing around, I found that if $r$ is of order $n$, the max and min absolute values of submatrix eigenvalues are still close within $O(1)$ to $n$ and $0$, respectively $\endgroup$
    – Leonid Petrov
    Commented Nov 20, 2016 at 14:33
  • $\begingroup$ @LeonidPetrov Suppose $r=\gamma n$ where $0\le \gamma\le 1/2$. The first (largest) eigenvalue is at least $2n\sqrt{\gamma(1-\gamma)}+o(n)$. Does this constant $2\sqrt{\gamma(1-\gamma)}$ match with your calculation? $\endgroup$
    – Sihuang Hu
    Commented Nov 21, 2016 at 7:06
  • $\begingroup$ I'm not sure about the asymptotics since my computer can do only matrices of size about 200, but this looks correct if your $o(n)$ is negative $\endgroup$
    – Leonid Petrov
    Commented Nov 21, 2016 at 14:09

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