1. Introduction

In my first several posts, I will discuss about the replica method. The replica method is a non-rigorous but powerful approach that originated within the statistical physics literature. It has been widely applied in other fields, including coding theory and high dimensional statistics. I used this method several times in my research. It helped me quickly figure out the answer of numerous problems.

Most physicists know the replica method better than me. Here I would like to introduce the replica method to theorists outside the physics community. I will not assume knowledges of statistical physics from the audience, but I will occasionally use terminology from physics. For a more sophisticated introduction to the replica method, I recommend Chapter 8 of this fantastic book [1].

In this post, I will discuss how to use the replica method to calculate the spectral norm of certain random matrices. In the next few posts, I will discuss how to use it to calculate the spectral density (Stieltjes transforms) of certain random matrices.

2. A motivating example: spiked GOE matrix

We consider a symmetric random matrix , \[ \bA= \lambda \bu \bu^\sT + \bW, \] where is the signal to noise ratio, is a spike vector, and : meaning that is a symmetric matrix, for , and for .

We are interested in the largest eigenvalue and the corresponding eigenvector of , which are respectively denoted by and . The result below is called the BBP phase transition phenomenon (see [2] for spiked Wishart matrix and [3] for spiked Wigner matrix).

  • The largest eigenvalue :
    For , we have \[ \lim_{n \to \infty} \E[\lambda_{\max} ({\bA})] = 2; \] For , we have \[ \lim_{n \to \infty} \E[\lambda_{\max} ({\bA})] = \lambda + 1/\lambda. \]

  • The correlation of top eigenvector with :
    For , we have \[ \lim_{n \to \infty} \E[\langle \bv_{\max}(\bA), \bu \rangle^2] = 0; \] For , we have \[ \lim_{n \to \infty} \mathbb{E}[\langle {\boldsymbol v}_{\max}({\boldsymbol A}), {\boldsymbol u}\rangle^2] = 1 - 1/\lambda^2. \]

In the rest of this blog post, we will derive this result using the replica method.

3. Tricks in statistical physics

3.1. The free energy trick

Let and be two functions. We would like to calculate the following quantities \[ \begin{aligned} &\max_{\bsigma \in \Omega} H(\bsigma),
&f(\arg \max_{\bsigma} H(\bsigma)). \end{aligned} \]

The free energy trick works as follows. Define a Gibbs measure \[ \mu_{\beta, h}(\de \bsigma) = \frac{1}{Z(\beta, h)} \exp\{ \beta H(\bsigma) + h f(\bsigma)\} \nu_0(\de \bsigma), \] where denotes the normalizing constant \[ \begin{align}\tag{Def Z} \label{eqn:def_Z} Z(\beta, h) = \int_{\Omega} \exp\{ \beta H(\bsigma) + h f(\bsigma)\} \nu_0(\de \bsigma), \end{align} \] and denotes a reference measure on . In physics, stands for the inverse temperature, and stands for the strength of external field. Define the free energy function by \[ \Psi(\beta, h) = \log Z(\beta, h). \] The following lemma shows that, the free energy function can generate many interesting quantities, including and .

Denote the ensemble average operator by \[ \langle g \rangle_{\beta, h} = \int_\Omega g(\bsigma) \mu_{\beta, h}(\de \bsigma). \] Then we have \begin{align}\tag{1}\label{eq:1} \partial_\beta \Psi(\beta, h) =& \langle H \rangle_{\beta, h},\\ \partial_h \Psi(\beta, h) =& \langle f \rangle_{\beta, h},\\ \end{align} and \begin{align}\tag{2}\label{eq:2} \lim_{\beta \to \infty} \partial_\beta \Psi(\beta, h) =& \max_{\bsigma \in \Omega} H(\bsigma),\\ \lim_{\beta \to \infty} \partial_h \Psi(\beta, h) =& f(\arg\max_{\bsigma \in \Omega} H(\bsigma) ). \end{align}

Equation \eqref{eq:1} follows from basic calculus of exponential family. The intuition for Equation \eqref{eq:2} gives the following: for and large , the measure concentrates at the maxima of .

In the spiked GOE model, the Hamiltonian associated with the random matrix is defined as \[ H_{n, \lambda}(\bsigma) = \langle \bsigma, \bA \bsigma \rangle = \lambda \langle \bu, \bsigma\rangle^2 + \langle \bsigma, \bW \bsigma\rangle. \] Denote the (random) partition function associated with the Hamiltonian by \[ Z(\beta, \lambda, n) = \int_{\S^{n-1}} \exp\{ \beta n H_{n, \lambda}( \bsigma ) \} \nu_0(\de \bsigma). \] Here is the uniform probability measure on , and the factor in the exponent serves for proper normalization.

Define \[ \begin{align}\tag{3}\label{eq:3} \vphi_n(\lambda) =& \lim_{\beta \to \infty} \frac{1}{\beta n}\E[\log Z(\beta, \lambda, n)],\\ \vphi(\lambda) =& \lim_{n\to \infty} \vphi_n(\lambda) = \lim_{n \to \infty} \lim_{\beta \to \infty} \frac{1}{\beta n}\E[\log Z(\beta, \lambda, n)]. \end{align} \] The function describes the asymptotics of and , as shown in the lemma below.

For , we have \[ \begin{aligned} \vphi_n(\lambda) =& \E[\sup_{\bsigma} H_{n, \lambda}(\bsigma)]= \E[\lambda_{\max} (\bA)],\\ \vphi_n'(\lambda) =& \E[\langle \bv_{\max}(\bA), \bu \rangle^2]. \end{aligned} \]

We leave the proof of this lemma to readers.

3.2. The replica trick

The difficulty of calculating originated from the expectation of as in Eq. \eqref{eq:3}. The replica trick can convert the problem of calculating the expectation of into calculating the moments of . Calculating the moments of is an easier problem.

In the language of mathematics, the replica trick gives the following formula \[\tag{4} \label{eq:4} \E[\log Z] = \lim_{k \to 0} \frac{1}{k}\log \E[Z^k ]. \] Here is a simple proof of this equality: \[ \begin{aligned} \E[\log Z] =& \E[(\log Z^k) / k] = \lim_{k \to 0} \E [ \log(1 + Z^k - 1) / k] = \lim_{k \to 0} \E [ ( Z^k - 1 ) / k ] \\ =&\lim_{k \to 0} (\E [ Z^k] - 1)/k = \lim_{k \to 0} \frac{1}{k} \log (1 + \E[Z^k] - 1) = \lim_{k \to 0} \frac{1}{k}\log \E[Z^k ]. \end{aligned} \]

Using the definition of in Eq. \eqref{eq:3} and the replica trick in Eq. \eqref{eq:4}, we can rewrite as \[ \begin{aligned} \vphi(\lambda) \equiv& \lim_{n\to \infty} \lim_{\beta \to \infty} \frac{1}{\beta n}\E[\log Z(\beta, \lambda, n)] = \lim_{n \to \infty} \lim_{\beta \to \infty} \lim_{k \to 0}\frac{1}{\beta k n}\log \E[Z(\beta, \lambda, n)^k]. \\ \end{aligned} \] With some heuristic change of limit (which we will not justify), we get \[ \vphi(\lambda) \stackrel{\cdot}{=} \lim_{\beta \to \infty} \lim_{k \to 0} \lim_{n \to \infty} \frac{1}{\beta k n}\log \E[Z(\beta, \lambda, n)^k]. \] There are three limits in the right hand side of the above equation. Define \[ \begin{align}\tag{Def S} \label{eqn:defS} S(k, \beta, \lambda) = \lim_{n \to \infty} \frac{1}{n}\log \E[Z(\beta, \lambda, n)^k], \end{align} \] and \[ \begin{align}\tag{Def $\Psi$} \label{eqn:defPsi} \Psi(\beta, \lambda) = \lim_{k \to 0} \frac{1}{k} S(k, \beta, \lambda). \end{align} \] This gives \[ \begin{align}\tag{Def $\vphi$} \label{eqn:defvphi} \vphi(\lambda) = \lim_{\beta \to \infty} \frac{1}{\beta}\Psi(\beta, \lambda). \end{align} \] In the following section, we calculate , , and functions sequentially.

4. The replica calculations

4.1. The large limit: moments calculation

4.1.1. A variational formula for function

Recall the definition of function given in Eq. \eqref{eqn:defS}. We claim the following equality for \begin{align}\tag{5}\label{eqn:S_function} S(k, \beta, \lambda) \equiv \lim_{n \to \infty} \frac{1}{n}\log\E[Z(\beta, \lambda, n)^k]= \sup_{\bQ \in \R^{(k+1) \times (k+1)}, \diag(\bQ) = 1, \bQ \succeq \bzero} {U(\bQ) }, \end{align} where \begin{align}\tag{6}\label{eqn:U_function} U(\bQ) = \beta \lambda \sum_{i=1}^k q_{0i}^2 + \beta^2 \sum_{ij = 1}^k q_{ij}^2 + \frac{1}{2} \log(\det(\bQ)), \end{align} and is symmetric, with (the index of starts from and ends at ) \[ \begin{align}\tag{Def $\bQ$} \label{eqn:def_Q} \bQ = \begin{bmatrix} 1& q_{01} & q_{02} & \ldots & q_{0k}\\ q_{01} & 1 & q_{12} & \ldots & q_{1k}\\ q_{02} & q_{12} & 1 & \ldots & q_{2k} \\ \ldots & \ldots & \ldots & \ldots & \ldots\\ q_{0k} & q_{1k} & q_{2k}& \ldots & 1\\ \end{bmatrix}. \end{align} \] Equation \eqref{eqn:S_function} is exact and rigorous, but the computation is involved. By calculating the moments of directly (the calculation is given in Section 4.1.2), we get \begin{align}\tag{7} \label{eqn:moments_result} \E[Z(\beta, \lambda, n)^k] = \int \exp\{ n U(\bQ) + o(n)\} \de \bQ. \end{align} Eq. \eqref{eqn:S_function} follows from Eq. \eqref{eqn:moments_result} and Laplace method. We suggest the readers skipping the derivation of Eq. \eqref{eqn:moments_result} at the first time reading, and continue to read Section 4.2.

4.1.2. The replica calculations

In this section, we derive Eq. \eqref{eqn:moments_result}. At the first time reading, we suggest the readers skipping this section and continue to read Section 4.2.

Recall the definition of given in Eq. \eqref{eqn:def_Z}. The ‘th moment of gives \[ \begin{aligned} \E[Z(\beta, \lambda, n)^k] = & \E\Big[\Big(\int_{\S^{n-1}} \exp\{ \beta n H_{n, \lambda}( \bsigma ) \} \nu_0(\de \bsigma)\Big)^k \Big].
\end{aligned} \] Here, the expectation is taken with respect to the randomness of matrix in . The first trick is to rewrite the ‘th power of integration into integration with respect to replicas of . In this way, we can exchange the expectation with integrals. \[ \begin{aligned} \E[Z(\beta, \lambda, n)^k] =&\E\Big[\Big(\int_{\S^{n-1}} \exp\{ \beta n H_{n, \lambda}( \bsigma ) \} \nu_0(\de \bsigma)\Big)^k \Big] = \E\Big[\int_{(\S^{n-1})^k} \exp\Big\{ \beta n \sum_{i = 1}^k H_{n, \lambda}( \bsigma_i ) \Big\} \prod_{i \in [k]} \nu_0(\de \bsigma_i) \Big]\\ =&\int_{(\S^{n-1})^k} \E\Big[ \exp\Big\{ \beta n \sum_{i = 1}^k [\lambda \langle \bu, \bsigma_i\rangle^2 + \langle \bsigma_i, \bW \bsigma_i\rangle ] \Big\} \Big] \prod_{i \in [k]} \nu_0( \de \bsigma_i). \end{aligned} \] Let with . Noting that the GOE matrix shares the same distribution with matrix , we have \[ \begin{aligned} \E[Z(\beta, \lambda, n)^k] =& \int_{(\S^{n-1})^k} \E_{G_{ij} \sim \cN(0,1)}\Big[ \exp\Big\{ \beta n \lambda \sum_{i = 1}^k \langle \bu, \bsigma_i\rangle^2 + \beta n \Big\langle (\bG + \bG^\sT) / \sqrt{2n}, \sum_{i=1}^k \bsigma_i\bsigma_i^\sT \Big\rangle \Big\} \Big] \prod_{i \in [k]} \nu_0(\de \bsigma_i) \\ =& \int_{(\S^{n-1})^k} \exp\Big\{ \beta n \lambda \sum_{i = 1}^k \langle \bu, \bsigma_i\rangle^2 \Big\} \times \E_{G_{ij} \sim \cN(0,1)}\Big[ \exp\Big\{ \beta \sqrt{2 n} \Big\langle \bG, \sum_{i=1}^k \bsigma_i\bsigma_i^\sT \Big\rangle \Big\} \Big] \prod_{i \in [k]} \nu_0(\de \bsigma_i). \end{aligned} \] Using the formula of Gaussian moment generating function , we get \[ \begin{aligned} \E[Z(\beta, \lambda, n)^k] =& \int_{(\S^{n-1})^k} \exp\Big\{ \beta n \lambda \sum_{i = 1}^k \langle \bu, \bsigma_i \rangle^2 + n \beta^2 \Big \Vert \sum_{i=1}^k \bsigma_i \bsigma_i^\sT \Big\Vert_F^2 \Big\} \prod_{i \in [k]} \nu_0(\de \bsigma_i) \\ =& \int_{(\S^{n-1})^k} \exp\Big\{ \beta n \lambda \sum_{i = 1}^k \langle \bu, \bsigma_i \rangle^2 + n \beta^2 \sum_{i, j=1}^k\Big\langle \bsigma_i\bsigma_i^\sT, \bsigma_j \bsigma_j^\sT\Big \rangle \Big\} \prod_{i \in [k]} \nu_0(\de \bsigma_i) \\ =& \int_{(\S^{n-1})^k} \exp\Big\{ \beta n \lambda \sum_{i = 1}^k \langle \bu, \bsigma_i \rangle^2 + n \beta^2 \sum_{i, j=1}^k \langle \bsigma_i, \bsigma_j \rangle^2 \Big\} \prod_{i \in [k]} \nu_0(\de \bsigma_i). \end{aligned} \]

In the following, we make use a heuristic argument (using Dirac delta function). Note we have \[ 1 = \int \prod_{i=1}^k \delta \Big( \langle \bu,\bsigma_i\rangle - n q_{0i} \Big) \prod_{1 \le i < j\le k} \delta\Big( \langle\bsigma_i, \bsigma_j\rangle - n q_{ij} \Big) \de \bQ \] where . Using this equality, we get \[ \begin{aligned} \E[Z(\beta, \lambda, n)^k] =& \int \de \bQ \exp\Big\{\beta n \lambda \sum_{i = 1}^k q_{0i}^2 + n \beta^2 \sum_{i, j=1}^k q_{ij}^2 \Big\} \cdot \int_{(\R^n)^k} \prod_{i=1}^k \delta \Big( \langle \bu,\bsigma_i\rangle - n q_{0i} \Big) \prod_{1 \le i < j\le k} \delta\Big( \langle\bsigma_i, \bsigma_j\rangle - n q_{ij} \Big) \prod_{i \in [k]} \nu_0(\de \bsigma_i) \\ =& \int \de \bQ \exp\Big\{\beta n \lambda \sum_{i = 1}^k q_{0i}^2 + n \beta^2 \sum_{i, j=1}^k q_{ij}^2 + n H_n(\bQ) \Big\},
\end{aligned} \] where is given by Eq. \eqref{eqn:def_Q} and \[ \begin{aligned} H_n(\bQ) =& \frac{1}{n} \log \int_{(\R^n)^k} \prod_{i=1}^k \delta \Big( \langle \bu,\bsigma_i\rangle - n q_{0i} \Big) \prod_{1 \le i < j\le k} \delta\Big( \langle\bsigma_i, \bsigma_j\rangle - n q_{ij} \Big) \prod_{i \in [k]} \nu_0(\de \bsigma_i) \\ =& \frac{1}{n} \log \int_{(\R^n)^{k+1}} \prod_{0 \le i < j\le k} \delta\Big( \langle\bsigma_i, \bsigma_j\rangle - n q_{ij} \Big) \prod_{0 \le i \le k} \nu_0( \bsigma_i), \end{aligned} \] which serves as the rate function of the random matrix when . This rate function can be found in standard textbooks \[ \lim_{n \to \infty} H_n(\bQ) = \frac{1}{2}\log \det(\bQ). \] Here we give a heuristic argument to calculate the entropy term , (here the argument is handwaving, but can be made rigorous)

\[ \begin{aligned} H_n(\bQ) \stackrel{\cdot}{=}& \frac{1}{n} \P_{(\bsigma_i)_{i \in [k]} \sim \Unif(\S^{n-1})} \Big( \big(\langle \bsigma_i, \bsigma_j \rangle / n \big)_{i, j \in [k]}\approx \bQ \Big)\\\ \stackrel{\cdot}{=}& \inf_{\lambda_{ij}} \frac{1}{n} \log \int_{(\R^n)^{k+1}} \prod_{0 \le i \le j\le k} \exp\Big\{ - \lambda_{ij}\langle\bsigma_i, \bsigma_j\rangle / 2 + n q_{ij} \lambda_{ij} / 2 \Big\} \prod_{0 \le i \le k} \de \bsigma_i + \const\\\ =& \inf_{\lambda_{ij}} \log \int_{\R^{k+1}} \prod_{0 \le i \le j\le k} \exp\Big\{ - \lambda_{ij} \sigma_i\sigma_j / 2 + q_{ij} \lambda_{ij}/2 \Big\} \prod_{0 \le i \le k} \de \sigma_i + \const\\\ =& \inf_{\bLambda = (\lambda_{ij})_{0 \le i \le j \le k}} \Big[ \langle \bQ, \bLambda\rangle / 2 - \log(\det(\bLambda))/2 \Big] + \const\\\ \stackrel{\cdot}{=}& \frac{1}{2} \log \det(\bQ), \end{aligned} \]

where we use the fact that, for , we have \[ \lim_{n \to \infty} \frac{1}{n} \log \P_{\bX_i}\Big( \frac{1}{n} \sum_{i=1}^n \bX_i \approx \bx \Big) = \inf_{\blambda} \log \E_\bX \Big[\exp\{ \langle\blambda, \bX - \bx\rangle \} \Big]. \] This gives Eq. \eqref{eqn:moments_result}.

4.2. The small limit: replica symmetric ansatz

4.2.1. Replica symmetric ansatz

Next we calculate (recall Eq. \eqref{eqn:defPsi}) \[ \Psi(\beta, \lambda) = \lim_{k \to 0} S(\beta, \lambda, k)/k. \] We gave the expression of when is integral (c.f. Eq. \eqref{eqn:S_function}), where serves as the dimension of variable in the variational formula. However, to take the limit , we need the expression for when takes real numbers. This is the difficulty to calculate the small limit using Eq. \eqref{eqn:S_function}.

The physicists’ trick is to plug in an ansatz for the matrix to simplify the expression of . Note that defined in Eq. \eqref{eqn:U_function} has some symmetry: if we permute the ‘st to ‘th rows of , and perform the same permutation to its columns, the function stays the same. This motivates us to consider the “replica symmetric ansatz”: \[ \bQ = \begin{bmatrix} 1& \mu & \mu & \ldots & \mu\\ \mu & 1 & q& \ldots & q\\ \mu & q & 1 & \ldots & q \\ \ldots & \ldots &\ldots & \ldots & \ldots\\ \mu & q & q & \ldots & 1\\ \end{bmatrix}, \] or equivalently \[ q_{0i} = \mu, ~~ \forall i \in [k], ~~~~~~~~~~~~~~ q_{ij} = q, ~~ \forall 1 \le i < j \le k. \] Plugging this ansats into Eq. \eqref{eqn:U_function}, and simplifying the log determinant term using the matrix determinant lemma \[ \begin{aligned} &\log(\det(\bQ)) = \log \det( (1-q) \id_k + (q - \mu^2) \ones \ones^\sT) \\ =& \log \det( \id_k + [(q - \mu^2)/(1-q) ] \ones \ones^\sT) + k \log(1-q)\\ =& \log(1 + k [(q - \mu^2)/(1-q) ]) + k \log(1-q)\\ =& \log\Big( 1 - \frac{\mu^2 k}{1 + (k - 1)q} \Big) + \log \Big(1 + \frac{kq}{1-q} \Big) + k \log(1 - q), \end{aligned} \] we get \[ U(\bQ) = \beta \lambda k \mu^2 + \beta^2 k + \beta^2 k (k-1) q^2 + \frac{1}{2} \Big[ \log\Big( 1 - \frac{\mu^2 k}{1 + (k - 1)q} \Big) + \log \Big(1 + \frac{kq}{1-q} \Big) + k \log(1 - q) \Big] \equiv U(\mu, q, k). \]

Assuming that the maximum of is taken at a “replica symmetric” form of , we get \[ S(\beta, \lambda, k) \stackrel{\cdot}{=} \sup_{\mu, q} U(\mu, q, k). \]

4.2.2. The small limit

Using the replica symmetric ansatz, we gave an expression for for any . Next we would like to calculate the limit (recall the definition of given in Eq. \eqref{eqn:defPsi}) \[ \Psi(\beta, \lambda) = \lim_{k \to 0} S(\beta, \lambda, k) / k = \lim_{k \to 0} \sup_{\mu, q} \frac{1}{k} U(\mu, q, k). \] Define \[ u(\mu, q) \equiv \lim_{k \to 0} \frac{1}{k} U(\mu, q, k) = \beta \lambda \mu^2 + \beta^2 (1 - q^2) - \frac{\mu^2}{2(1 - q)} + \frac{q}{2 (1 - q)} + \frac{1}{2}\log(1 - q). \] We need to exchange the operations and in some way. Another heuristic physicists’ convention comes here: the operation becomes the operation, where is a set defined as \[ \ext_{\bx} f(\bx) = \Big\{ f(\bx_\star): \nabla f(\bx_\star) = \bzero\Big\}. \] Using this convention, we get \[ \Psi(\beta, \lambda) = \lim_{k \to 0} S(\beta, \lambda, k) / k \in \ext_{\mu, q}u(\mu, q), \]

Next, we calculate all the stationary points of , and collect the value of evaluated at these stationary points.

By basic calculus, we have \[ \begin{aligned} \partial_\mu u(\mu, q) =& 2 \Big( \beta \lambda - \frac{1}{2 (1-q)} \Big) \mu, \\ \partial_q u(\mu, q) =& -2 \beta^2 q - \frac{\mu^2}{2(1 - q)^2} + \frac{1}{2 (1 - q)^2}. \end{aligned} \] When , by basic algebra, we find two solutions of .

  • One branch of solution of gives \[ \begin{aligned} \mu_1 =& 0, \\ q_1 =& 1 - \frac{1}{2 \beta}. \end{aligned} \] At this solution, we have \[ \Psi_1(\beta, \lambda) = u(\mu_1, q_1) = 2 \beta - \frac{3}{4} - \frac{1}{2} \log(2 \beta). \]
  • Another branch of solution of gives \[ \begin{aligned} \mu_2 =& \Big( \Big(1 - \frac{1}{\lambda^2}\Big) \Big(1 - \frac{1}{2 \beta \lambda}\Big) \Big)^{1/2},\\ q_2 =& 1 - \frac{1}{2 \lambda \beta}, \end{aligned} \] At this solution, we have \[ \Psi_2(\beta, \lambda) = u(\mu_2, q_2) = \beta\Big( \lambda + \frac{1}{\lambda}\Big) - \Big( \frac{1}{4 \lambda^2} + \frac{1}{2} \Big) - \frac{1}{2}\log(2 \lambda \beta). \]

Hence we get \[ \Psi \in \ext_{\mu, q}u(\mu, q) = \{\Psi_1, \Psi_2 \}. \]

4.3. The large limit

Recall the definition of in Eq. \eqref{eqn:defvphi}. For the first branch , the corresponding function gives \[ \vphi_1(\lambda) = \lim_{\beta \to \infty} \frac{1}{\beta} \Psi_1(\beta, \lambda) = 2. \] For the second branch , the corresponding function gives \[ \vphi_2(\lambda) = \lim_{\beta \to \infty} \frac{1}{\beta} \Psi_2(\beta, \lambda) = \lambda + \frac{1}{\lambda}. \]

4.4. Select a branch

There are two branches of solutions. We need to select the branch that makes sense: we hope that gives an expression for , hence it should be non-decreasing in , and it should satisfy .

  • When , the branch is decreasing. Hence, we should select the branch .
  • When , the branch stays constant. Hence, we should select the branch .

The discussion above gives the following prediction \[ \lim_{n \to \infty} \E[\lambda_{\max}(\bA)]= \vphi(\lambda) = \begin{cases} 2, ~~~& \lambda \le 1, \\ \lambda + \frac{1}{\lambda}, ~~~& \lambda > 1. \end{cases} \] This turns out to be the correct answer!

5. Summary

In summary, we calculated the largest eigenvalue of the spiked GOE matrix . Using the free energy trick and the replica trick, the problem becomes calculating three limits sequentially: the large limit, the small limit, and the large limit. The large limit calculation is rigorous, but is also complicated. The small limit calculation is non-rigorous: we pluged in the replica symmetric ansatz, and changed the operation to the operation. The large limit calculation is straightforward. Finally, we get two branches of solutions. We used some simple properties of the largest eigenvalue of to select the branch when is in different regime. The prediction formula turns out to be the correct answer.


[1] Mezard, Marc, and Andrea Montanari. Information, physics, and computation. Oxford University Press, 2009.

[2] Baik, Jinho, Gérard Ben Arous, and Sandrine Péché. “Phase transition of the largest eigenvalue for nonnull complex sample covariance matrices.” The Annals of Probability 33.5 (2005): 1643-1697.

[3] Péché, Sandrine. “The largest eigenvalue of small rank perturbations of Hermitian random matrices.” Probability Theory and Related Fields 134.1 (2006): 127-173.