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Bounding the mixing time via Spectral Gap Graph Random Walk Seminar Fall 2009

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.

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Bounding the mixing time via Spectral Gap Graph Random Walk Seminar Fall 2009

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  1. Bounding the mixing time via Spectral GapGraph Random Walk Seminar Fall 2009 Ilan Ben-Bassat Omri Weinstein

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

  3. 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.

  4. 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?

  5. 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.

  6. So?...

  7. Intuition How could eigenvalues and mixing time be connected?

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

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

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

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

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

  13. 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:

  14. According to spectral decomposition of symmetric matrices: Note:

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

  16. So, we can get

  17. Now, we can bound the mixing time:

  18. 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).

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

  20. Entry (i,j) 

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

  22. 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:

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