arXiv:1901.09257v3 [math.PR] 1 Oct 2019 if ento 2. Definition and 1 eoetesto symmetric of set the denote ento .[asinOtooa nebe(GOE)]. Ensemble matrix Orthogonal real [Gaussian 3. Definition I ewl eoeby denote will We I Background 2 for project final Matem´at Investigaci´on en a de as Centro the done at was course work matrice graduate This the of characterization Ensemble. main Orthogonal the of one prove will We Introduction 1 ento 1. Definition ∗ d † ≤ negaut tdn.Uiesddd unjao victor Guanajuato. de Universidad student. Undergraduate negaut tdn.Uiesddd unjao jose.s Guanajuato. de Universidad student. Undergraduate × O d k < j O ∈ ildnt h dniymti fteset the of matrix identity the denote will nqeeso h asinOrthogonal Gaussian the of Uniqueness lmnaypoete ftecaatrsi ucino ra of function characteristic the of properties elementary ymti admmti uhta ti nain ne orth under invariant matrix converse, is such the it then prove that tions, to such is matrix work random this symmetric of goal The conjugations. o inrmatrix Wigner dom GE,te uhmatrix such then (GOE), nw euti admmti hoysae h following: the states theory matrix random in result known A ( n ≤ B Jos´e then ) B ( n ,k j, n r needn admrno variables random random independent are ymti matrix symmetric A eoeby Denote M ∈ = ) ne S´anchez G´omezAngel ´ O M ⊺ B O n n × n × = ( m n ,j k, ( ( OO F R O X ,testo arcso size of matrices of set the ), ) B si h asinOtooa nebeif, Ensemble Orthogonal Gaussian the in is ) B , ( n ⊺ n n X hc eog oteGusa rhgnlEnsemble Orthogonal Gaussian the to belongs which × = h e fotooa arcsof matrices orthogonal of set the ) : = eog oteGE ewl rv hsuigsome using this prove will We GOE. the to belongs X n I n eebr 2017. December,    n ( arcsby matrices Ensemble × ,k j, a nivratdsrbto ne orthogonal under distribution invariant an has B B n X . n n ) ( (1 . . . n, N ∼ sasur arxsc that such matrix square a is , 1) 1) 1 (0 itrAaaCarvajal Amaya Victor , ∗ B ...... B . . . M , S 1) n d . j , × d [email protected] [email protected] ( n n 6= F ( (1 ,n n, . . . ). n ,B k, n , × ) ) m    cs ..(IA) M´exico. (CIMAT), A.C. icas, htbln oteGaussian the to belong that s vrtefield the over n dmvariables. ndom ( esyta symmetric a that say We ,j j, M ) h admMatrices Random the hti,if is, that gnlconjuga- ogonal n N ∼ × X ie ran- a Given n ( ⊺ R (0 = .Ti means, This ). F , B 2) h matrix The . n X X . † ( . ,k j, ewill We . sa is ,with ), Definition 4. Denote by GOE(n) the set of n n matrices that belong to the Gaussian × Orthogonal Ensemble.

Definition 5. Let X n n(R) be a random matrix. We say that X has invariant ∈ M × distribution (or just invariant) under orthogonal conjugations if for every non-random orthogonal matrix O (n), we have that OXO⊺ =L X. ∈ O

Observation 1. It follows from the definition that, if we take Z n n(R) whose ∈ M × entries are independent random variables zi,j (0, 1), then ∼N 1 X = (Z + Z⊺) GOE(n). √2 ∈ n Observation 2. Note that, given any Bn GOE(n) there exists a matrix Z = [zi,j] , ∈ i,j=1 1 ⊺ with zi,j (0, 1), such that Bn =L (Z + Z ). ∼N √2 Now, lets state the theorem whose converge we would like to prove. Theorem 2.1. ([Invariance under conjurations]) For any given O (n) (non-random) ∈ O and Bn GOE(n), then ∈ ⊺ OXO =L X.

Proof. By the observation 2, we can write matrix B =L 1 (Z + Z⊺), with Z such that n √2 Zi,j (0, 1). Now, observe that, ∼N ⊺ 1 ⊺ ⊺ ⊺ 1 ⊺ ⊺ 1 ⊺ OBnO =L (OZO + OZ O ) =L (OZ + Z O ) =L (Z + Z ) =L Bn √2 √2 √2

3 Background II

We define the characteristic function of a R to be a function ϕR : R C → given by, itR ϕR(t)= E e , t R. ∀ ∈ In particular, given a random variable R  (µ,σ2), its characteristic function is given ∼N by: itR iµt 1 σ2t2 ϕR(t)= E[e ]= e − 2 .

Another important thing to remember is that if two random variables R1,R2 are such that their characteristic function coincide in every point, then their distributions are the same. In other words, if ϕX (t)= ϕY (t), for every t R, then X Y . ∈ ∼

Given a symmetric random matrix X Sd its characteristic function is defined by: ∈

CX : Sd C → ⊺ CX (M)= E [exp i Tr(X M) ] , M Sd. { } ∀ ∈ Where Tr( ) is the trace operator. Recall that the trace is invariant under cyclic permu- · tations. From this, if A d d(R) and for all M Sd. ∈ M × ∈ ⊺ ⊺ ⊺ CX (A MA) = E [exp i Tr(X A MA) ] { } = E [exp i Tr(A⊺X⊺AM) ] { ⊺ ⊺ ⊺ } = E [exp i Tr([A X A] M) ]= CA⊺XA(M). { }

2 Notice that, if A, B Sd, ∈ d d d ⊺ ⊺ ⊺ Tr(A B)= (A B)jj = (A )jkBkj = AkjBkj. Xj=1 j,kX=1 j,kX=1

Furthermore, if A, B Sd, ∈ d ⊺ Tr(A B)= AjjBjj + 2 AjkBjk. Xj=1 1 Xj

d ⊺ CX (M)= E [exp i Tr(X M) ]= E exp i MjjXjj + 2i MkjXkj . { }  Xj=1 1 Xj

d CX (M) = E [exp iMjjXjj ] E [exp 2iMkjXkj ] . { } · { } jY=1 1 j

4 The problem

Our aim is to prove the following theorem using nothing more than the characteristic function of a random matrix.

Theorem 4.1. Let X be a non-zero random symmetric d d matrix such that it is × invariant under orthogonal conjugations. Suppose that all the entries of X are independent with finite variance. Then, there exist a matrix belonging to the Gaussian Orthogonal 2 2 Ensemble Y , and real numbers µ R, σ 0 such that X =L µId d + σ Y . ∈ ≥ × Proof : We will divided our proof into three steps: 1. Show that the elements inside the diagonal share the same distribution, and that all the elements outside the main diagonal share the same distribution.

2. Show that it is enough to prove the result for square 2 2 matrices. × 3. We will prove that the characteristic functions of the elements of the matrix X correspond to normally distributed random variables.

t Step one: Let us define the matrix A(k,j) as the matrix that has the value t in the positions (k, j) and (j, k) and zeros everywhere else, for 1 k < j d and t R. If we t ≤ ≤ ∈ calculate the the characteristic function of A(k,j) we obtain:

E t ⊺ CX (A(k,j)(t)) = exp i Tr((A(k,j)) X) h n oi = E [exp 2itXkj ] { } = φXkj (2t).

3 Now, let k = j and l = g. There exists a permutation matrix P (d) such that 6 6 ∈ O for all t R, P ⊺At P = At . From the invariance of X with respect to orthogonal ∈ (k,j) (l,g) conjugations:

t ⊺ t t ϕX (2t)= CX (A )= CX (P A P )= CX (A )= ϕX (2t), t R. kj (k,j) (k,j) (l,g) lg ∀ ∈

It follows that Xkj Xlg. It means that all entries outside the main diagonal have the ∼ same distribution. In an analogous way, by conjugating we can prove that if 1 j < k d, ≤ ≤ Xkj Xkj. This implies the distribution of the entries outside of the main diagonal is ∼ − symmetric. This is because all entries of the matrix have finite second moment, E(Xkj)= t 0. Now, let A = diag(t ej), i.e. a matrix that has as only non-zero entry t in the position i · (j, j), for 1 j d and t R. From this, ≤ ≤ ∈ t E t ⊺ CX (Aj) = exp i Tr((Aj ) X) = E [exp itXjj ]  { } = ϕXjj (t), for all 1 j d and t R. Now, given j = k, there exists a permutation matrix Q (d) ≤ ⊺≤ t t∈ 6 ∈ O such that Q AjQ = Ak.From this,

t ⊺ t t ϕX (t)= CX (A )= CX (Q A Q)= CX (A )= ϕX (t) t R. jj j j k kk ∈

Thus Xjj Xkk for all j = k. Then, all entries in the diagonal have the same distribution. ∼ 6 Step two: In the last step, we have proved that all entries in the main diagonal have the same distribution, and the entries outside the diagonal share the same distribution. From this, it is possible to find the distribution of all entries of the matrix X by finding the distribution of the entries of the sub-matrix,

X11 X12 X2 = . X12 X22

It can be proved that X is invariant with respect to rotations in R2. For this, let θ R 2 ∈ and,

Mθ 0 R Qθ′ = d d( ),  0 I(d 2) (d 2) ∈ M × − × − where, cos(θ) sin(θ) Qθ = − 2 2(R). sin(θ) cos(θ).  ∈ M ×

The matrix Qθ′ is an orthogonal matrix that rotates the first two coordinates and keep fixed the others. Let M , where ∈ S2 a b M = S . b d ∈ 2

We can embed M into Sd by filling all entries outside the 2 2 first sub-matrix with zeros: × M 0 M ′ = d.  0 0 ∈ S

Now lets observe that,

4 ⊺ CX (M ′)= E [exp i Tr(M X) ]= E [exp i(aX + 2bX + dX ) ]= CX2 (M). { } { 11 12 22 }

By evaluating the characteristic function of X on M ′ and using the orthogonal in- variance, and considering that all entries outside the 2 2 matrix in M are zero, × ′ ⊺ ⊺ CX2 (M)= CX (M ′)= CX ((Qθ′ ) M ′Qθ′ )= CX2 (QθMQθ). Step three: Since we have reduced the problem to the case of 2 2 matrices, we will × ⊺ keep M and Qθ to be the matrices defined in the previous step. We take Mθ = QθMQθ, a 2 2 symmetric matrix. Name the entries of Mθ as follow × A B M = , θ B D we can compute explicitly the values of A, B, D in function the entries of M. We get the following:

a + d a d A = + − cos(2θ) b sin(2θ), 2 2 − a d B = − sin(2θ)+ b cos(2θ), 2 a + d a d D = − cos(2θ)+ b sin(2θ). 2 − 2 These expressions are valid for every θ R. If we calculate the derivatives of the last ∈ expressions with respect to θ, what we get is the following: dA dB dD = 2B, = A D, = 2B (1) dθ − dθ − dθ

Now, since X2 is invariant under rotations, we have that:

CX2 (Mθ)= CX2 (M).

Computing the characteristic function on both matrices we obtain that,

ϕ1(a)ϕ2(2b)ϕ1(d)= ϕ1(A)ϕ2(2B)ϕ1(D), (2) where ϕ1 represents the characteristic function of the random variable X11 (which is the same of X22 since the have the same distribution). And ϕ2 is the characteristic function of X12. Since entries of X2 have finite variance ϕ1 and ϕ2 are at least two times derivable at E zero and φj′ (0) = i (X1j ) for j = 1, 2. We can derive equation (1) with respect to θ, we get:

dA dB dD 0= ϕ′ (A)ϕ (2B)ϕ (D) + ϕ (A)ϕ′ (2B)ϕ (D)2 + ϕ (A)ϕ (2B)ϕ′ (D) (3) 1 2 1 dθ 1 2 1 dθ 1 2 1 dθ

Note that since ϕ1(0) = ϕ2(0) = 1, and the are continuous at zero, there exist an open set around zero where these functions are not zero. So, we can rewrite equation (3) in the following way:

5 ϕ1′ (A) dA ϕ2′ (2B) dB 0= ϕ1(A)ϕ2(2B)ϕ1(D) + ϕ1(A)ϕ2(2B)ϕ1(D)2 + ϕ1(A) dθ ϕ2(2B) dθ

ϕ1′ (D) dD + ϕ1(A)ϕ2(2B)ϕ1(D) . ϕ1(D) dθ Now, replacing the values we found in (1) and dividing by B(A D) (for those values − − B = 0, A = D), we get that 6 6 1 ϕ (2B) 1 ϕ (A) ϕ (D) 2′ = 1′ 1′ (4) B ϕ (2B) A D ϕ (A) − ϕ (D) 2 − 1 1 We observed that the left hand side of (4) only depends on B, whereas the right hand side depends in both A and D. By this observation we can conclude that it has to be constant. Then, there exists k R such that: ∈ 1 ϕ (2B) 2′ = k, B ϕ2(2B) −

Observe that, since ϕ2 is the characteristic function of a symmetric random variable, the co-domain of ϕ2 is contained in R. This implies k R. Now, lets take x = 2B. We xk ∈ have that ϕ2′ (x)= ϕ(x). This is an easy real-valued ODE whose solution is given by 2 2 kx − ϕ (x)= B e− 2 , for some B R. Now, as ϕ is a characteristic function it is true that 2 0 0 ∈ 2 ϕ2(0) = 1. This give us as a result that B0 = 1. So, we have that k x2 ϕ (x) = exp( ). 2 − 2 2 At first, we have this is true at a neighborhood of 0 R, but we can extend this solution ∈ on R. Observe that if k < 0 then ϕ (t) when t . This leads to a contra- | 2 |→∞ | |→∞ diction since characteristic functions have bounded image. Then, if k > 0 and ϕ2 is the characteristic function of a normal random variable with parameters (0, k/2). Given N k = 0, ϕ2 is the characteristic function of a singular distribution at 0. On the other hand, taking D = 0 in equation (4), it follows that 1 ϕ (A) ϕ (0) 1′ 1′ . A ϕ1(A) − ϕ1(0)

We know that ϕ1(0) = 1 and that ϕ′(0) = iµ, where µ = E(X11). So,

ϕ1′ (A) = Ak + iµ ϕ′ (A) = ( Ak + iµ)ϕ (A). ϕ(A) − ⇒ 1 − 1 2 iµx kx The solution to this ODE is ϕ1(x)= e − 2 . which is nothing more than the charac- teristic function of a random variable with distribution (µ, k). N Since k is non-negative we can write it as k = 2σ2. So we can write that characteristic functions in the following way: 2 2 2 2 x iµx σ x σ 2 ϕ1(x)= e − , ϕ2(x)= e− . which means that X (µ, 2σ2) and X (0,σ2). 11 ∼N 12 ∼N 2 2 In summary, we have proved that Xjj (µ, 2σ ) and Xkj (0,σ ). Now, if the ∼ N ∼ N take a GOE matrix Y , i.e., Yii (0, 2) and Yjk (0, 1), for every j = k, we conclude ∼N ∼N 6 that:

2 X =L σ Y + µId. We see that X is clearly invariant under orthogonal conjugations. 

6 5 Conclusions

There are important remarks that are worth stressing out about this proof. First, we find interesting that it unfolded by using elementary knowledge of basic properties of characteristic functions, differential equations and . It is a simple proof for this result. Second, it states a generalization of the standard result. This result is proved usually for matrices that have zero-mean entries. This proof allow us to study general matrices that are invariant with respect to orthogonal conjugations. The independence of the entries is a key requirement for this proof. One example of a distribution that is invariant under orthogonal conjugation is the Haar distribution on (d). This distribution does not hold the entry-wise independence since orthogonal O matrices satisfy relations between them. In terms of this work, it is important to write the characteristic function of the matrix in terms of the product of the characteristic function of each entry. One possible extension of this project is to find the distribution of Wishart matrices by applying the same characteristic-function approach.

References

[1] Yi-Kai Liu. Junior Seminar, Spring 2001. Princeton University. http://web.math.princeton.edu/mathlab/projects/ranmatrices/yl/randmtx.PDF.

[2] Measure Theory and . Krishna B. Athreya and Soumendra N. Lahiri. Springer Text in .

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