This question is motivated by a talk I went to earlier today.

Suppose we have a $d$-regular graph $G$ with $n$ vertices, with adjacency matrix $A$.

Let $$\lambda_1\geq \lambda_2 \geq\dots \geq \lambda_n$$ be the eigenvalues of $A$, so in particular $\lambda_1=d$. If the first two eigenvalues are the same, that is $\lambda_2=\lambda_1$, then it tells us a lot about the structure of the graph. In particular, the graph must be disconnected. (This is an if and only if condition)

What if the second and third eigenvalues are equal? That is, suppose that $\lambda_1>\lambda_2=\lambda_3$. What does that tell us (if anything) about the structure of the graph?

**Additional questions:** If $\lambda_1=\lambda_2=\cdots=\lambda_k<\lambda_{k+1}$, then the graph will have exactly $k$ connected components. What can we say about $G$ if $\lambda_1<\lambda_2=\cdots=\lambda_{k+1}<\lambda_{k+2}$? That is, the second eigenvalue has multiplicity $k$.

What if the $n^{th}$ eigenvalue has multiplicity $k$?

directmeaning. However, large multiplicities of eigenvalues in general suggest (but don't imply) that $G$ has nontrivial symmetries. – Qiaochu Yuan Jul 4 '12 at 19:52