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I look at sparsly populated matrices $W_1$ with entries $0$ and $1$ only. $W_1$ has a graph-theoretical background. I'll give the Definition 2.2:

We define the 0,1 edge matrix $W_1$ by orienting the $m$ edges of [the graph] $X$ and labeling them as in formula (2.1). Then $W_1$ is the $2m×2m$ matrix with $ij$ entry $1$ if edge $e_i$ feeds into $e_j$ provided that $e_j\neq e^{−1}_i$, and $ij$ entry $0$ otherwise. By “$a$ feeds into $b$,” we mean that the terminal vertex of edge $a$is the same as the initial vertex of edge $b$.

My graphs under inspection have $n$ vertices and are bicubic, so $W_1$ will have two entries per row. I gave a procedure to calculate $W_1$ here and now I'm trying to simplify the determination of the eigenvalues of $W_1$.

From numerical experiments, I assume that all my eigenvalues $\lambda_k$ are:

  1. $0$ and $\pm1$ or
  2. $\pm\sqrt2\exp(\pm i2\pi c)$ or
  3. $\pm\sqrt2\exp(\pm 2\pi c)$,

where $|c|\leq \frac{\log2}{4\pi}$. Can one make use of these facts? Maybe like a symmetry? Can we, for example group all complex eigenvalues on the unit circle, such that it gives a scaled unitary block diagonal structure?

Further, all four possible values of the second class are always available. This is due to the fact that the characteristic polynomial is real and of even degree, as Paul pointed out in a comment.

Fun fact(?): The eigenvalues of the third class, only appeared in example graph s, that were not ramanujan.

Since the number of eigenvalues $+1$ and $-1$ are at least $k_1=r(G)-1$ (it is the inverse Ihara $\zeta$ function), I know that the characteristic polynomial will be of the form: $$ {x^{k_0}} {(x^2-1)^{k_1}} \prod_\color{cyan}{\text{r}}(x^4-4x^2\cos(4\pi c_\color{cyan}{\text{r}}) +4) \underbrace{\prod_\color{red}{\text{n}}(x^4-4x^2\cosh(4\pi c_\color{red}{\text{n}})+4)}_{\text{only for $\color{red}{\text{n}}$on $\color{cyan}{\text{r}}$amanujan graphs?}}, $$

with $k_0=n^2-2m=n(n-3)$.

Any help or hint appreciated...

draks ...
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  • You don't have an answerable question here. You've given a bunch of facts (some of which seem dubious to me without more information about the particular matrices you are dealing with), but you have not said what it is you are trying to accomplish. So we cannot say whether these fact help, or can be made use of. You have not given enough information for us to tell if there is a symmetry. The only thing I can tell you is that because your matrices are real, so is the characteristic polynomial, therefore its complex-valued roots - the eigenvalues - must come in conjugate pairs. – Paul Sinclair Aug 07 '19 at 21:58
  • @PaulSinclair the matrices I'm dealing with the 0,1 edge matrices $W_1$ as defined in the linked paper. I'm trying to reduce the complexity of the determination of the eigenvalues. Does that make it clearer? – draks ... Aug 07 '19 at 22:38
  • You should edit the post to say you are trying to simplify determination of the eigenvalues. I hadn't read the link because you had indicated it was only background information, but it does indeed indicate a strong symmetry in $W_1$. Since every edge $i$ has $i+n$ as its inverse edge, $w_{ij} = 1$ if and only if $w_{(j+n)(i+n)} = 1$ (index additions all performed modulo $2n$). This probably explains your observation that the eigenvalues can be paired off in pairs that are opposites It may also explain the other eigenvalues being on a circle of radius $\sqrt 2$, – Paul Sinclair Aug 07 '19 at 23:42
  • The formula you mention (Ihara zeta) doesn't guarantee that the multiplicity of +1 and -1 are the same. It guarantees that they are both at least $r(G) - 1$. – Leo Aug 12 '19 at 20:25
  • @Leo thanks I edited my post... – draks ... Aug 29 '19 at 10:27

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