Introduction: RMT deals with matrices with random variables as its entries. RMT is a fascinating field which is expanding rapidly in recent years. In this series of posts, I will be proving the vague convergence of the celebrated Tracy-Widom Law. I will be mostly following the book ‘An Introduction to Random Matrices’ by Anderson et.al.
Let be independent variables. Define to be a complex random variable. We say that follows standard complex normal distribution and denote it as , since and . Let be a random matrix with iid entries and be a random matrix with iid entries. Define
Let denote the law of . The random matrix is said to belong to the Gaussian orthogonal ensemble  (GOE(n)). Observe that is symmetric and is a random variable while is random variable.
Let denote the law of . Then the random matrix is said to belong to the Gaussian unitary ensemble  (GUE(n)). Observe that is hermitian and is a random variable while is random variable.
Why the law special?
Because the density at is proportional to . Thus for the law , the density at and is same for any orthogonal matrix and for law , the density at and is same for any unitary matrix (and hence the names GOE and GUE). This also enables us to find the joint distribution of the eigenvalues exactly!
Target: Let be the eigenvalues of a matrix from GUE/GOE. Then it is known that the empirical distribution of the eigenvalues converges weakly, in probability to the semicircle distribution with density . We also have that . What we are interested in the limiting distribution of this largest eigenvalue. In fact what we intend to show is that converges weakly. In the initial series of posts we will be only showing the vague convergence of above expression.
Let us adapt this convention: If we are using the notation for eigenvalues then they are ordered. If we use for eigenvalues then they are unordered.
Notations: Let be a square matrix of order . The determinant of will be denoted as
By laplace expansion we also have the following formula for the determinants.
Theorem 1: Let be a random matrix with law . Let denotes the ordered eigenvalues of the random matrix . The joint distribution of the ordered eigenvalues has density with respect to Lebesgue measure which equals
where is the normalizing constant and is Vandermonde determinant associated with which equals to .
Proof: The proof will be skipped. Essentially the proof involves the change of variable formula. It takes a certain suitable transformation with a smooth inverse. Then we obtain the required density by integrating out the extra variables. The part emerges as the jacobian of the transformation and appears by making the change of variables. However to justify it clearly, lot more mathematics is needed.
Remark: What we are interested in, is the distribution of the unordered eigenvalues which is obtained by symmetrizing the density . We say the joint density of the unordered eigenvalues is given by
Technically speaking, this is just a symmetrized valid density. This should not be called as the joint density of the unordered eigenvalues as integrating this joint density times gives us a marginal density. But that marginal density is the density of which eigenvalue? The best we can say is that if we choose one eigenvalue randomly, it is the density of that randomly chosen eigenvalue.
From now on, we will concentrate only on GUE. We now introduce Hermite Polynomials and associated wave function.
The -th Hermite polynomial is defined as . Hermite polynomials are actually polynomials! In fact is a monic polynomial of degree . The associated -th normalized wave function is defined as
That weird scalling factor is used so that forms an orthonormal basis for , that is, we have
Let us define to be the Hermite Kernel. With this in hand, we have a different expression for .
Theorem 2: Consider the density stated in the remark of the previous theorem. We have
Proof: We need the following lemma to prove this theorem.
Lemma 1: For any square-integrable functions and on the real line, we have
Proof of Lemma: The first equality follows easily due to the fact that . Enough to prove the last equality. By laplace expansion we have
In the penultimate line, we permutated each product such that the ‘s are in order and then we take the over which gives rise to . Note that the is different from the one in above line.
Proof of Theorem 2: The Vandermonde matrix is given by
Note that if we replace -th column ( say) of the matrix by , then the value of the determinant will remain same. So, we can replace -th column by a monic polynomial of degree . We will replace the columns with Hermite polynomials.
So, we have .
where the penultimate equality follows from the fact that .
Now if we plug in the lemma, then we have
Hence is the normalising constant. This completes the proof of theorem 2.
Theorem 3: For any measurable subset of ,
where ‘s are the unordered eigenvalues of belonging to GUE. [Note that it doesn’t matter whether it is ordered or unordered.]
Proof: By theorem 2, we have
The last equality above follows from Lemma 1.
Let us now pause for a bit. We digress a bit and discuss a formula involving determinants. Consider a square matrix . Suppose we want write in increasing degrees ‘s. Indeed, is an inhomogeneous polynomial in the . The term of degree is obtained from the term of degree in the product of the diagonal elements of , therefore it is .
The term of degree is obtained by taking the from diagonal entries and multiply them by the variable in the remaining diagonal entry, in all the possible ways (that are ); this gives rise to
The term of degree is obtained by taking diagonal ‘s multiplied by the minor obtained from the remaining two rows/columns; you have possible ways to do that, as many as the minors centered on the diagonal; this gives rise to
In general, the term of degree is obtained by taking diagonal ‘s multiplied by the complementary minor centered on the diagonal. This gives rise to
So we have the following formula
Coming back to the proof of the theorem, let us observe that using the above formula we have
where the last equality follows again from Lemma 1.
Note that . The above expression inside the integral is just the Cauchy-Binet’s expansion of the . Hence we have
where the sum can be taken to infinity because the matrix is of rank at most . Hence the determinant is zero for . This completes the proof.
Why this nasty expression of the Theorem 3 is useful at all? It turns out that the right hand side of the theorem 2 is very closely related to the Fredholm determinant of which has been studied extensively. We will look into some of the key properties of Fredholm determinant in next post.