Eigenvalue Statistics for Toeplitz and Circulant Ensembles

Transcription

1 Eigenvalue Statistics for Toeplitz and Circulant Ensembles Murat Koloğlu 1, Gene Kopp 2, Steven J. Miller 1, and Karen Shen 3 1 Williams College 2 University of Michigan 3 Stanford University IMS-APRM2012 meeting Tsukuba, Japan, July 3, 2012

2 Goals See how the structure of the ensembles affects limiting behavior. Discuss the tools and techniques needed to prove the results.

3 Real Symmetric Toeplitz Matrices Chris Hammond and Steven J. Miller

4 Toeplitz Ensembles Toeplitz matrix is of the form b 0 b 1 b 2 b N 1 b 1 b 0 b 1 b N 2 b 2 b 1 b 0 b N b 1 N b 2 N b 3 N b 0 Will consider Real Symmetric Toeplitz matrices. Main diagonal zero, N 1 independent parameters. Normalize Eigenvalues by N.

5 5 Toeplitz PT HPT Block Circulant Weighted Toeplitz Refs Appendices Eigenvalue Density Measure µ A,N (x)dx = 1 N N i=1 ( δ x λ ) i(a) dx. N The k th moment of µ A,N (x) is Let M k (A, N) = 1 N k 2 +1 M k N i=1 have M 2 = 1 and M 2k+1 = 0. λ k i (A) = Trace(Ak ) N k = lim N E A [M k (A, N)];

6 Even Moments M 2k (N) = 1 N k+1 1 i 1,,i 2k N Main Term: b j s matched in pairs, say E(b i1 i 2 b i2 i 3 b i2k i 1 ). b im i m+1 = b in i n+1, x m = i m i m+1 = i n i n+1. Two possibilities: i m i m+1 = i n i n+1 or i m i m+1 = (i n i n+1 ). (2k 1)!! ways to pair, 2 k choices of sign.

7 7 Toeplitz PT HPT Block Circulant Weighted Toeplitz Refs Appendices Main Term: All Signs Negative (else lower order contribution) M 2k (N) = 1 N k+1 1 i 1,,i 2k N E(b i1 i 2 b i2 i 3 b i2k i 1 ). Let x 1,...,x k be the values of the i j i j+1 s, ǫ 1,...,ǫ k the choices of sign. Define x 1 = i 1 i 2, x 2 = i 2 i 3,.... i 2 = i 1 x 1 i 3 = i 1 x 1 x 2. i 1 = i 1 x 1 x 2k k x x 2k = (1+ǫ j )η j x j = 0, η j = ±1. j=1

8 Even Moments: Summary Main Term: paired, all signs negative. M 2k (N) (2k 1)!!+O k ( 1 N ). Bounded by Gaussian.

9 The Fourth Moment ij ij ij li jk jk jk li li kl kl kl M 4 (N) = 1 E(b N 3 i1 i 2 b i2 i 3 b i3 i 4 b i4 i 1 ) Let x j = i j i j+1. 1 i 1,i 2,i 3,i 4 N

10 The Fourth Moment Case One: x 1 = x 2, x 3 = x 4 : i 1 i 2 = (i 2 i 3 ) and i 3 i 4 = (i 4 i 1 ). Implies i 1 = i 3, i 2 and i 4 arbitrary. Left with E[b 2 x 1 b 2 x 3 ]: N 3 N times get 1, N times get p 4 = E[b 4 x 1 ]. Contributes 1 in the limit.

11 The Fourth Moment M 4 (N) = 1 E(b N 3 i1 i 2 b i2 i 3 b i3 i 4 b i4 i 1 ) 1 i 1,i 2,i 3,i 4 N Case Two: Diophantine Obstruction: x 1 = x 3 and x 2 = x 4. i 1 i 2 = (i 3 i 4 ) and i 2 i 3 = (i 4 i 1 ). This yields i 1 If i 2, i 4 2N 3 choices. = i 2 + i 4 i 3, i 1, i 2, i 3, i 4 {1,...,N}. and i 3 < N 3, i 1 > N: at most ( )N3 valid

12 The Fourth Moment Theorem: Fourth Moment: Let p 4 be the fourth moment of p. Then M 4 (N) = 2 2 ( ) O p 4. N 500 Toeplitz Matrices,

13 Main Result Theorem: HM 05 For real symmetric Toeplitz matrices, the limiting spectral measure converges in probability to a unique measure of unbounded support which is not the Gaussian. If p is even have strong convergence).

14 Not rescaled. Looking at middle 11 spacings, 1000 Toeplitz matrices ( ), entries iidrv from the standard normal. Poissonian Behavior?

15 Real Symmetric Palindromic Toeplitz Matrices Adam Massey, Steven J. Miller, Jon Sinsheimer

16 Real Symmetric Palindromic Toeplitz matrices b 0 b 1 b 2 b 3 b 3 b 2 b 1 b 0 b 1 b 0 b 1 b 2 b 4 b 3 b 2 b 1 b 2 b 1 b 0 b 1 b 5 b 4 b 3 b 2 b 3 b 2 b 1 b 0 b 6 b 5 b 4 b b 3 b 4 b 5 b 6 b 0 b 1 b 2 b 3 b 2 b 3 b 4 b 5 b 1 b 0 b 1 b 2 b 1 b 2 b 3 b 4 b 2 b 1 b 0 b 1 b 0 b 1 b 2 b 3 b 3 b 2 b 1 b 0 Extra symmetry fixes Diophantine Obstructions. Always have eigenvalue at 0.

17 Results Theorem: MMS 07 For real symmetric palindromic matrices, converge in probability to the Gaussian (if p is even have strong convergence).

18 Results Theorem: MMS 07 Let X 0,...,X N 1 be iidrv (with X j = X N j ) from a distribution p with mean 0, variance 1, and finite higher moments. For ω = (x 0, x 1,...) set X l (ω) = x l, and S (k) N (ω) = 1 N 1 N l=0 X l (ω) cos(2πkl/n). Then as n ({ Prob ω Ω : sup x R 1 N N 1 I (k) S Φ(x) N (ω) x k=0 }) 0 = 1; I the indicator fn, Φ CDF of standard normal.

19 Real Symmetric Highly Palindromic Toeplitz Matrices Steven Jackson, Victor Luo, Steven J. Miller, Vincent Pham, Nicholas George Triantafillou

20 Notation: Real Symmetric Highly Palindromic Toeplitz matrices For fixed n, we consider N N real symmetric Toeplitz matrices in which the first row is 2 n copies of a palindrome, entries are iidrv from a p with mean 0, variance 1 and finite higher moments. For instance, a doubly palindromic Toeplitz matrix is of the form: b 0 b 1 b 1 b 0 b 0 b 1 b 1 b 0 b 1 b 0 b 2 b 1 b 0 b 0 b 2 b 1 b 2 b 1 b 3 b 2 b 1 b 0 b 3 b 2 A N = b 2 b 3 b 0 b 1 b 2 b 3 b 1 b 2 b 1 b 2 b 0 b 0 b 1 b 2 b 0 b 1 b 0 b 1 b 1 b 0 b 0 b 1 b 1 b 0

21 Main Results Theorem: JMP 12 Let n be a fixed positive integer, N a multiple of 2 n, consider the ensemble of real symmetric N N palindromic Toeplitz matrices whose first row is 2 n copies of a fixed palindrome (independent entries iidrv from p with mean 0, variance 1 and finite higher moments). 1 As N the measures µ n,an converge in probability to a limiting spectral measure which is even and has unbounded support. 2 If p is even, then converges almost surely. 3 The limiting measure has fatter tails than the Gaussian (or any previously seen distribution).

22 Work in Progress (with Victor Luo and Nicholas Triantafillou) Highly Palindromic Real Symmetric: all matchings contribute equally for fourth moment, conjectured equally in general. Highly Palindromic Hermitian: matchings do not contribute equally: fourth moment non-adjacent case is 1 3 (2n + 2 n ), while the adjacent case is 1 2 (2n + 2 n ).

23 Block Circulant Matrices Murat Koloğlu, Gene Kopp and Steven J. Miller

24 Choose distinct entries i.i.d.r.v. The Ensemble of m-block Circulant Matrices Study symmetric matrices periodic with period m on wrapped diagonals, i.e., symmetric block circulant matrices. 8-by-8 real symmetric 2-block circulant matrix: c 0 c 1 c 2 c 3 c 4 d 3 c 2 d 1 c 1 d 0 d 1 d 2 d 3 d 4 c 3 d 2 c 2 d 1 c 0 c 1 c 2 c 3 c 4 d 3 c 3 d 2 c 1 d 0 d 1 d 2 d 3 d 4. c 4 d 3 c 2 d 1 c 0 c 1 c 2 c 3 d 3 d 4 c 3 d 2 c 1 d 0 d 1 d 2 c 2 c 3 c 4 d 3 c 2 d 1 c 0 c 1 d 1 d 2 d 3 d 4 c 3 d 2 c 1 d 0

25 Oriented Matchings and Dualization Compute moments of eigenvalue distribution (as m stays fixed and N ) using the combinatorics of pairings. Rewrite: M n (N) = = 1 N n i 1,...,i n N E(a i1 i 2 a i2 i 3 a ini 1 ) 1 η( )m N n 2 +1 d1 ( ) m dl ( ). where the sum is over oriented matchings on the edges {(1, 2),(2, 3),...,(n, 1)} of a regular n-gon.

26 Oriented Matchings and Dualization c 0 c 1 c 2 c 3 c 4 d 3 c 2 d 1 c 1 d 0 d 1 d 2 d 3 d 4 c 3 d 2 c 2 d 1 c 0 c 1 c 2 c 3 c 4 d 3 c 3 d 2 c 1 d 0 d 1 d 2 d 3 d 4 c 4 d 3 c 2 d 1 c 0 c 1 c 2 c 3 d 3 d 4 c 3 d 2 c 1 d 0 d 1 d 2 c 2 c 3 c 4 d 3 c 2 d 1 c 0 c 1 d 1 d 2 d 3 d 4 c 3 d 2 c 1 d 0 Figure: A matching in the expansion for M n (N) = M 6 (8).

27 Oriented Matchings and Dualization c 0 c 1 c 2 c 3 c 4 d 3 c 2 d 1 c 1 d 0 d 1 d 2 d 3 d 4 c 3 d 2 c 2 d 1 c 0 c 1 c 2 c 3 c 4 d 3 c 3 d 2 c 1 d 0 d 1 d 2 d 3 d 4 c 4 d 3 c 2 d 1 c 0 c 1 c 2 c 3 d 3 d 4 c 3 d 2 c 1 d 0 d 1 d 2 c 2 c 3 c 4 d 3 c 2 d 1 c 0 c 1 d 1 d 2 d 3 d 4 c 3 d 2 c 1 d 0 Figure: An oriented matching in the expansion for M n (N) = M 6 (8).

28 Contributing Terms As N, the only terms that contribute to this sum are those in which the entries are matched in pairs and with opposite orientation.

29 Only Topology Matters Think of pairings as topological identifications; the contributing ones give rise to orientable surfaces. Contribution from such a pairing is m 2g, where g is the genus (number of holes) of the surface. Proof: combinatorial argument involving Euler characteristic.

30 Computing the Even Moments Theorem: Even Moment Formula M 2k = k/2 g=0 ε g (k)m 2g + O k ( 1 N ), with ε g (k) the number of pairings of the edges of a (2k)-gon giving rise to a genus g surface. J. Harer and D. Zagier (1986) gave generating functions for the ε g (k).

31 Harer and Zagier where k/2 g=0 1+2 ε g (k)r k+1 2g = (2k 1)!! c(k, r) c(k, r)x k+1 = k=0 ( ) r 1+x. 1 x Thus, we write M 2k = m (k+1) (2k 1)!! c(k, m).

32 A multiplicative convolution and Cauchy s residue formula yield the characteristic function of the distribution. φ(t) = k=0 (it) 2k M 2k (2k)!

33 A multiplicative convolution and Cauchy s residue formula yield the characteristic function of the distribution. φ(t) = k=0 (it) 2k M 2k (2k)! = 1 m ( t 2 /2m) k c(k, m) k! k=0

34 A multiplicative convolution and Cauchy s residue formula yield the characteristic function of the distribution. φ(t) = = (it) 2k M 2k (2k)! k=0 1 2πim z =2 = 1 ( t 2 /2m) k c(k, m) m k! k=0 ( (1+z ) 1 m 1) 1 2z 1 1 z 1 e t2 z/2m dz z

35 A multiplicative convolution and Cauchy s residue formula yield the characteristic function of the distribution. φ(t) = = (it) 2k M 2k (2k)! k=0 1 2πim = 1 m e t2 2m z =2 2z 1 m l=1 ( m l = 1 ( t 2 /2m) k c(k, m) m k! k=0 ( (1+z ) 1 m 1 1) 1 z 1 ) ( ) 1 t 2 l 1 (l 1)! m e t2 z/2m dz z

36 Results Fourier transform and algebra yields Theorem: Koloğlu, Kopp and Miller The limiting spectral density function f m (x) of the real symmetric m-block circulant ensemble is given by the formula f m (x) = mx 2 e 2 m 2πm r=0 1 m r (2r)! (2r + 2s)! (r + s)!s! s=0 ( m r + s + 1 ) ( 1 2) s (mx 2 ) r. As m, the limiting spectral densities approach the semicircle distribution.

37 Results (continued) Figure: Plot for f 1 and histogram of eigenvalues of 100 circulant matrices of size

38 Results (continued) Figure: Plot for f 2 and histogram of eigenvalues of block circulant matrices of size

39 Results (continued) Figure: Plot for f 3 and histogram of eigenvalues of block circulant matrices of size

40 Results (continued) Figure: Plot for f 4 and histogram of eigenvalues of block circulant matrices of size

41 Results (continued) Figure: Plot for f 8 and histogram of eigenvalues of block circulant matrices of size

42 Results (continued) Figure: Plot for f 20 and histogram of eigenvalues of block circulant matrices of size

43 Results (continued) Figure: Plot of convergence to the semi-circle.

44 Weighted Real Symmetric Toeplitz Matrices Olivia Beckwith, Steven J. Miller and Karen Shen

45 New Ensemble: Signed Toeplitz and Palindromic Toeplitz Matrices For each entry, multiply by a randomly chosen ǫ ij = {1, 1} with p = P(ǫ ij = 1) such that ǫ ij = ǫ ji.

46 New Ensemble: Signed Toeplitz and Palindromic Toeplitz Matrices For each entry, multiply by a randomly chosen ǫ ij = {1, 1} with p = P(ǫ ij = 1) such that ǫ ij = ǫ ji. Varying p allows us to continuously interpolate between: Real Symmetric at p = 1 2 (less structured) Unsigned Toeplitz/Palindromic Toeplitz at p = 1 (more structured)

47 New Ensemble: Signed Toeplitz and Palindromic Toeplitz Matrices For each entry, multiply by a randomly chosen ǫ ij = {1, 1} with p = P(ǫ ij = 1) such that ǫ ij = ǫ ji. Varying p allows us to continuously interpolate between: Real Symmetric at p = 1 (less structured) 2 Unsigned Toeplitz/Palindromic Toeplitz at p = 1 (more structured) What is the eigenvalue distribution of these signed ensembles?

48 Weighted Contributions Theorem: Each configuration weighted by (2p 1) 2m, where 2m is the number of points on the circle whose edge crosses another edge. Example: 2m = 4 2m=6 2m=8

49 Proof of Weighted Contributions Theorem We compute the average k th moment to be: 1 N k i 1,...,i k N where the b s are matched in pairs. E ( ) ǫ i1 i 2 b i1 i 2 ǫ i2 i 3 b i2 i 3...ǫ ik i 1 b ik i 1

50 Proof of Weighted Contributions Theorem We compute the average k th moment to be: 1 N k i 1,...,i k N where the b s are matched in pairs. E ( ) ǫ i1 i 2 b i1 i 2 ǫ i2 i 3 b i2 i 3...ǫ ik i 1 b ik i 1 If ǫ ij is matched with some ǫ kl, then E(ǫ ij ǫ kl ) = 1.

51 Proof of Weighted Contributions Theorem We compute the average k th moment to be: 1 N k i 1,...,i k N where the b s are matched in pairs. E ( ) ǫ i1 i 2 b i1 i 2 ǫ i2 i 3 b i2 i 3...ǫ ik i 1 b ik i 1 If ǫ ij is matched with some ǫ kl, then E(ǫ ij ǫ kl ) = 1. If ǫ ij is not matched with any ǫ kl, then E(ǫ ij ) = (2p 1).

52 Proof of Weighted Contributions Theorem We compute the average k th moment to be: 1 N k i 1,...,i k N where the b s are matched in pairs. E ( ) ǫ i1 i 2 b i1 i 2 ǫ i2 i 3 b i2 i 3...ǫ ik i 1 b ik i 1 If ǫ ij is matched with some ǫ kl, then E(ǫ ij ǫ kl ) = 1. If ǫ ij is not matched with any ǫ kl, then E(ǫ ij ) = (2p 1). Can show two ǫ s are matched if and only if their b s are not in a crossing.

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

54 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

55 55 Toeplitz PT HPT Block Circulant Weighted Toeplitz Refs Appendices 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.

56 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?

57 57 Toeplitz PT HPT Block Circulant Weighted Toeplitz Refs Appendices Counting Crossing Configurations To calculate Cross 2k,2m, we write it as the following sum: m 4 Cross 2k,2m = P 2k,2m,p. p=1 where P 2k,2m,p is the number of configurations of 2k vertices with 2m vertices crossing in p partitions.

58 Counting Crossing Configurations To calculate Cross 2k,2m, we write it as the following sum: m 4 Cross 2k,2m = P 2k,2m,p. p=1 where P 2k,2m,p is the number of configurations of 2k vertices with 2m vertices crossing in p partitions. For example: p = 1 p = 2

59 Non-Crossing Regions Theorem: If 2m vertices are already paired, the number of ways to pair and place the remaining 2k 2m vertices as non-crossing non-partitioning edges is ( 2k k m). Example: ( 8 2) = 28 pairings with 4 crossing vertices. x 8 x 4 x 8 x 8

60 Non-Crossing Regions Theorem: If 2m vertices are already paired, the number of ways to pair and place the remaining 2k 2m vertices as non-crossing non-partitioning edges is ( 2k k m). Example: ( 8 2) = 28 pairings with 4 crossing vertices. x 8 x 4 x 8 x 8 Lemma: P 2k,2m,1 = Cross 2m,2m ( 2k k m).

61 Proof of Non-Crossing Regions Theorem We showed the following equivalence: s 1 +s 2 + +s 2m =2k 2m C s1 C s2 C s2m = ( ) 2k. k m s 6 s 1 s 5 s 2 s 4 s 3

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

63 Summary of Results p = 1 : Semicircle Distribution (Bounded Support) 2 p 1 : Unbounded Support 2 Some progress towards exact formulas for the moments, from which we can recover the distribution

64 Summary of Results p = 1 : Semicircle Distribution (Bounded Support) 2 p 1 : Unbounded Support 2 Some progress towards exact formulas for the moments, from which we can recover the distribution Weight of each configuration as a function of p and the number of vertices in a crossing (2m): (2p 1) 2m

65 Summary of Results p = 1 : Semicircle Distribution (Bounded Support) 2 p 1 : Unbounded Support 2 Some progress towards exact formulas for the moments, from which we can recover the distribution Weight of each configuration as a function of p and the number of vertices in a crossing (2m): (2p 1) 2m A way to count the number of configurations with 2m vertices crossing for small m

66 Summary of Results p = 1 : Semicircle Distribution (Bounded Support) 2 p 1 : Unbounded Support 2 Some progress towards exact formulas for the moments, from which we can recover the distribution Weight of each configuration as a function of p and the number of vertices in a crossing (2m): (2p 1) 2m A way to count the number of configurations with 2m vertices crossing for small m Tight bounds on the moments in the limit The expected number of vertices involved in a crossing is ( 2k 2k 2 2 F 1 (1, 3 2, 5 2 k; 1) ) (2k 1) 2 F 1 (1, 1 2k 1 2k k, 3 2 ; 1), which is 2k 2 2 k + O( 1 k 2 ) as k. The variance tends to 4 as k.

67 References

68 C. Hammond and S. J. Miller, Distribution of eigenvalues for the ensemble of real symmetric Toeplitz matrices, Journal of Theoretical Probability 18 (2005), no. 3, A. Massey, S. J. Miller and J. Sinsheimer, Distribution of eigenvalues of real symmetric palindromic Toeplitz matrices and circulant matrices, Journal of Theoretical Probability 20 (2007), no. 3, S. Jackson, S. J. Miller and V. Pham, Distribution of eigenvalues for highly palindromic real symmetric Toeplitz matrices, Journal of Theoretical Probability 25 (2012), M. Koloğlu, G. S. Kopp, S. J. Miller, F. W. Strauch and W. Xiong, The Limiting Spectral Measure for Ensembles of Symmetric Block Circulant Matrices, to appear in the Journal of Theoretical Probability.

69 Appendices

70 Higher Toeplitz Moments: Brute Force For sixth moment, five configurations occurring (respectively) 2, 6, 3, 3 and 1 times. M 6 (N) = 11 (Gaussian s is 15). M 8 (N) = 64 4 (Gaussian s is 105). 15 Lemma: For 2k 4, lim N M 2k (N) < (2k 1)!!.

71 Higher Toeplitz Moments: Unbounded support Lemma: Moments growth implies unbounded support. Proof: Main idea: i 2 = i 1 x 1 i 3 = i 1 x 1 x 2. i 2k = i 1 x 1 x 2k. Once specify i 1 and x 1 through x 2k, all indices fixed. If matched in pairs and each i j {1,...,N}, have a valid configuration, contributes +1. Problem: a running sum i 1 x 1 x m {1,...,N}. Lots of freedom in locating positive and negative signs, use CLT to show most configurations are valid.

72 Real Symmetric Palindromic Toeplitz Real Symmetric Palindromic Toeplitz, Note the bump at the zeroth bin is due to the forced eigenvalues at 0.

73 Palindromic Toeplitz: Effects of Palindromicity on Matchings a imi m+1 paired with a ini n+1 implies one of the following hold: i m+1 i m = ±(i n+1 i n ) i m+1 i m = ±(i n+1 i n )+(N 1) i m+1 i m = ±(i n+1 i n ) (N 1). Concisely: There is a C {0, ±(N 1)} such that i m+1 i m = ±(i n+1 i n )+C.

74 Palindromic Toeplitz: Fourth Moment Highlights the effect of palindromicity. Still matched in pairs, but more diagonals now lead to valid matchings.

75 75 Toeplitz PT HPT Block Circulant Weighted Toeplitz Refs Appendices Palindromic Toeplitz: Fourth Moment Highlights the effect of palindromicity. Still matched in pairs, but more diagonals now lead to valid matchings. Non-adjacent case was x 1 = x 3 and x 2 = x 4 : This yields i 1 i 2 = (i 3 i 4 ) and i 2 i 3 = (i 4 i 1 ). i 1 = i 2 + i 4 i 3, i 1, i 2, i 3, i 4 {1,...,N}.

76 Palindromic Toeplitz: Fourth Moment Highlights the effect of palindromicity. Still matched in pairs, but more diagonals now lead to valid matchings. Non-adjacent case now x 1 = x 3 and x 2 = x 4 : j i = (l k)+c 1 k j = (i l)+c 2, or equivalently j = i + k l + C 1 = i + k l C 2. We see that C 1 = C 2, or C 1 + C 2 = 0.

77 77 Toeplitz PT HPT Block Circulant Weighted Toeplitz Refs Appendices Highly Palindromic: Key Lemmas Much of analysis similar to previous ensembles (though combinatorics more involved). For the fourth moment: both the adjacent and non-adjacent matchings contribute the same. Lemma: As N the fourth moment tends to M 4,n = 2 n n. Note: Number of palindromes is 2 n ; thus smallest is 2 0 = 1 (and do recover 3 for palindromic Toeplitz).

78 Highly Palindromic: Conjectures Conjecture In the limit, all matchings contribute equally. Very hard to test; numerics hard to analyze. To avoid simulating ever-larger matrices, noticed Diophantine analysis suggests average 2m th moment of N N matrices should satisfy M 2m,n;N = M 2m,n + C 1,n N + C 2,n N + + C m,n 2 N. m Instead of simulating prohibitively large matrices, simulate large numbers of several sizes of smaller matrices, do a least squares analysis to estimate M 2m,n.

79 Highly Palindromic: Conjectures Table: Conjectured and observed moments for 1000 real symmetric doubly palindromic Toeplitz matrices. The conjectured values come from assuming Conjecture. Moment Conjectured Observed Observed/Predicted

80 Highly Palindromic: Conjectures Table: Observed moments for doubly palindromic Toeplitz matrices. Conjectured values from assuming Conjecture. N #sims 2nd 4th 6th 8th 10th 8 1,000, ,000, ,000, ,000, ,000, ,000, , , , , , , , , , , , , Conjectured Best Fit M 2m,

81 Weighted Toeplitz: Proof of Weighted Contributions Theorem A non-crossing pair of b s must have matched ǫs: Assume b ir i r+1 and b ip i p+1 are a non-crossing pair. i r i r 1 p k=r (i k i k+1 ) = 0 = i r i r+1 + i r+1 +i p i p+1 = i r i p+1 i p i p 1 This implies that i r = i p+1. Similarly, i r+1 = i p Thus, ǫ ir i r+1 = ǫ ipi p+1.

82 Weighted Toeplitz: Proof of Weighted Contributions Theorem A matched pair of ǫ s must not be in a crossing: Suppose ǫ iai a+1 = ǫ ib i b+1, with a < b. b (i k i k+1 ) = i a i b+1 = 0 k=a = d δ k i k i k+1 k=b where δ k = 0 if and only if the vertex k is paired with is between a and b. Need N k+1 degrees of freedom, so δ k = 0 for all k. Thus, ǫ iai a+1 and ǫ ib i b+1 are not in a crossing.