# 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.