8.1 Density of Eigenvalues in Classical Ensembles of Random Matrices
We begin our tour from an important type of matrix G\(\beta \)E (\(\beta =1,2,4\)). Let \(X\) be \(N\times N\) matrix with i.i.d. entries: \[ \begin {cases} a) {\mathcal{N}(0,1)} \\ b) {\mathcal{N}(0,1)} + i{\mathcal{N}(0,1)} \\ c) {\mathcal{N}(0,1)} + i{\mathcal{N}(0,1)} + j{\mathcal{N}(0,1)} + k{\mathcal{N}(0,1)} \end {cases} \] Set \(M = \frac {1}{2}(X+X^*)\), and we have the main theorem.
Proof. We prove \(\beta = 1\) here and leave \(\beta = 2\) to the homework.
Step 1 Claim density of \(M = \frac {1}{2}(X+X^*) \sim \exp (-\frac {1}{2}{\operatorname{tr}} (M^2))\).
Indeed, \[ {\operatorname{tr}} (M^2) = \sum _{i,j} m_{ij}^2 = \sum _{i=1}^N \underbrace {x_{ii}^2}_{\mathcal {N}(0,1)} + \frac {1}{2}\sum _{i<j} \underbrace {(x_{ij}+x_{ji})^2}_{\mathcal {N}(0,2)} \] implies that the density of \(M\propto \exp (-\frac {1}{2}{\operatorname{tr}} (M^2))\)
Step 2 We can calculate that \(\exp (-\frac {1}{2}{\operatorname{tr}} (M^2)) = \prod _{i=1}^N \exp (-\frac {\lambda _i^2}{2})\).
We derive it immediately by diagonalizing the the matrix \(M\).
Step 3 Each symmetric matrix is determined by its eigenvalues and eigenvectors, i.e. there exists an almost bijection \(\pi \) \[ \pi : \underbrace {\mathcal {W}_N}_{\lambda _1<\lambda _2<...<\lambda _N} \times \underbrace {O(N)}_{\text {Orthogonal Bases}} \to \underbrace {\mathcal {H}_N}_{\text {Symmetric Matrix}} \] We say "almost" because indeed the map is not injective: we can multiply the eigenvalue by \(\pm 1\). In other words, if the eigenvalues are unique, \(\pi ^{-1}(M)\) has exactly \(2^N\) elements.
Now we give the key proposition of the proof.
Proof. It is sufficient to calculate by taking \(O = {\operatorname{Id}} \), as when we transform \(O\) to \(A\cdot O\) (where \(A\) is orthogonal), the uniform measure on \(O(N)\) and the Lebesgue measure on \(\mathcal {H}(N)\) remain unchanged.
Then we take \(O = \exp (B) = {\operatorname{Id}} + B + ...\) and we can deduce that \(B + B^* = 0\) as \(OO^* = {\operatorname{Id}} \). Then the map \(\pi \) can be written as \begin{align*} \left ((\lambda _1,...,\lambda _N), \exp (B)\right ) &\mapsto \exp (B) \begin {pmatrix} \lambda _1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda _2 & 0 & \cdots & 0 \\ 0 & 0 & \lambda _3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda _n \end {pmatrix} \exp (-B)\\ &\mapsto ({\operatorname{Id}} +B) \begin {pmatrix} \lambda _1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda _2 & 0 & \cdots & 0 \\ 0 & 0 & \lambda _3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda _n \end {pmatrix} (Id-B) + o(B)\\ &\mapsto \begin {pmatrix} \lambda _1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda _2 & 0 & \cdots & 0 \\ 0 & 0 & \lambda _3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda _n \end {pmatrix}\\ &+ \begin {pmatrix} 0 & b_{12}(\lambda _2 - \lambda _1) & b_{13}(\lambda _3 - \lambda _1) & \cdots & b_{1n}(\lambda _n - \lambda _1) \\ b_{21}(\lambda _1 - \lambda _2) & 0 & b_{23}(\lambda _3 - \lambda _2) & \cdots & b_{2n}(\lambda _n - \lambda _2) \\ b_{31}(\lambda _1 - \lambda _3) & b_{32}(\lambda _2 - \lambda _3) & 0 & \cdots & b_{3n}(\lambda _n - \lambda _3) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b_{n1}(\lambda _1 - \lambda _n) & b_{n2}(\lambda _2 - \lambda _n) & b_{n3}(\lambda _3 - \lambda _n) & \cdots & 0 \end {pmatrix} + o(B) \end{align*}
As \(B\) has only \(\frac {n(n-1)}{2}\) free parameter, so the Jacobian of the map is \(\prod _{i<j}|\lambda _j-\lambda _i|\). □
Step 4 We now calculate \(Z\): \begin{align*} Z = \int _{\mathbb {R}^N} \prod _{i<j}|\lambda _j-\lambda _i|^\beta \cdot \prod _{i=1}^N\exp (-\frac {\lambda _i^2}{2}) {\mathop{}\!\mathrm{d}} \lambda _{1}...{\mathop{}\!\mathrm{d}} \lambda _{N} = \frac {(2\pi )^{N/2}}{N!} \prod _{j=0}^N \frac {\Gamma (1+(j+1)\beta /2)}{\Gamma (1+\beta /2)} \end{align*} □