1. Ilan Ben-Bassat Omri Weinstein
Bounding the mixing time via Spectral Gap Graph 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. Larger Spectral Gap = Rapid Mixing
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)
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
(0,0)
(0,1)
(1,1)

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: