(i) Show that if $\mathbf{A}$ is a matrix and $\mathbf{v}, \mathbf{u}$ are vectors, then

$$ (\mathbf{A} + \mathbf{u}\mathbf{v}^T)^{-1} = \mathbf{A}^{-1} - \frac{\mathbf{A}^{-1}\mathbf{u}\mathbf{v}^T\mathbf{A}^{-1}}{1 + \mathbf{v} \cdot \mathbf{A}^{-1}\mathbf{u}}. $$

Use this formula (Shermann-Morrison formula) to get an expression for the resolvent operator $\mathbf{G}_M(z)$.

(ii) The isolated eigenvalue, when it exists, is a pole of the resolvent operator $\mathbf{G}_M(z)$, which is real and such that $\lambda^{\text{iso}} > 2\sigma$. Using that $\lambda^{\text{iso}}$ does not belong to the spectrum of the unperturbed matrix $\mathbf{J}$, show that it solves the equation

$$ r\mathbf{w} \cdot \mathbf{G}_J(\lambda^{\text{iso}})\mathbf{w} = 1. $$

(iii) Using that $\mathbf{J}$ and $\mathbf{w}$ are independent (free) and that typically $\mathbf{w}$ is delocalized in the eigenbasis of $\mathbf{J}$, show that

$$ \mathbf{w} \cdot \mathbf{G}_J(\lambda^{\text{iso}})\mathbf{w} \xrightarrow{N \to \infty} \mathbf{g}_{\text{sc},\sigma}(\lambda^{\text{iso}}), $$

where $\mathbf{g}_{\text{sc},\sigma}(\lambda)$ is the Stieltjes transform of the GOE($\sigma^2$) matrix $\mathbf{J}$.

(iv) Using the self-consistent equation satisfied by $\mathbf{g}_{\text{sc},\sigma}(\lambda)$, derive the expression of the inverse function $\mathbf{g}_{\text{sc},\sigma}^{-1}$ and determine its domain; use it to show that

$$ \lambda^{\text{iso}} = \frac{\sigma^2}{r} + r, \quad r \ge \sigma. $$

(v) The eigenvectors projections $\xi^\alpha = (\mathbf{w} \cdot \mathbf{u}^\alpha)^2$ can be obtained from the resolvent as residues of the poles:

$$ \xi^\alpha = \lim_{\lambda \to \lambda^\alpha} (\lambda - \lambda^\alpha)\mathbf{w} \cdot \mathbf{G}_M(\lambda) \cdot \mathbf{w}. $$

Use this to show that

$$ \xi^{\text{iso}} = -\frac{1}{r^2 \mathbf{g}'_{\text{sc},\sigma}(\lambda^{\text{iso}})} = 1 - \frac{\sigma^2}{r^2}. $$

Hint. Use that if $\lim_{\lambda \to \lambda_0} f(\lambda) = 0 = \lim_{\lambda \to \lambda_0} g(\lambda)$, then $\lim_{\lambda \to \lambda_0} \frac{f(\lambda)}{g(\lambda)} = \lim_{\lambda \to \lambda_0} \frac{f'(\lambda)}{g'(\lambda)}$.

### A.3 Exercise 3: Equilibrium phase diagram

The goal of this exercise is to derive the equilibrium phase diagram and discuss the condensation transition in the $p = 2$ spherical model with interaction couplings given by matrices of the form $\mathbf{M} = \mathbf{J} + \mathbf{R}$.

(i) Call $\lambda^\alpha$ ($\lambda^1 \le \lambda^2 \le \cdots \lambda^N$) the eigenvalues of $\mathbf{M} = \mathbf{J} + \mathbf{R}$, and $\mathbf{u}^\alpha$ the corresponding eigenvectors. Call $s_\alpha = \mathbf{s} \cdot \mathbf{u}^\alpha$. Show that the partition function can be written as

$$ \mathcal{Z}_\beta = \int d\lambda \int \prod_{\alpha=1}^{N} ds_\alpha e^{\frac{\beta}{2} [\sum_\alpha \lambda^\alpha s_\alpha^2 - \lambda(\sum_\alpha s_\alpha^2 - N)]}. $$

(ii) Show that the thermal expectation value of the mode occupations is

$$ \langle s_{\gamma}^{2} \rangle_{\beta} = \frac{1}{\mathcal{Z}_{\beta}} \int d\lambda \int \prod_{\alpha=1}^{N} ds_{\alpha} s_{\gamma}^{2} e^{-\frac{\beta}{2} [-\sum_{\alpha} \lambda^{\alpha} s_{\alpha}^{2} + \lambda(\sum_{\alpha} s_{\alpha}^{2} - N)]} = \frac{1}{\beta(\lambda^{*} - \lambda_{\gamma})}, $$

where $\lambda^* > \lambda^\gamma$ for all $\gamma$ is fixed by the equation

$$ \sum_{\gamma=1}^{N} \langle s_{\gamma}^{2} \rangle_{\beta} = N = \sum_{\gamma=1}^{N} \frac{1}{\beta(\lambda^{*} - \lambda_{\gamma})}. $$
(iii) The matrix M is a spiked GOE. Take $r < r_c = \sigma$. Justify why for large N the equation for $\lambda^*$ becomes:

$$ \beta = g_{sc,\sigma}(\lambda^*), \quad \lambda^* > 2\sigma, $$

where $g_{sc,\sigma}(\lambda^*)$ is the Stieltjes transform of the GOE($\sigma^2$); show that there is a critical temperature $\beta_c = \sigma^{-1}$ and compute the solution $\lambda^*$ for $\beta < \beta_c$. Show that at $\beta_c$, $\lambda^*$ attains its maximal possible value. Show that at low temperature $\beta > \beta_c$ the equation can be solved assuming condensation of the fluctuations in the lowest-energy mode:

$$ \frac{1}{N} \langle s_N^2 \rangle_{\beta} = 1 - \frac{1}{\beta \sigma}. $$

This condensation transition corresponds also to a transition between a paramagnet at high temperature, and a “ferromagnet in disguise” at low temperature.

(iv) Consider now $r > r_c = \sigma$, when the maximal eigenvalue is $\lambda^N = \lambda^{\text{iso}} = \frac{\sigma^2}{r} + r$; justify why now the critical temperature is $\beta_c = 1/r$, and a solution of the equation for $\lambda^*$ (with $\lambda^* > \lambda^\gamma$) exists for $\beta < \beta_c$. Show that for $\beta > \beta_c$ it must hold

$$ \frac{1}{N} \langle s_N^2 \rangle_{\beta} = \frac{1}{N} \langle s_{\text{iso}}^2 \rangle_{\beta} = 1 - \frac{1}{\beta r}. $$

In this regime, the condensation transition coincides with a transition between a paramagnet at high temperature, and a ferromagnet at low temperature.

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