Expanders, Property Testing and the PCP theorem

Lecture 7: Random walks on expanders are rapidly mixing

Posted in lectures by HQN on September 19, 2008

Let \hat{\mathbf A} be the normalized adjacency matrix of a d-regular graph, 1=\hat \lambda_1 \geq \dots \geq \hat \lambda_n be the eigenvalues of this matrix, and define \hat\lambda(G) = \max\{|\hat\lambda_2|, |\hat\lambda_n|\}.

Suppose we take a random walk on G starting from an initial distribution \pi^{(0)}. Then, the distribution after t steps is \pi^{(t)} = \hat{\mathbf A}^t\pi^{(0)}. Using the spectral theorem, it is not difficult to prove that, if \hat\lambda(G) <1 then \lim_{t\to \infty} \pi^{(t)} = \mathbf u, where \mathbf u is the uniform distribution. The more difficult question is “how fast is the walk converging to this limit?”

To answer this question, we need to define a notion of distance between two distributions. The total variation distance is probably the most commonly used. Since we are working on a finite probability space, the total variation distance is half the l_1-norm. Thus, we would like a theorem bounding the rate at which \|\pi^{(t)}-\mathbf u\|_1 tends to zero. We proved that

Theorem 1 (Convergence in l_1) \|\pi^{(t)}-\mathbf u\|_1 \leq \sqrt n \hat\lambda^t

Theorem 2 (Convergence in l_2) \|\pi^{(t)}-\mathbf u\|_2 \leq \hat\lambda^t

Theorem 1 follows from Theorem 2 by the Cauchy-Schwarz inequality. To prove Theorem 2, note that

\|\pi^{(t)}-\mathbf u\|_2 = \|\hat{\mathbf A}^t\pi^{(0)} - \mathbf u\|_2 =  \|\hat{\mathbf A}^t(\pi^{(0)} - \mathbf u)\|_2 =  \|\hat{\mathbf A}^t\mathbf v\|_2

Where, \mathbf v = \pi^{(0)} - \mathbf u is perpendicular to \mathbf u and \|\mathbf v\|_2 \leq 1. Each time we apply \hat{\mathbf A} to a vector \mathbf y perpendicular to \mathbf u (its first eigenvector), we shrink the length of \mathbf y by a factor of at least \hat\lambda. Hence the theorem follows.

An (n,d,\alpha)-spectral expander is a d-regular graph on n vertices such that \hat \lambda(G) \leq \alpha. If \alpha < 1 then an (n,d,\alpha)-spectral expander certainly has spectral gap bounded away from zero. Conversely, a graph whose spectral gap is bounded away from zero (plus self-loops) is a spectral expander.

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