Let $G$ be a finite, undirected $d$-regular graph and $A(G)$ its normalized adjacency matrix, with eigenvalues $1 = \lambda_1(A)\geq \dots \ge \lambda_n \ge -1$. It is a classical fact that $\lambda_n = -1$ if and only if $G$ is bipartite. Our main result provides a quantitative separation of $\lambda_n$ from $-1$ in the case of Cayley graphs, in terms of their expansion. Denoting $h_{out}$ by the (outer boundary) vertex expansion of $G$, we show that if $G$ is a non-bipartite Cayley graph (constructed using a group and a symmetric generating set of size $d$) then $\lambda_n \ge -1 + ch_{out}^2/d^2\,,$ for $c$ an absolute constant. We exhibit graphs for which this result is tight up to a factor depending on $d$. This improves upon a recent result by Biswas and Saha who showed $\lambda_n \ge -1 + h_{out}^4/(2^9d^8)\,.$ We also note that such a result could not be true for general non-bipartite graphs.

翻译：让G$变成一个限定的、非方向的固定美元图表和A(G)$(美元)的普通对称矩阵,其普通对称矩阵值为1美元=\lambda_1(A)\geq\dots\ge\ge\ g@lambda_n\ge -1美元。一个典型的事实是,$\lambda_n= -1美元(如果和只有当美元为双倍)时,我们的主要结果为Cayley图表的扩展提供了从$-1美元中减去$\lambda_n(G)的量化的分隔。如果(外部边界)的顶端值扩展为$1(A)\geq\d\d_1(G)$(美元),我们的主要结果是:$_(lumbda_n) 美元(美元) 而不是美元(美元),那么这个绝对的结果就是- 美元(我们这个绝对值) 。这个绝对的图表是最新的结果。这个结果是- b__xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx