Asymptotic moments of random Vandermonde matrix March Boedihardjo - PowerPoint PPT Presentation

Asymptotic moments of random Vandermonde matrix March Boedihardjo Joint work with Ken Dykema Texas A&M University March 2016

Vandermonde matrix x n − 1  x 2  1 x 1 . . . 1 1 x n − 1 x 2 1 x 2 . . .   2 2  . . . .  ... . . . .   . . . .   x 2 x n − 1 1 x m . . . m m If m = n then the determinant is � ( x j − x i ) . 1 ≤ i < j ≤ n Vandermonde matrix can be used to find the interpolating polynomial with given data.

Random Vandermonde matrix ζ 2 ζ N − 1   1 ζ 1 . . . 1 1 ζ N − 1 ζ 2 1 ζ 2 . . . 1  2  2 X N = √  ,  . . . .  ... . . . .   N . . . .  ζ N − 1 ζ 2 1 ζ N . . . N N where ζ 1 , . . . , ζ N are i.i.d. random variables uniformly distributed on the unit circle.

The rows of X N are i.i.d. copies of  1  ζ     ζ 2 v = ,     . .   .   ζ N − 1 where ζ is uniformly distributed on the unit circle. The random vector is isotropic: E |� v , x �| 2 = � x � 2 x ∈ C N . 2 ,

f 1 , . . . , f N are i.i.d. real random variables with mean 0, variance 1 and uniformly bounded moments. 2 2 � � � � N N � � � � � � ζ k E f k = N E = N � � � � � � � � � k =1 � � k =1 � 4 4 � N � � N � ∼ 2 � � � � � � ∼ 3 N 2 ζ k 3 N 3 E f k E � � � � � � � � � � � � k =1 k =1 2 p 2 p � � � � N N � � � � � � = O ( N 2 p − 1 ) = O ( N p ) ζ k E f k E � � � � � � � � � � � � k =1 k =1

To compute 4 � N � � � � f k , E � � � � � � k =1 we expand it as N � E f k 1 f k 2 f k 3 f k 4 . k 1 , k 2 , k 3 , k 4 =1 We consider all partitions on { 1 , 2 , 3 , 4 } . Only pair partitions contribute. There are 3 pair partitions so 4 � � N � � � ∼ 3 N 2 . E f k � � � � � � k =1

To compute 4 � � N � � � ζ k E , � � � � � � k =1 we expand it as N N E ζ k 1 ζ − k 2 ζ k 3 ζ − k 4 = � � E ζ k 1 − k 2 + k 3 − k 4 . k 1 , k 2 , k 3 , k 4 =1 k 1 , k 2 , k 3 , k 4 =1 � 1 , k 1 − k 2 + k 3 − k 4 = 0 E ζ k 1 − k 2 + k 3 − k 4 = . 0 , Otherwise

So N E ζ k 1 − k 2 + k 3 − k 4 = |{ ( k 1 , k 2 , k 3 , k 4 ) ∈ { 1 , . . . , N } 4 : � k 1 , k 2 , k 3 , k 4 =1 k 1 − k 2 + k 3 − k 4 = 0 }| . The limit 1 N 3 |{ ( k 1 , k 2 , k 3 , k 4 ) ∈ { 1 , . . . , N } 4 : lim N →∞ k 1 − k 2 + k 3 − k 4 = 0 }| is given by Vol 3 { ( t 1 , t 2 , t 3 , t 4 ) ∈ [0 , 1] 4 : t 1 − t 2 + t 3 − t 4 = 0 } = 2 3 .

Therefore, N E ζ k 1 − k 2 + k 3 − k 4 ∼ 2 � 3 N 3 . k 1 , k 2 , k 3 , k 4 =1 So 4 � � N ∼ 2 � � � 3 N 3 ζ k E � � � � � � k =1

Random Vandermonde matrix ζ 2 ζ N − 1   1 ζ 1 . . . 1 1 ζ N − 1 ζ 2 1 ζ 2 . . . 1  2  2 X N = √  ,  . . . .  ... . . . .   N . . . .  ζ N − 1 ζ 2 1 ζ N . . . N N where ζ 1 , . . . , ζ N are i.i.d. random variables uniformly distributed on the unit circle.

First moment 1 ζ j ( X N ) i , j = √ i . N 1 ( X ∗ ζ − i N ) i , j = √ j N Thus N 1 � E ◦ tr X ∗ E ( X ∗ = N ) i (1) , i (2) ( X N ) i (2) , i (1) N X N N i (1) , i (2)=1 N 1 E ζ − i (1) i (2) ζ i (1) � = i (2) = 1 . N 2 i (1) , i (2)=1 Here tr means normalized trace.

Second moment N 1 � E ζ − i (1) i (2) ζ i (3) i (2) ζ − i (3) i (4) ζ i (1) E ◦ tr ( X ∗ N X N ) 2 = i (4) N 3 i (1) , i (2) , i (3) , i (4)=1 N 1 E ζ i (3) − i (1) ζ i (1) − i (3) � = . i (2) i (4) N 3 i (1) , i (2) , i (3) , i (4)=1 Summing over i (2) = i (4), we get 1. If i (2) � = i (4) then i (1) = i (3). Summing over i (2) � = i (4), we get N ( N − 1) N = 1 − 1 N . N 3

Third moment So N X N ) 2 = 2 − 1 E ◦ tr ( X ∗ N . So N X N ) 2 → 2 E ◦ tr ( X ∗ Using the same method, we obtain N X N ) 3 → 5 . E ◦ tr ( X ∗ because there are 5 partitions on { 2 , 4 , 6 } .

Fourth moment N X N ) 4 E ◦ tr ( X ∗ N = 1 E ζ i (3) − i (1) ζ i (5) − i (3) ζ i (7) − i (5) ζ i (1) − i (7) � . N 5 i (2) i (4) i (6) i (8) i (1) ,..., i (8)=1 For each noncrossing partition π on { 2 , 4 , 6 , 8 } , summing over i (2) , i (4) , i (6) , i (8) that respect π , we get 1. There are 14 noncrossing partitions on { 2 , 4 , 6 , 8 } . So noncrossing partitions give 14.

For the crossing partition i (2) = i (6) � = i (4) = i (8), E ζ i (3) − i (1) ζ i (5) − i (3) ζ i (7) − i (5) ζ i (1) − i (7) i (2) i (4) i (6) i (8) = E ζ i (7) − i (5)+ i (3) − i (1) E ζ i (1) − i (7)+ i (5) − i (3) i (2) i (4) � 1 , i (7) − i (5) + i (3) − i (1) = 0 = . 0 , Otherwise Same as before: this gives 2 3. N X N ) 4 → 14 + 2 Therefore, E ◦ tr ( X ∗ 3.

General moments Let N X N ) p . N →∞ E ◦ tr ( X ∗ m p = lim 1. m 1 = 1, m 2 = 2, m 3 = 5, m 4 = 14 + 2 3 (Ryan, Debbah 09) 2. c p ≤ m p ≤ B p (Ryan, Debbah 09) 3. ∃ measure µ on [0 , ∞ ) of unbounded support such that (Tucci, Whiting 11) � x p d µ ( x ) , m p = p ≥ 0 .

∗ -moments Question Compute N →∞ E ◦ tr P ( X N , X ∗ lim N ) for all polynomial P in two noncommuting variables.

R -diagonality Let ( A , φ ) be a tracial ∗ -probability space. Definition (Nica and Speicher 97) a ∈ A is R-diagonal if a has the same ∗ -distribution as up where 1. u and p are ∗ -free in some ∗ -probability space ( A ′ , τ ), 2. u is a Haar unitary, i.e., τ ( u n ) = δ n =0 .

Maximal alternating interval partition Definition Let ǫ = ( ǫ 1 , . . . , ǫ p ) ∈ { 1 , ∗} p . Then σ ( ǫ ) is the interval partition on { 1 , . . . , p } determined by σ ( ǫ ) j ∼ j + 1 ⇐ ⇒ ǫ j � = ǫ j +1 . If ǫ = (1 , 1 , ∗ , 1 , ∗ , ∗ ) then σ ( ǫ ) = {{ 1 } , { 2 , 3 , 4 , 5 } , { 6 }} .

Equivalent definition Lemma (Nica, Shlyakhtenko, Speicher 01) a is R-diagonal if and only if 1. φ ( aa ∗ aa ∗ . . . aa ∗ a ) = 0 and 2. ∀ ǫ 1 , . . . , ǫ p ∈ { 1 , ∗} ,  �� �� ��  � a ǫ k − φ  = 0 . a ǫ k φ B ∈ σ ( ǫ ) k ∈ B k ∈ B Observations 1. R -diagonality completely determines the ∗ -distribution of a in terms of the distribution of a ∗ a . 2. a ∗ a and aa ∗ are free.

Is Vandermonde R-diagonal Question (Tucci) Is the asymptotic ∗ -distribution of X N R -diagonal? Affirmative reason: X N has i.i.d. rows so X N ∼ H N X N where H N is the N × N random permutation matrix independent of X N .

Question Are X ∗ N X N and X N X ∗ N asymptotically free?

By hand computation: Low moments of X ∗ N X N and X N X ∗ N coincide as if they were asymptotically free. By Matlab: When N = 10 , 000, N X N ) 4 ( X N X ∗ N ) 2 ( X ∗ N X N ) 4 ( X N X ∗ N ) 2 | E ◦ tr ( X ∗ − Value computed as if they were asymptotically free | < 0 . 05 .

X ∗ N X N and X N X ∗ N are not asymptotically free N →∞ E ◦ tr ( X ∗ N X N ) 4 ( X N X ∗ N ) 2 ( X ∗ N X N ) 4 ( X N X ∗ N ) 2 lim 1 − Value computed as if they were asymptotically free = 270 .

R -diagonality with amalgamation Let A be a unital ∗ -algebra. Let E : A → B be a conditional expectation onto a ∗ -subalgebra B . Definition (´ Sniady and Speicher 01) a ∈ A is R-diagonal with amalgamation over B if 1. E ( ab 1 a ∗ b 2 ab 3 a ∗ . . . b 2 p a ) = 0 , 2. ∀ ǫ 1 , . . . , ǫ p ∈ { 1 , ∗} ,  �� �� �� b k a ǫ k − E  �  = 0 , E b k a ǫ k B ∈ σ ( ǫ ) k ∈ B k ∈ B

Main result Theorem (B. and Dykema) ∃ ∗ -algebra A containing C [0 , 1], a conditional expectation E : A → C [0 , 1] and X ∈ A such that 1. X is R -diagonal with amalgamation over C [0 , 1] and 2. ∀ b 1 , . . . , b p ∈ C [0 , 1] and ǫ 1 , . . . , ǫ p ∈ { 1 , ∗} � p � p � 1 � � b ( N ) � � X ǫ k b k X ǫ k N →∞ E ◦ tr lim = E d λ, k N 0 k =1 k =1 where λ is the Lebesgue measure on [0 , 1] and = diag ( b k ( 1 N ) , b k ( 2 N ) , . . . , b k ( N b ( N ) N )) . k

Some C [0 , 1]-valued moments E ( XX ∗ ) = 1 , E ( XX ∗ ) 2 = 2 , E ( XX ∗ ) 3 = 5 E ( XX ∗ ) 4 = 14 + 2 3 E ( X ∗ X ) = 1 , E ( X ∗ X ) 2 = 2 , E ( X ∗ X ) 3 = 5 E ( X ∗ X ) 4 = 14 + 3 4 − ( t − 1 2) 2 , t ∈ [0 , 1] .

THANK YOU