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


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


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


N=1

N=2

N=3

N=4

N=8

N=12

N=16

N=20


N = 200

N = 400

N = 600

N = 800


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


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...


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


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



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

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:


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)


N=200

1000 experiments

(10000)


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)


What does one want to calculate

as N ! 1?


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

�


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

1

#

N

e

x

p

⇢ �NT

r

1

2

M2

�� dM,



1

#

Ne

x

p

{�NT

r

[

MM

⇤ ]}dM

,



1

#Nexp

⇢�N Tr

MM⇤

+

T 2

2

⇣M2

+ (M⇤)

2⌘��

dM,


1

#Nexp

n

�N Tr

h

M2(M⇤

)

2io

dM,

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


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.


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


1

ˆ

ZN

e

�Nt

PNj=1 Q(zj ,zj)

Y

1j<kN

|zj � zk|2NY

j=1

dxjdyj


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)


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 .


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


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 .


Y

1j<kN

(�k � �j) = det

0

BBB@

1 · · · 1�1 · · · �N...

. . ....

�N�11 · · · �N�1

N

1

CCCA.


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.


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.


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) =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


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

(�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


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


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


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,


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,


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 .


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



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


KN (z, w) = e�1N (Q(z,z)+Q(w,w)) 1

N

N�1X

k=0


Verify thatZ

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

andZ

RKN (z, z)d2z = N


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!


introduction to random matrix theorymath.as.uky.edu/sites/default/files/mclaughlin-1.pdf1. random...

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