Let $G$ be any graph with at least one edge and let $e$ be any edge of $G$. Let $G-e$ denote the subgraph of $G$ obtained by deletion of the edge $e$. Assume that $G$ has $n$ vertices. 
Suppose that $\lambda\_1(G)\geq\cdots\geq \lambda\_n(G)$ and $\lambda\_1(G-e)\geq\cdots\geq \lambda\_n(G-e)$ are eigenvalues of $G$ and $G-e$, respectively.
It is true that $\lambda\_i(G)\cdot\lambda\_i(G-e)\geq 0$ for all $i=1,\dots,n$?



Eigenvalues of a graph and its one-edge-delation graph
I think this is false and there are counterexamples.


Here is a sage session.



```
sage: ed,e=([(0, 2), (0, 4), (0, 5), (1, 3), (1, 5), (2, 4), (3, 5)], (0, 2))
sage: G=Graph(ed);F=Graph(G);F.delete_edge(e)
sage: eg=G.adjacency_matrix().eigenvalues();eg.sort(reverse=1);eg
[2.414213562373095?, 1.732050807568878?, -0.4142135623730951?, -1, -1, -1.732050807568878?]
sage: ef=F.adjacency_matrix().eigenvalues();ef.sort(reverse=1);ef
[2.228328215946636?, 1.360409337131395?, 0.1858853464264596?, -1, -1, -1.774622899504490?]

```

