The eigenvalues associated to a graph's adjacency matrix are necessarily algebraic integers, because the adjacency matrix itself is entirely integer. I'm curious as to whether it's possible to have all irrational eigenvalues $\lambda$ simultaneously obey $\lambda > 1$. This is not possible for *all* eigenvalues, as the trace is zero, so there must be negative eigenvalues. There are many graphs whose integer eigenvalues are all negative, so intuitively we might feel that if we put enough negative integers in the spectrum, then we could a significant positive set of algebraic integers all conjugate to each other. The "obstruction" is that these would then need to have a norm that's a significantly large integer, and it's difficult to find these.


A natural starting place to look, perhaps, would be for a graph that has $x^2-5x+5$ as one of the factors of its characteristic polynomial.


I'm sorry if many of these statements are somewhat elementary, I'm just trying to provide the little ideas I have on the question. I couldn't find anything elsewhere online.



Are there graphs with irrational eigenvalues which are all $>1$?
Yes, the graph with adjacency matrix



```
 0     0     1     0     1     1     1     1
 0     0     0     1     0     0     0     1
 1     0     0     0     1     1     1     1
 0     1     0     0     0     0     0     0
 1     0     1     0     0     1     1     1
 1     0     1     0     1     0     1     1
 1     0     1     0     1     1     0     0
 1     1     1     0     1     1     0     0 

```

and graph6 string



```
GQil^W

```

is such a graph, as the characteristic polynomial is $x(x+2)(x+1)^4(x^2-6x+6)$, and the two roots of $(x^2-6x+6)$ are both larger than $1$.


