Introduction to Random Matrix Theorymath.as.uky.edu/sites/default/files/McLaughlin-1.pdf1. Random...

1. Random matrices: intro

2. Orthogonal polynomial connection

3. Density of eigenvalues as N grows

Ken McLaughlin, Univ. of Arizona

Introduction to Random Matrix Theory

Thursday, May 15, 14

Homework:

1. Enjoy numerical simulations of random matrices using matlab

2. Understand the connection between Random Matrix Theory and Orthogonal Polynomials

3. Work out the OPs and mean density in a simple example

Thursday, May 15, 14

The basic example of random matrices: N ⇥N Hermitian matrices:

M

jj

: Gaussian random variable: Prob {Mjj

< x} = 1p2⇡

R

x

�1 e

� 12 t

2

dt

M

jk

: Complex Gaussian random variable, Mjk

= M

(R)jk

+ iM

(I)jk

, with

Probn

M

(R)jk

< x

o

= 1p⇡

R

x

�1 e

�t

2

dt

Probn

M

(I)jk

< x

o

= 1p⇡

R

x

�1 e

�t

2

dt

0

B

B

B

B

@

M11 M

(R)12 + iM

(I)12 · · · M

(R)1N + iM

(I)1N

M

(R)12 � iM

(I)12 M22 · · · M

(R)2N + iM

(I)2N

.... . .

. . ....

M

(R)1N � iM

(I)1N M

(R)2N � iM

(I)2N · · · M

NN

1

C

C

C

C

A

This is referred to as the Gaussian Unitary Ensemble

Thursday, May 15, 14

The basic example of random matrices: N ⇥N Hermitian matrices:

M

jj

: Gaussian random variable: Prob {Mjj

< x} = 1p2⇡

R

x

�1 e

� 12 t

2

dt

M

jk

: Complex Gaussian random variable, Mjk

= M

(R)jk

+ iM

(I)jk

, with

Probn

M

(R)jk

< x

o

= 1p⇡

R

x

�1 e

�t

2

dt

Probn

M

(I)jk

< x

o

= 1p⇡

R

x

�1 e

�t

2

dt

0

B

B

B

B

@

M11 M

(R)12 + iM

(I)12 · · · M

(R)1N + iM

(I)1N

M

(R)12 � iM

(I)12 M22 · · · M

(R)2N + iM

(I)2N

.... . .

. . ....

M

(R)1N � iM

(I)1N M

(R)2N � iM

(I)2N · · · M

NN

1

C

C

C

C

A

This is referred to as the Gaussian Unitary Ensemble

Thursday, May 15, 14

The basic example of random matrices: N ⇥N Hermitian matrices:

M

jj

: Gaussian random variable: Prob {Mjj

< x} = 1p2⇡

R

x

�1 e

� 12 t

2

dt

M

jk

: Complex Gaussian random variable, Mjk

= M

(R)jk

+ iM

(I)jk

, with

Probn

M

(R)jk

< x

o

= 1p⇡

R

x

�1 e

�t

2

dt

Probn

M

(I)jk

< x

o

= 1p⇡

R

x

�1 e

�t

2

dt

0

B

B

B

B

@

M11 M

(R)12 + iM

(I)12 · · · M

(R)1N + iM

(I)1N

M

(R)12 � iM

(I)12 M22 · · · M

(R)2N + iM

(I)2N

.... . .

. . ....

M

(R)1N � iM

(I)1N M

(R)2N � iM

(I)2N · · · M

NN

1

C

C

C

C

A

This is referred to as the Gaussian Unitary Ensemble

Thursday, May 15, 14

Using Matlab, we can generate random matrices and compute their eigenvalues

Thursday, May 15, 14

Using Matlab, we can generate random matrices and compute their eigenvalues

Thursday, May 15, 14

Using Matlab, we can generate random matrices and compute their eigenvalues

Thursday, May 15, 14

Using Matlab, we can generate random matrices and compute their eigenvalues

Thursday, May 15, 14

Using Matlab, we can generate random matrices and compute their eigenvalues

Thursday, May 15, 14

N=1

N=2

N=3

N=4

N=8

N=12

N=16

N=20

Thursday, May 15, 14

N = 200

N = 400

N = 600

N = 800

Thursday, May 15, 14

The basic example of random matrices: N ⇥N Hermitian matrices:

Mjj : Gaussian random variable, 1⇤2�

e�12 M2

jj dMjj

Mjk: Complex Gaussian random variable, 1� e�|Mjk|2dM (R)

jk dM (I)jk

Joint prob. measure:1

#Nexp

⇤�Tr

�12M2

⇥⌅dM,

dM =⇧

j<k

dM (R)jk dM (I)

jk

N⇧

j=1

dMjj

This is referred to as the Gaussian Unitary Ensemble

Thursday, May 15, 14

Support seems to grow like

pN ...

Rescale the matrices M 7! N1/2M :

Joint prob. measure:

1

#Nexp

⇢�N Tr

1

2

M2

��dM,

dM =

Y

j<k

dM (R)jk dM (I)

jk

NY

j=1

dMjj

Still unitarily invarant...

Thursday, May 15, 14

Thursday, May 15, 14

There are many di↵erent matrix models.

1

ˆZN

exp {�NTr [V (M)]} dM

1

ˆZN

exp

⇢�NTr

1

2

A2+B2

+ C2 � it (ABC +ACB)

��dA dB dC

Example 3: A coupled “Matrix Model”

Example 2: N ⇥N Symmetric matrices

Thursday, May 15, 14

Example 4: Normal Matrix Model:

A weaker symmetry requirement: [M,M

⇤] = 0.

Given Q = Q(x, y) = Q(z, z), one defines

1

ZNexp

�N

t

Tr (Q (M,M

⇤))

�dM

Induced measure on eigenvalues:

1

ˆ

ZN

e

�Nt

PNj=1 Q(zj ,zj)

Y

1j<kN

|zj � zk|2NY

j=1

dxjdyj

Thursday, May 15, 14

Induced measure on eigenvalues:

1

ˆZN

exp {�NTr [V (M)]} dMExample 1:

1

ZN

e�NPN

j=1 V (�j)Y

1j<kN

|�j � �k|2NY

j=1

d�j

Thursday, May 15, 14

Induced measure on eigenvalues:

1

ˆZN

exp {�NTr [V (M)]} dMExample 1:

1

ZN

e�NPN

j=1 V (�j)Y

1j<kN

|�j � �k|2NY

j=1

d�j

1

ZNexp

�N

tTr (Q (M,M⇤

))

�dM

1

ZN

e

�Nt

PNj=1 Q(zj ,zj)

Y

1j<kN

|zj � zk|2NY

j=1

dxjdyj

Example 4:

Thursday, May 15, 14

A Physical Interpretation

The asymptotic behavior for N ! 1:

low temperature and many particle limit

1

ˆ

ZN

exp

8<

:�N

t

NX

j=1

Q(zj , zj) +

X

j 6=k

log |zj � zk|

9=

;

NY

j=1

dxjdyj

Eigenvalues are a system of particles

Logarithmic interaction potential,log |zj � zk|In the presence of an external field with potential Q(z, z)

Thursday, May 15, 14

Thursday, May 15, 14

N=200

1000 experiments

(10000)

Thursday, May 15, 14

Theoretical Physics, String Theory, Combinatorics:

Finance/Genetics/Statistics: Detecting correlations in data sets

Applications of random matrix theory

Form large matrices from huge data sets, compare to our benchmark.

Cellular networks: theoretical predictions of capacity

Information Theory: based on random strings of data

Feynman diagrammatic expansions + explicit asymptoticsyield predictions that can even be PROVEN!

Random growth models / Stochastic interface models

Probabilistic models of interfacial growth discrete matrix models

Hele-Shaw / Laplacian Growth / IntegrabilityNormal Matrices: Support set obeys Laplacian growth (deforming parameters)

Thursday, May 15, 14

What does one want to calculate

as N ! 1?

Thursday, May 15, 14

Mean density of eigenvalues

What is its average behavior?

Consider the random variable 1N # { �j � B}.

E�

1N

# { �j � B}⇥

�(N)1 is called the mean density of eigenvalues

Question: behavior when N �⇥???

=

Z

B⇢(N)1

⇢dx

dxdy

�

Thursday, May 15, 14

Movie of �(N)1 for N = 1 through N = 50.

1

#

N

e

x

p

⇢ �NT

r

1

2

M2

�� dM,

Thursday, May 15, 14

Movie of �(N)1 for N = 1 through N = 50.

1

#

N

e

x

p

⇢ �NT

r

1

2

M2

�� dM,

Thursday, May 15, 14

Movie of �(N)1 for N = 1 through N = 50.

1

#

Ne

x

p

{�NT

r

[

MM

⇤ ]}dM

,

Thursday, May 15, 14

Movie of �(N)1 for N = 1 through N = 50.

1

#Nexp

⇢�N Tr

MM⇤

+

T 2

2

⇣M2

+ (M⇤)

2⌘��

dM,

Thursday, May 15, 14

1

#Nexp

n

�N Tr

h

M2(M⇤

)

2io

dM,

Movie of ⇢(N)1 for N = 1 through N = 60.

Thursday, May 15, 14

Question: behavior when N �⇥???

Occupation Probabilities

Prob {B has k eigenvalues}

F (B, z) =1X

k=0

Prob {B has k eigenvalues} zk

F (B, z) is called the occupation probability generating function.

Thursday, May 15, 14

Example 4: Normal Matrix Model:

n⇥ n matrices with [M,M

⇤] = 0.

Given Q = Q(x, y) = Q(z, z), one defines

1

ZNexp

�N

t

Tr (Q (M,M

⇤))

�dM

Induced measure on eigenvalues:

1

ˆ

ZN

e

�Nt

PNj=1 Q(zj ,zj)

Y

1j<kN

|zj � zk|2NY

j=1

dxjdyj

Thursday, May 15, 14

Orthogonal Polynomials

Their asymptotic behavior is wide open.

Z

CPk,N (z)Pl,N (z)e�NQ(z,z)

dA(x, y) = hk,N�kl

⇢(N)1 (z, z) = e�NQ(z,z)

N�1X

`=0

h�1`,N |P`,N (z)|2

KN (z, w) = e�1N (Q(z)+Q(w)) 1

N

N�1X

k=0

h�1k Pk,N (z)Pk,N (w)

Thursday, May 15, 14

1

ZN

e�NPN

j=1 V (�j)Y

1j<kN

|�j � �k|2NY

j=1

d�j

Random Hermitian Matrices

Orthgonal Polynomials

Z

Rp

j

(x)p

k

(k)e

�NV (x)dx = �

jk

reproducing kernel,

K

N

(x, y) = e

�N2 (V (x)+V (y))

N�1X

`=0

p

`

(x)p

`

(y).

We want to establish (first) the following formula:

1

ˆ

Z

N

e

�N

PNj=1 V (�j)

Y

1j<kN

|�j

� �

k

|2 = det

0

B@K

N

(�1,�1) · · · K

N

(�1,�N

)

.

.

.

.

.

.

.

.

.

K

N

(�

N

,�1) · · · K

N

(�

N

,�

N

)

1

CA .

Thursday, May 15, 14

1

ZN

e

�Nt

PNj=1 Q(zj ,zj)

Y

1j<kN

|zj � zk|2NY

j=1

dxjdyj

Random Normal Matrices

KN (z, w) = e�1N (Q(z)+Q(w)) 1

N

N�1X

k=0

h�1k Pk,N (z)Pk,N (w)

1

ZN

e�NPN

j=1 Q(zj ,zj)Y

1j<kN

|zj � zk|2

= det

0

B@KN (z1, z1) · · · KN (z1, zN )

.... . .

...KN (zN , z1) · · · KN (zN , zN )

1

CA .

Thursday, May 15, 14

Y

1j<kN

(�k � �j) = det

0

BBB@

1 · · · 1�1 · · · �N...

. . ....

�N�11 · · · �N�1

N

1

CCCA.

Thursday, May 15, 14

Y

1j<kN

(�k � �j) = det

0

BBB@

1 · · · 1⇡1 (�1) · · · ⇡1 (�N )

.... . .

...⇡N�1 (�1) · · · ⇡N�1 (�N )

1

CCCA.

Y

1j<kN

(�k � �j) = det

0

BBB@

1 · · · 1�1 · · · �N...

. . ....

�N�11 · · · �N�1

N

1

CCCA.

Thursday, May 15, 14

Y

1j<kN

(�k � �j) = det

0

BBB@

1 · · · 1⇡1 (�1) · · · ⇡1 (�N )

.... . .

...⇡N�1 (�1) · · · ⇡N�1 (�N )

1

CCCA.

Y

1j<kN

(�k � �j) =1

QN�1j=0 j

det

0

BBB@

0 · · · 0

1⇡1 (�1) · · · 1⇡1 (�N )...

. . ....

N�1⇡N�1 (�1) · · · N�1⇡N�1 (�N )

1

CCCA.

Y

1j<kN

(�k � �j) = det

0

BBB@

1 · · · 1�1 · · · �N...

. . ....

�N�11 · · · �N�1

N

1

CCCA.

Thursday, May 15, 14

Y

1j<kN

(�k � �j) =1

QN�1j=0 j

det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA.

Thursday, May 15, 14

Y

1j<kN

(�k � �j) =1

QN�1j=0 j

det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA.

Y

1j<kN

(�k � �j)2 =

1QN�1

j=0 2j

det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

T

·

· det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

Thursday, May 15, 14

Y

1j<kN

(�k � �j)2 =

1QN�1

j=0 2j

det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

T

·

· det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

Thursday, May 15, 14

Y

1j<kN

(�k � �j)2 =

1QN�1

j=0 2j

det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

T

·

· det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

Y

1j<kN

|zk � zj |2 =1

QN�1j=0 |j |2

det

0

BBB@

p0 (z1) · · · p0 (zN )p1 (z1) · · · p1 (zN )

.... . .

...pN�1 (z1) · · · pN�1 (zN )

1

CCCA

T

·

· det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

Thursday, May 15, 14

Y

1j<kN

(�k � �j)2 =

1QN�1

j=0 2j

det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

T

·

· det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

P(�1, . . . ,�N ) =1

ZN

e�NPN

j=1 V (�j)Y

1j<kN

|�j � �k|2NY

j=1

d�j

Thursday, May 15, 14

Y

1j<kN

(�k � �j)2 =

1QN�1

j=0 2j

det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

T

·

· det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

P(�1, . . . ,�N ) =1

ZN

e�NPN

j=1 V (�j)Y

1j<kN

|�j � �k|2NY

j=1

d�j

P(�1, . . . ,�N ) =1

ZNQN�1

j=0 2j

det

0

BBB@

e�N2 V (�1)p0 (�1) · · · e�

N2 V (�N )p0 (�N )

e�N2 V (�1)p1 (�1) · · · e�

N2 V (�N )p1 (�N )

.... . .

...

e�N2 V (�1)pN�1 (�1) · · · e�

N2 V (�N )pN�1 (�N )

1

CCCA

T

·

· det

0

BBB@

e�N2 V (�1)p0 (�1) · · · e�

N2 V (�N )p0 (�N )

e�N2 V (�1)p1 (�1) · · · e�

N2 V (�N )p1 (�N )

.... . .

...

e�N2 V (�1)pN�1 (�1) · · · e�

N2 V (�N )pN�1 (�N )

1

CCCA

Thursday, May 15, 14

Y

1j<kN

(�k � �j)2 =

1QN�1

j=0 2j

det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

T

·

· det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

P(�1, . . . ,�N ) =1

ZN

e�NPN

j=1 V (�j)Y

1j<kN

|�j � �k|2NY

j=1

d�j

P(�1, . . . ,�N ) =1

ZNQN�1

j=0 2j

det

0

BBB@

e�N2 V (�1)p0 (�1) · · · e�

N2 V (�N )p0 (�N )

e�N2 V (�1)p1 (�1) · · · e�

N2 V (�N )p1 (�N )

.... . .

...

e�N2 V (�1)pN�1 (�1) · · · e�

N2 V (�N )pN�1 (�N )

1

CCCA

T

·

· det

0

BBB@

e�N2 V (�1)p0 (�1) · · · e�

N2 V (�N )p0 (�N )

e�N2 V (�1)p1 (�1) · · · e�

N2 V (�N )p1 (�N )

.... . .

...

e�N2 V (�1)pN�1 (�1) · · · e�

N2 V (�N )pN�1 (�N )

1

CCCA

=1

ZNQN�1

j=0 2j

⇥ det

0

BBB@

e�N2 (V (�1)+V (�1)))

PN�1`=0 p`(�1)p`(�1) · · · e�

N2 (V (�1)+V (�N ))

PN�1`=0 p`(�1)p`(�N )

e�N2 (V (�2)+V (�1))

PN�1`=0 p`(�2)p`(�1) · · · e�

N2 (V (�2)+V (�N ))

PN�1`=0 p`(�2)p`(�N )

.... . .

...

e�N2 (V (�N )+V (�1))

PN�1`=0 p`(�N )p`(�1) · · · e�

N2 (V ((�N )+V (�N ))

PN�1`=0 p`(�N )p`(�N )

1

CCCA,

Thursday, May 15, 14

Y

1j<kN

(�k � �j)2 =

1QN�1

j=0 2j

det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

T

·

· det

0

BBB@

p0 (�1) · · · p0 (�N )p1 (�1) · · · p1 (�N )

.... . .

...pN�1 (�1) · · · pN�1 (�N )

1

CCCA

P(�1, . . . ,�N ) =1

ZN

e�NPN

j=1 V (�j)Y

1j<kN

|�j � �k|2NY

j=1

d�j

P(�1, . . . ,�N ) =1

ZNQN�1

j=0 2j

det

0

BBB@

e�N2 V (�1)p0 (�1) · · · e�

N2 V (�N )p0 (�N )

e�N2 V (�1)p1 (�1) · · · e�

N2 V (�N )p1 (�N )

.... . .

...

e�N2 V (�1)pN�1 (�1) · · · e�

N2 V (�N )pN�1 (�N )

1

CCCA

T

·

· det

0

BBB@

e�N2 V (�1)p0 (�1) · · · e�

N2 V (�N )p0 (�N )

e�N2 V (�1)p1 (�1) · · · e�

N2 V (�N )p1 (�N )

.... . .

...

e�N2 V (�1)pN�1 (�1) · · · e�

N2 V (�N )pN�1 (�N )

1

CCCA

K

N

(x,

y

) =e

�N2(V

(x)+

V

(y))

N

�1X

`

=0

p

`

(x)p `

(y).

=1

ZNQN�1

j=0 2j

⇥ det

0

BBB@

e�N2 (V (�1)+V (�1)))

PN�1`=0 p`(�1)p`(�1) · · · e�

N2 (V (�1)+V (�N ))

PN�1`=0 p`(�1)p`(�N )

e�N2 (V (�2)+V (�1))

PN�1`=0 p`(�2)p`(�1) · · · e�

N2 (V (�2)+V (�N ))

PN�1`=0 p`(�2)p`(�N )

.... . .

...

e�N2 (V (�N )+V (�1))

PN�1`=0 p`(�N )p`(�1) · · · e�

N2 (V ((�N )+V (�N ))

PN�1`=0 p`(�N )p`(�N )

1

CCCA,

Thursday, May 15, 14

1

ZN

e�NPN

j=1 V (�j)Y

1j<kN

|�j � �k|2NY

j=1

d�j

Random Hermitian Matrices

Orthgonal Polynomials

Z

Rp

j

(x)p

k

(k)e

�NV (x)dx = �

jk

reproducing kernel,

K

N

(x, y) = e

�N2 (V (x)+V (y))

N�1X

`=0

p

`

(x)p

`

(y).

We want to establish (first) the following formula:

1

ˆ

Z

N

e

�N

PNj=1 V (�j)

Y

1j<kN

|�j

� �

k

|2 = det

0

B@K

N

(�1,�1) · · · K

N

(�1,�N

)

.

.

.

.

.

.

.

.

.

K

N

(�

N

,�1) · · · K

N

(�

N

,�

N

)

1

CA .

Thursday, May 15, 14

1

ZN

e

�Nt

PNj=1 Q(zj ,zj)

Y

1j<kN

|zj � zk|2NY

j=1

dxjdyj

Random Normal Matrices

1

ZN

e�NPN

j=1 Q(zj ,zj)Y

1j<kN

|zj � zk|2

= det

0

B@KN (z1, z1) · · · KN (z1, zN )

.... . .

...KN (zN , z1) · · · KN (zN , zN )

1

CA .

KN (z, w) = e�1N (Q(z,z)+Q(w,w)) 1

N

N�1X

k=0

h�1k Pk,N (z)Pk,N (w)

Thursday, May 15, 14

K

N

(x, y) = e

�N2 (V (x)+V (y))

N�1X

`=0

p

`

(x)p`

(y).

Verify thatZ

RK

N

(x, y)KN

(y, z)dy = K

N

(x, z),

andZ

RK

N

(x, x)dx = N

Thursday, May 15, 14

KN (z, w) = e�1N (Q(z,z)+Q(w,w)) 1

N

N�1X

k=0

h�1k Pk,N (z)Pk,N (w)

Verify thatZ

RKN (z, w)KN (w, s)d2w = KN (z, w),

andZ

RKN (z, z)d2z = N

Thursday, May 15, 14

Small collection of examples for which calculations are explicit

• Q(x, y) = Q(x

2+ y

2): Show: Pn = cn,Nz

n, and if Q(x, y) = x

2+ y

2, then

KN (z, w) =

1

⇡

e

�N(

|z|2+|w|2)

/2N�1X

j=0

(Nzw)

j

j!

and (still assuming Q = x

2+ y

2) then

⇢N =

1

⇡

e

�N |z|2N�1X

j=0

�N |z|2

�j

j!

Thursday, May 15, 14