TITLE: Second eigenvalue of suspension of a graph 
QUESTION [5 upvotes]: Suppose I have some $d$-regular graph $G$. Let $\lambda = \max\{\lambda_2(G), |\lambda_n(G)|\}$ be the second largest eigenvalue of the adjacency matrix of $G$. Now take $\tilde{G}$, the suspension of $G$, obtained by adding a new vertex $v$ and connecting $v$ to all vertices of $G$. Is it true that $\lambda(\tilde{G}) \leq \lambda(G)$? (or more generally, is it true if instead of suspending with one vertex we suspend with a clique of size $n$)
The intuition is that $\tilde{G}$ should exhibit better mixing (hence, smaller $\lambda$), since it's easier for a random walk to get from one vertex to any other.

REPLY [6 votes]: The eigenvalues of $G$ determine the eigenvalues of $\tilde G$ in the regular case.  If I calculated correctly, just replace the eigenvalue $d$ by the two zeros of $x^2-dx-n$, and keep the other eigenvalues the same. This is for the adjacency matrix. I agree with Alon that you might be looking at the wrong matrix, and I would also call this the cone of $G$.