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I am looking for symmetric matrices $A \in \{-1,0,1\}^{n \times n}$ that have an "interesting" spectrum.

What do I mean by that?

Suppose we have the following matrix:

$$A_1 := \left( \begin{array}{rr}0 & 1 \\1 & 0 \\\end{array}\right), \qquad A_n := \left( \begin{array}{rr} A_{n-1} & I_{n-1} \\I_{n-1} & - A_{n-1} \\\end{array}\right)$$

$A_n $ has spectrum $ \{\sqrt{n}, -\sqrt{n}\}$ each with multiplicity $2^{n-1}$.

(Because $A_n^2 = n \cdot I_n$ and $\mbox{tr} (A_n) = 0$, and since $A_n$ is symmetric its trace is equal to the sum of its eigenvalues each with its corresponding multiplicity).

I am looking for matrices, that have "comparably" interesting spectra, thanks in advance for any creative propositions!

EDIT: for clarifying purposes - any spectrum not solely consisting of $\{-1,1\}$ I'd consider interesting.

peterparker
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1 Answers1

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A canonical example for this is $$ S=\begin{bmatrix} 0&1&0&0&\cdots&0 \\ 1&0&1&0 &\cdots&\vdots\\ 0&1&0&1&0 &\vdots\\ \vdots&0&\ddots&\ddots&\ddots &\vdots\\ 0&\vdots&0&1&0&1 \\ 0&\cdots&\cdots&0&1&0 \end{bmatrix} . $$ It is well-known (since Lagrange, I understand) that the eigenvalues of $S$ are $$ 2\cos\frac{k\pi}{n+1},\ \ \ \ k=1,\ldots,n. $$ It is actually easy to check, if you are told to use the eigenvectors $$ x_k=\left(\sin\frac{kj\pi}{n+1}\right)_j. $$

Martin Argerami
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