Distribution of Eigenvalues of Weighted, Structured Matrix Ensembles

Transcription

1 Distribution of Eigenvalues of Weighted, Structured Matrix Ensembles Olivia Beckwith 1, Steven J. Miller 2, and Karen Shen 3 1 Harvey Mudd College 2 Williams College 3 Stanford University Joint Meetings of the AMS/MAA Boston, January 2012

2 Random Matrices and their Limiting Spectral Measure random matrix: a matrix whose entries are chosen randomly according to some probability distribution where E (a ij ) = 0 and Var (a ij ) = 1

3 Random Matrices and their Limiting Spectral Measure random matrix: a matrix whose entries are chosen randomly according to some probability distribution where E (a ij ) = 0 and Var (a ij ) = 1 Interested in the distribution of eigenvalues µ A,N (x) of an N N random matrix A as N

4 Random Matrices and their Limiting Spectral Measure random matrix: a matrix whose entries are chosen randomly according to some probability distribution where E (a ij ) = 0 and Var (a ij ) = 1 Interested in the distribution of eigenvalues µ A,N (x) of an N N random matrix A as N Applications: Nuclear Physics Number Theory

5 Random Matrix Ensembles Question: What happens when we impose structure on the entries of a matrix?

6 Random Matrix Ensembles Question: What happens when we impose structure on the entries of a matrix? To answer, study "families" or "ensembles" of random matrices:

7 Random Matrix Ensembles Question: What happens when we impose structure on the entries of a matrix? To answer, study "families" or "ensembles" of random matrices: 0.35 Real Symmetric Semicircle [Wigner, 1955]

8 Random Matrix Ensembles Question: What happens when we impose structure on the entries of a matrix? To answer, study "families" or "ensembles" of random matrices: Real Symmetric 0.4 Toeplitz Semicircle Almost Gaussian [Wigner, 1955] [HM, 05, BDJ 06]

9 Random Matrix Ensembles Question: What happens when we impose structure on the entries of a matrix? To answer, study "families" or "ensembles" of random matrices: Real Symmetric Toeplitz Palindromic Toeplitz Semicircle Almost Gaussian Gaussian [Wigner, 1955] [HM, 05, BDJ 06] [MMS, 07]

10 Our Ensemble: Signed Toeplitz and Palindromic Toeplitz Matrices Multiply each { entry of a (Palindromic) Toeplitz matrix by 1 with prob. p ɛ ij = ɛ ji = 1 with prob. 1-p so a ij = a ji = ɛ ij b i j

11 Our Ensemble: Signed Toeplitz and Palindromic Toeplitz Matrices Multiply each { entry of a (Palindromic) Toeplitz matrix by 1 with prob. p ɛ ij = ɛ ji = 1 with prob. 1-p so a ij = a ji = ɛ ij b i j Varying p allows us to continuously interpolate between: Real Symmetric at p = 1 2 (less structured) Toeplitz/Palindromic Toeplitz at p = 1 (more structured)

12 Our Ensemble: Signed Toeplitz and Palindromic Toeplitz Matrices Multiply each { entry of a (Palindromic) Toeplitz matrix by 1 with prob. p ɛ ij = ɛ ji = 1 with prob. 1-p so a ij = a ji = ɛ ij b i j Varying p allows us to continuously interpolate between: Real Symmetric at p = 1 (less structured) 2 Toeplitz/Palindromic Toeplitz at p = 1 (more structured) What is the eigenvalue distribution of these signed ensembles?

13 Markov s Method of Moments We show µ A,N (x) converges on average to a probability distribution P by controlling convergence of average moments of the measures to the moments of P.

14 Markov s Method of Moments We show µ A,N (x) converges on average to a probability distribution P by controlling convergence of average moments of the measures to the moments of P. By putting a unit point mass δ(x x 0 ) at x 0 so f (x)δ(x x0 )dx = f (x 0 ), we have: µ A,N (x) = 1 N N i=1 ( δ x λ ) i(a) 2 N

15 Markov s Method of Moments We show µ A,N (x) converges on average to a probability distribution P by controlling convergence of average moments of the measures to the moments of P. By putting a unit point mass δ(x x 0 ) at x 0 so f (x)δ(x x0 )dx = f (x 0 ), we have: µ A,N (x) = 1 N N i=1 which will then have k th moment: x k µ A,N (x) dx = 1 N λ i (A) k ( N i=1 2 N ( δ x λ ) i(a) 2 N ) k

16 Markov s Method of Moments We show µ A,N (x) converges on average to a probability distribution P by controlling convergence of average moments of the measures to the moments of P. By putting a unit point mass δ(x x 0 ) at x 0 so f (x)δ(x x0 )dx = f (x 0 ), we have: µ A,N (x) = 1 N N i=1 ( δ x λ ) i(a) 2 N which will then have k th moment: x k µ A,N (x) dx = 1 N λ i (A) k ( N i=1 2 ) k = Trace ( A k) N 2 k N k 2 +1

17 The average k th moment, M k (N) = E [M N,k (A N )] is thus: 1 N k i 1,...,i k N E ( ) ɛ i1 i 2 b i1 i 2 ɛ i2 i 3 b i2 i 3... ɛ ik i 1 b ik i 1 We look at groups of the N k terms in the sum, "configurations," that all have the same contribution.

18 The average k th moment, M k (N) = E [M N,k (A N )] is thus: 1 N k i 1,...,i k N E ( ) ɛ i1 i 2 b i1 i 2 ɛ i2 i 3 b i2 i 3... ɛ ik i 1 b ik i 1 We look at groups of the N k terms in the sum, "configurations," that all have the same contribution. What is their contribution? How many terms have this configuration?

19 The average k th moment, M k (N) = E [M N,k (A N )] is thus: 1 N k i 1,...,i k N E ( ) ɛ i1 i 2 b i1 i 2 ɛ i2 i 3 b i2 i 3... ɛ ik i 1 b ik i 1 We look at groups of the N k terms in the sum, "configurations," that all have the same contribution. What is their contribution? How many terms have this configuration? Preliminary Result: The b s must be matched in pairs to contribute in the limit.

20 Thus:

21 Thus: Odd moments vanish.

22 Thus: Odd moments vanish. For the even moments M 2k we can represent each contributing term as a pairing of 2k vertices on a circle as follows: i 5 i 6 i 6 i 1 i 4 i 5 i 1 i 2 i 3 i 4 i 2 i 3

23 Weighted Contributions Theorem: Each configuration contributes its unsigned case contribution weighted by (2p 1) 2m, where 2m is the number of vertices involved in at least one crossing.

24 Weighted Contributions Theorem: Each configuration contributes its unsigned case contribution weighted by (2p 1) 2m, where 2m is the number of vertices involved in at least one crossing. Example: 2m = 4 2m=6 2m=8

25 Weighted Contributions Theorem: Each configuration contributes its unsigned case contribution weighted by (2p 1) 2m, where 2m is the number of vertices involved in at least one crossing. Example: 2m = 4 2m=6 2m=8 Semicircle: Only non-crossing configurations contribute 1 Gaussian: All configurations contribute 1

26 Counting Crossing Configurations Problem: Out of the (2k 1)!! ways to pair 2k vertices, how many will have 2m vertices crossing (Cross 2k,2m )?

27 Counting Crossing Configurations Problem: Out of the (2k 1)!! ways to pair 2k vertices, how many will have 2m vertices crossing (Cross 2k,2m )? Example: Cross 8,4 = 28 x 8 x 4 x 8 x 8

28 Counting Crossing Configurations Problem: Out of the (2k 1)!! ways to pair 2k vertices, how many will have 2m vertices crossing (Cross 2k,2m )? Example: Cross 8,4 = 28 x 8 x 4 x 8 x 8 Fact: Cross 2k,0 = C k, the k th Catalan number.

29 Counting Crossing Configurations Problem: Out of the (2k 1)!! ways to pair 2k vertices, how many will have 2m vertices crossing (Cross 2k,2m )? Example: Cross 8,4 = 28 x 8 x 4 x 8 x 8 Fact: Cross 2k,0 = C k, the k th Catalan number. What about for higher m?

30 Counting Crossing Configurations: Non-Crossing Regions Theorem: Suppose 2m vertices are already paired in some configuration. The number of ways to pair and place the remaining 2k 2m vertices such that none of them are involved in a crossing is ( 2k k m).

31 Counting Crossing Configurations: Non-Crossing Regions Theorem: Suppose 2m vertices are already paired in some configuration. The number of ways to pair and place the remaining 2k 2m vertices such that none of them are involved in a crossing is ( 2k k m). s 6 s 1 s 5 s 2 s 4 s 3

32 Counting Crossing Configurations: Non-Crossing Regions Theorem: Suppose 2m vertices are already paired in some configuration. The number of ways to pair and place the remaining 2k 2m vertices such that none of them are involved in a crossing is ( 2k k m). s 6 s 1 s 5 s 2 s 4 s 3 Proof: s 1 + +s 2m =2k 2m C s 1 C s2 C s2m = ( 2k k m).

33 Counting Crossing Configurations Open Question: If we require 2m crossing vertices, how many ways can we divide up the 2k vertices into non-crossing regions?

34 Counting Crossing Configurations Open Question: If we require 2m crossing vertices, how many ways can we divide up the 2k vertices into non-crossing regions? We solved for small m, then applying our Non-Crossing Regions Theorem gives: Cross 2k,4 = ( ) 2k k 2 Cross 2k,6 = 4 ( ) 2k k 3 Cross 2k,8 = 31 ( ) 2k k k 5 ( 2k ) 2 i=0 i (2k 2i) Cross 2k,10 = 288 ( ) 2k k k 6 ( 2k ) i=0 i (2k 2i)

35 Summary of Results p = 1 : Semicircle Distribution (Bounded Support) 2 p 1 : Unbounded Support 2

36 Summary of Results p = 1 : Semicircle Distribution (Bounded Support) 2 p 1 : Unbounded Support 2 Some progress toward exact moment formulas:

37 Summary of Results p = 1 : Semicircle Distribution (Bounded Support) 2 p 1 : Unbounded Support 2 Some progress toward exact moment formulas: Weight of each configuration as a function of p and the number of vertices in a crossing (2m): (2p 1) 2m

38 Summary of Results p = 1 : Semicircle Distribution (Bounded Support) 2 p 1 : Unbounded Support 2 Some progress toward exact moment formulas: Weight of each configuration as a function of p and the number of vertices in a crossing (2m): (2p 1) 2m Formulas for the number of configurations with 2m vertices crossing for small m

39 Summary of Results p = 1 : Semicircle Distribution (Bounded Support) 2 p 1 : Unbounded Support 2 Some progress toward exact moment formulas: Weight of each configuration as a function of p and the number of vertices in a crossing (2m): (2p 1) 2m Formulas for the number of configurations with 2m vertices crossing for small m Limiting behavior of the mean and variance of the moments, giving bounds for the moments

40 Many thanks to: AMS/MAA Williams College, SMALL 2011 National Science Foundation Olivia Beckwith, Professor Steven J Miller