The question I am wondering about is: 


Can the [discrete Laplacian](https://en.wikipedia.org/wiki/Laplacian_matrix#Laplacian_matrix_for_simple_graphs) have complex eigenvalues on a graph?


Clearly, there are two cases where it is obvious that this is impossible.


1.) The graph is finite 
2.) The underlying space is $\ell^2$, since then the discrete Laplacian is self-adjoint.


Thus, my question requires us to look at an infinite graph and a large space.


Hence: **Does there exist an infinite graph such that the discrete Laplacian on $\ell^{\infty}$ has complex eigenvalues?**


Thank you very much


BTW: My casual use of complex in the above text refers to $\mathbb C \backslash \mathbb R$ 



Graph with complex eigenvalues
Take the infinite binary tree $T\_2 = (V,E)$, viewed as a bi-infinite `backbone' $B \approx \mathbb Z$ with binary trees dangling off $B$. For every vertex $v \in V$ on the tree, there then is a closest element $\pi(v) \in \mathbb Z$ on the backbone and we write $d(v)$ for the distance from $v$ to $\pi(v)$. We then set $A(v) = \exp(i\theta (\pi(v) + d(v)))$ and check that $\Delta A = \lambda A$ with $\lambda = 2 e^{i\theta} + e^{-i\theta} - 2$.


