TITLE: Stochastic Matrix: Second largest eigenvalue and second largest absolute value of eigen value
QUESTION [6 upvotes]: Setup
Let $A$ be a stochastic matrix.
Let the eigenvalues of $A$ be $1 = \lambda_1 \geq \lambda_2 \geq \lambda_3 ... \geq -1$.
Let $\lambda = \max_{x: x \perp 1} \frac{||Ax||}{|| x ||}$
Question:
Besides $\lambda_2 \leq \lambda$, is there any relation between $\lambda$ and $\lambda_2$? In particular, I would love to see something of the form $\lambda \leq \lambda_2$.
Context:
Reading about expanders. Many of the proofs appears to prove upper bounds on $\lambda_2$, but I want upper bounds on $\lambda$, and it's not obvious to me:
(1) how an upper bound on $\lambda_2$ becomes an upper bound on $\lambda$
or
(2) how to generalize some of these proofs.
Thanks!

REPLY [5 votes]: The answer is due to Boyd, Diaconis, Sun & Xiao. If $A$ is a symmetric and bi-stochastic, then $\mu:=\max(\lambda_2,-\lambda_n)$ satisfies
$$\mu\ge\cos\frac\pi{n}.$$
In addition, there exists such a matrix for which the equality holds. See Exercise 164 of my list http:\www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf .
P.S. Because you assume that the eigenvalues are real, I presume that you have in mind that the matrix is symmetric.