When we apply an entrywise nonlinear function $\sigma$ to a matrix $X$, the resulting matrix $\sigma(X)$ no longer has independent entries. However, if $X$ is a Wigner matrix such as Gaussian Orthogonal Ensemble (GOE), then, in the large-$N$ limit, the spectrum of $\sigma(X)$ is equivalent to that of a simple affine model. Concretely, \[ \mu_{\sigma(X)}\;\Longrightarrow\;\mu_{\alpha X + \beta Z} \quad(N\to\infty), \] where $\mu_M$ denotes the empirical spectral distribution (ESD) of $M$, $Z$ is an independent GOE, and $\alpha,\beta$ depend only on $\sigma$. This asymptotic behavior is the Gaussian Equivalence Principle (GEP).
Empirical Spectral Distribution
The empirical spectral distribution ESD of $N \times N$ matrix $M$ is, loosely speaking, the histogram (or probability measure) that places equal weight on each of its $N$ eigenvalues $\lambda_{i}$. Concretely, the ESD is the measure \[ \mu_M = \frac1N\sum_{i=1}^N\delta_{\lambda_i}. \] Like \[ \int_{\mathbb{R}} f(x)\,\delta_{x_0}(\mathrm{d}x) = f(x_0) \] for for a continuous test‐function $f$, the Dirac delta $\delta_{\lambda_i}$ picks out the value at $\lambda_i$ when we integrate against it. Dividing by $N$ makes sure that $\mu_{M}$ becomes a probability measure on $\mathbb{R}$, \[ \mu_{M}(\mathbb{R}) = \frac{1}{N}\sum_{i=1}^{N}1 = 1. \]
Entrywise vs. Spectral Function
There are two ways to apply a scalar function $\sigma$ to a matrix $X\in\mathbb{R}^{N\times N}$:
-
Spectral (functional) calculus
If $X$ is diagonalizable as $X=Q\Lambda Q^T$, then
\[ \sigma(X) :=Q\,\sigma(\Lambda)\,Q^T, \quad \sigma(\Lambda)={\rm diag} \bigl(\sigma(\lambda_i)\bigr). \] By construction $\sigma(UXU^T)=U\sigma(X)U^T$ exactly for any orthogonal $U$. No randomness or invariance is needed. -
Entrywise application
If we define
\[ \bigl[\sigma(X)\bigr]_{ij} :=\sigma\bigl(X_{ij}\bigr), \] then, in general,
\[ \sigma(UXU^T)\neq U\,\sigma(X)\,U^T.
\] But if $X$ is orthogonally invariant ($X\overset{d}{=}U X U^T$), then
\[ \sigma(X) \;\overset{d}{=}\; \sigma(U X U^T) \;\overset{d}{=}\; U\,\sigma(X)\,U^T, \] so $\sigma(X)$ is distributionally invariant under conjugation, which suffices to control its ESD.
In the GEP, $\sigma$ is applied entrywise, but the orthogonal invariance of the underlying $X$ grants that $\sigma(X)$ is distributionally invariant under conjugation, allowing us to reduce spectral questions about the messy entrywise map to questions about diagonal eigenvalue transforms.
Hermite Expansion
In $\mathbb{R}^n$, two vectors $u,v$ are orthogonal if their dot‐product $u\cdot v=0$. A real function $f(x)$ can be thought of as an infinite‐dimensional “vector” of values. we give a notion of “length” and “angle” of functions by defining an inner product \[ \langle f,g\rangle =\int_{-\infty}^{\infty} f(x)\,g(x)\,\varphi(x)\,dx, \] where $\varphi(x)$ is a weight function (e.g., the standard Gaussian density). If $\langle f,g\rangle=0$, we call $f$ and $g$ orthogonal in this infinite‐dimensional function space—it does not mean $f(x)\,g(x)=0$ pointwise, but that their weighted integral vanishes.
Example (Fourier modes): On $[-\pi,\pi]$ with weight $\varphi(x)=1$, $f(x)=\sin(x)$ and $g(x)=\cos(x)$ satisfy \[ \langle f,g\rangle = \int_{-\pi}^\pi \sin(x)\cos(x)\,dx = \frac12\int_{-\pi}^\pi \sin(2x)\,dx =0. \]
Hermite Polynomials
We can write a periodic function $f(\theta)$ in a Fourier series $\displaystyle f(\theta)=\sum_{m=-\infty}^\infty c_m e^{im\theta}$ because the family ${e^{im\theta}}$ is orthogonal under the uniform weight on $[0,2\pi]$. Likewise, if $\sigma(z)$ is square–integrable under the Gaussian weight, i.e. \[ \int_{-\infty}^\infty \sigma(z)^2\,\varphi(z)\,dz <\infty, \quad\varphi(z)=\frac{e^{-z^2/2}}{\sqrt{2\pi}}, \] then $\sigma(z)$ admits a Hermite expansion in the orthogonal family ${He_k(z)}_{k=0}^\infty$, where the probabilists’ Hermite polynomials satisfy \[ \langle He_m,\,He_n\rangle = \int_{-\infty}^{\infty} He_m(z)\,He_n(z)\,\varphi(z)\,dz = n!\,\delta_{m n}. \] Hence ${He_k(z)/\sqrt{k!}}_{k\ge0}$ is an orthonormal basis of $L^2(\mathbb{R},\varphi)$, and by standard Hilbert‐space theory \[ \sigma(z) = \sum_{k=0}^{\infty} c_k\,\frac{He_k(z)}{\sqrt{k!}} \;=\; \sum_{k=0}^{\infty} a_k\,He_k(z), \quad a_k = \frac{c_k}{\sqrt{k!}}. \] Concretely the first few are \[ He_0(z)=1,\quad He_1(z)=z,\quad He_2(z)=z^2 - 1,\quad He_3(z)=z^3 - 3z,\quad z\sim\mathcal N(0,1). \] Because of orthonormality, \[ c_k = \bigl\langle\sigma,\;\frac{He_k}{\sqrt{k!}}\bigr\rangle = \int \sigma(z)\,\frac{He_k(z)}{\sqrt{k!}}\,\varphi(z)\,dz = \frac{1}{\sqrt{k!}}\,\mathbb{E}\bigl[\sigma(z)\,He_k(z)\bigr], \] so \[ a_k = \frac{1}{k!}\,\mathbb{E}\bigl[\sigma(z)\,He_k(z)\bigr]. \] In particular \[ a_1 = \mathbb{E}\bigl[\sigma(z)\,z\bigr], \quad a_2 = \tfrac1{2}\,\mathbb{E}\bigl[\sigma(z)\,(z^2-1)\bigr]. \]
Gaussian Equivalence Principle
Let $\sigma$ be the elementwise nonlinear function and $X \in \mathbb{R}^{N\times N}$ be a GOE. Setting $Y_N = \sigma(X)$ and thus \[ Y_N = \frac{1}{\sqrt N}\,\sigma\bigl(\sqrt N\,X_N\bigr), \] the GEP says that the ESD of $Y_N$ is equal to $\alpha X + \beta$ as $N \to \infty$. To show this, we use trace moments matching.
Trace‑Matching via Moments
A probability measure $\mu$ on a bounded interval is uniquely determined by its moments
\[
m_\ell=\int x^\ell\,d\mu(x).
\]
For an $N\times N$ matrix $M$ with ESD $\mu_M$,
\[
m_\ell \;=\;\int x^\ell\,d\mu_M(x)
= \frac1N\sum_{i=1}^N\lambda_i^\ell
= \frac{1}{N}\mathrm{tr}\bigl(M^\ell\bigr).
\]
Hence to show $\mu_{Y_N}$ and $\mu_{\alpha X_N+\beta Z_N}$ coincide, it suffices to prove for each $\ell$
\[
\frac{1}{N}\mathbb{E}\bigl[\mathrm{tr}(Y_N^\ell)\bigr]
\;-\;
\frac{1}{N}\mathbb{E}\bigl[\mathrm{tr}\bigl((\alpha X_N+\beta Z_N)^\ell\bigr)\bigr]
\;\xrightarrow[N\to\infty]{}\;0.
\]
The derivation is as follows.
1. Entrywise expansion
The entry of $Y_N$ is \[ Y_N(i,j) = \frac{1}{\sqrt N}\sum_{k=0}^\infty a_k\,He_k\bigl(\sqrt N\,X_{ij}\bigr). \] Grouping terms gives \[ Y_{N}(i,j) = \underbrace{\frac{a_0}{\sqrt N}}_{k=0} + \underbrace{a_1\,X_{ij}}_{k=1} + \underbrace{a_2\Bigl(\sqrt N\,X_{ij}^2-\frac1{\sqrt N}\Bigr)}_{k=2} \;+\;\sum_{k\ge3}\frac{a_k}{\sqrt N}\,He_k\bigl(\sqrt N\,X_{ij}\bigr). \]
The $k=0$ term vanishes. The $k=0$ contribution is $\tfrac{a_0}{\sqrt N}\,11^T$, where $1\,1^T$ is the all‑ones matrix. This rank-1 matrix has one eigenvalue $a_0\sqrt N$ since $1\,1^T$ has a single nonzero eigenvalue $N$ and $N-1$ zero eigenvalues. Thus it contributes a spike at $\lambda_i = a_0\sqrt N$ in the ESD $\displaystyle \mu_{Y_N}=\tfrac1N\sum_i\delta_{\lambda_i(Y_N)}$, but the single spike carries mass $1/N$, so it vanishes as $N\to\infty$. The ESD ignores a single spike.
The $k=1$ term contributes \[ a_1\,X_{ij} = \alpha\,X_{ij}, \quad \alpha = \mathbb{E}\bigl[\sigma(Z)\,Z\bigr], \quad Z\sim\mathcal N(0,1). \]
Defining
\[
Q_{ij}
:= \frac{1}{\sqrt N}\,He_2\bigl(\sqrt N\,X_{ij}\bigr)
= \sqrt N\,X_{ij}^2 - \frac{1}{\sqrt N},
\]
the $k=2$ term can be written
\[
a_2\,Q_{ij}.
\]
Hence
\[
Y_N = \alpha\,X + a_2\,Q + R,
\]
where $R$ collects the Hermite–$k$ pieces for $k\ge3$.
2. Matrix‐power formula
For any matrix $M$, \[ (M^\ell)_{i j} =\sum_{i_{2},\dots,i_{\ell}} M_{i\,i_{2}}\;M_{i_{2}\,i_{3}}\;\cdots\;M_{i_{\ell}\,j}, \] so \[ \mathrm{tr}(M^\ell) =\sum_{i_{1},i_{2},\dots,i_{\ell}} M_{i_{1}i_{2}}\;M_{i_{2}i_{3}}\;\cdots\;M_{i_{\ell}i_{1}}. \]
3. Expansion of $\mathrm{tr}(Y_N^\ell)$
\[ \mathrm{tr}\bigl(Y_N^\ell\bigr) =\sum_{i_{1},\dots,i_{\ell}} \bigl(\alpha X+a_2 Q+R\bigr)_{i_1 i_2}\, \cdots\, \bigl(\alpha X+a_2Q+R\bigr)_{i_\ell i_1}. \]
Taking expectations and dividing by $N$ kills almost every term. Only two types survive:
- Pure linear $(\alpha X)^\ell$.
- Pairings of the $Q$–factors in even numbers (Wick contractions).
4. Wigner‐pairing and combinatorics
By the definition of Wigner matrix, $\mathbb{E}[X_{ij}]=0,\;\text{Var}[X_{ij}]=1/N$, so even if the true law of $X_{ij}$ is not Gaussian, we treat it “as if” it were Gaussian and consider its fourth moment is $\mathbb{E}[X_{ij}^4]\approx3/N^2$. Using them, we know \[ \mathbb{E}[Q_{ij}] = \sqrt N \mathbb{E}[X_{ij}^2] - \frac{1}{\sqrt N} = \sqrt N \bigl( \text{Var}[X_{ij}] - \mathbb{E}[X_{ij}]^2 \bigr) - \frac1N = 0. \] Also, \[ \mathbb{E}[Q_{ij}]=0, \quad \mathbb{E}[Q_{ij}^2] =N\,\mathbb{E}[X_{ij}^4] \;-\;2\,\mathbb{E}[X_{ij}^2] \;+\;\frac1N =\frac{2}{N}, \] and if ${i,j}\neq{k,\ell}$, then $Q_{ij}$ and $Q_{k\ell}$ are independent so $\mathbb{E}[Q_{ij}Q_{k\ell}]= 0・0 = 0$.
-
Pairing rule: Only terms where each factor appears exactly twice can survive expectation. Each such pair forces one index‐identification, so $\ell$ factors → $\ell/2$ pairings → $\ell/2+1$ free sums → factor $N^{\ell/2+1}$.
Example ($\ell=4$): $\mathrm{tr}(X^4)=\sum_{i_1,i_2,i_3,i_4}X_{i_1i_2}X_{i_2i_3}X_{i_3i_4}X_{i_4i_1}.$ A nonzero pairing ${(1,2),(3,4)}$ sets $i_1=i_2$ and $i_3=i_4$, leaving 2 free indices plus the trace root → $N^3$.
-
Higher Hermite ($k\ge3$) require at least $k$ copies to form a nonzero cumulant because of the orthogonality of Hermite expansion, which costs an extra $(N^{-1/2})^{\,k-2}$ and yields at best \[ N^{\ell/2+1}\times(N^{-1/2})^{\ell}\times(N^{-1/2})^{\,k-2} =N^{1-\tfrac{k-2}{2}}. \] Thus any term with an $R$–factor is $o(N)$ and vanishes after dividing by $N$.
-
Scaling: The size of all the three terms, $X_{ij}$, $Q_{ij}$, and $R_{ij}$ is $O(N^{-1/2})$ in probability. The size of Wigner matrix entries is $X_{ij}$ = $O(N^{-1/2})$, since $\mathbb{E}[X_{ij}^2] = \text{Var}[X_{ij}] + \mathbb{E}[X_{ij}]^2 = 1/N + 0 = 1/N$. Similarly, $Q_{ij} = O(N^{-1/2})$, and from the definition $R_k = O(N^{-1/2})$.
5. Leading contribution
Only $\alpha X$ and $a_2 Q$ survive at order $O(N)$, giving \[ \frac{1}{N}\mathbb{E}[\mathrm{tr}(Y_N^\ell)] = \frac{1}{N}\mathbb{E}\bigl[\mathrm{tr}((\alpha X+a_2Q)^\ell)\bigr] +o(1). \]
6. Matching $Q$ to GOE noise
Define $\beta=a_2\sqrt2$ and let $Z_N\sim\mathrm{GOE}$. One checks $\text{Var}((\beta Z)_{ij})=a_2^2\text{Var}(Q_{ij})$ and zero off-diagonal covariances, so term–by–term the surviving pairings in $\mathrm{tr}((\alpha X+a_2Q)^\ell)$ match those in $\mathrm{tr}((\alpha X+\beta Z)^\ell)$.
7. Conclusion by method of moments
For each fixed $\ell$, \[ \frac{1}{N}\mathbb{E}\bigl[\mathrm{tr}(Y_N^\ell)\bigr] -\frac{1}{N}\mathbb{E}\bigl[\mathrm{tr}((\alpha X+\beta Z)^\ell)\bigr] \;\to\;0. \] Matching all moments implies $\mu_{Y_N}\xrightarrow{w}\mu_{\alpha X+\beta Z}$ as $N\to\infty$, which is the Gaussian Equivalence Principle.