Ilan Ben-Bassat Omri Weinstein

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Presentation transcript:

Ilan Ben-Bassat Omri Weinstein Bounding the mixing time via Spectral Gap Graph Random Walk Seminar Fall 2009 Ilan Ben-Bassat Omri Weinstein

Outline Why spectral gap? Undirected Regular Graphs. Directed Reversible Graphs. Example (Unit hypercube). Conductance. Reversible Chains.

P- Transition matrix (ergodic) of an undirected regular graph. P is real, stochastic and symmetric, thus: All eigenvalues are real. P has n real (orthogonal) eigenvectors.

P’s eigenvalues satisfy: Why do we have an eigenvalue 1? Why do all eigenvalues satisfy ? Why are there no more 1’s? Why are there no (-1)’s?

Laplacian Matrix L = I – P So, L is symmetric and positive semi definite. L has eigenvalue 0 with eigenvector 1v. Claim: The multiplicity of 0 is 1.

So?...

Intuition How could eigenvalues and mixing time be connected?

Larger Spectral Gap = Rapid Mixing Spectral Gap and Mixing Time The spectral gap determines the mixing rate: Larger Spectral Gap = Rapid Mixing

As for P’s spectral decomposition, P has an orthonormal basis of eigenvectors. We can bound by . So, the mixing time is bounded by:

Assume directed reversible graph (or general undirected graph) Assume directed reversible graph (or general undirected graph). We have no direct spectral analysis. But P is similar to a symmetric matrix!

Proof Let be a matrix with diagonal entries . Claim: is a symmetric matrix. From reversibility:

S and P have the same eigenvalues. What about eigenvectors?

Still: Why do all eigenvalues satisfy ? (same) Why do we have an eigenvalue 1? (same) Why is it unique? same for (-1). As for 1: Omri will prove:

According to spectral decomposition of symmetric matrices: Note:

Main Lemma: For every , we define: So, for every we get:

So, we can get

Now, we can bound the mixing time:

Summary We have bounded the mixing time for irreducible, a-periodic reversible graphs. Note: Reducible graphs have no unique eigenvalue. Periodic graphs – the same (bipartite graph).

Graph Product Let . The product Is defined by and 1 (0,0) (0,1) (1,1)

Entry (i,j) 

So, only the permutations that were counted for the determinant of AG1, will be counted here. We instead of we get So,

The eigenvectors of Qn are We now re-compute every eigenvalue by: Adding n (self loops) Dividing by 2n (to get a transition matrix). Now we get And the mixing time satisfies: