most recent 30 from http://mathoverflow.net 2013-05-22T06:12:17Z http://mathoverflow.net/feeds/question/80863 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80863/stochastic-matrix-second-largest-eigenvalue-and-second-largest-absolute-value-of anonymous coward 2011-11-14T03:41:42Z 2011-11-14T06:14:15Z

<h2>Setup</h2> <p>Let $A$ be a stochastic matrix.</p> <p>Let the eigenvalues of $A$ be $1 = \lambda_1 \geq \lambda_2 \geq \lambda_3 ... \geq -1$.</p> <p>Let $\lambda = \max_{x: x \perp 1} \frac{||Ax||}{|| x ||}$</p> <h2>Question:</h2> <p>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 $\lambda \leq \lambda_2$.</p> <h2>Context:</h2> <p>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:</p> <p>(1) how an upper bound on $\lambda_2$ becomes an upper bound on $\lambda$ or (2) how to generalize some of these proofs.</p> <p>Thanks!</p>

http://mathoverflow.net/questions/80863/stochastic-matrix-second-largest-eigenvalue-and-second-largest-absolute-value-of/80869#80869 Denis Serre 2011-11-14T06:12:37Z 2011-11-14T06:12:37Z

<p>The answer is due to Boyd, Diaconis, Sun &amp; 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> <p><strong>P.S.</strong> Because you assume that the eigenvalues are real, I presume that you have in mind that the matrix is symmetric.</p>