# A theory of singular values for finite free probability

Aurelien Gribinski
Princeton University
EPFL

###### Abstract

We introduce a finite version of free probability for rectangular matrices
that amounts to operations on singular values of polynomials. This study is
motivated by the companion papers [1] and [2], as well as the corresponding
paper dealing with the square case [8]. In the process we exhibit a canonic
bivariate operation on polynomials, seemingly more natural when singular
values are concerned. We show that we can replicate the transforms from free
probability, and that asymptotically there is convergence from rectangular
finite free probability to rectangular free probability. Lastly, we show that
classical distribution results such as a law of large numbers or a central
limit theorem can be made explicit in this new framework where random
variables are replaced by polynomials.

##  1 Introduction

Free probability is a recent field created by Dan Voiculescu (see [11] or
[12]) that studies operators on infinite dimensional spaces through the lens
of moment distributions and convolutions - both from the combinatorial and
analytic points of view. It led to many asymptotic results in random matrix
theory. On the other hand, finite free probability was introduced in a series
of papers by Marcus, Spielman and Srivastava (in particular [7], [6]) a few
years ago, and proved some surprising analogues of results in free probability
but for square matrices (operators on finite dimensional spaces) and their
associated characteristic polynomials (analogues of moment distributions).
They used their results to prove the existence of new large families of
Ramanujan graphs. The theory was further developed some time after by Marcus
([8]). Basically, equalities in the realm of free probability turn
experimentally into inequalities in the realm of finite free probability. This
transition, however, is not well understood yet. The goal of this paper is to
extend the theory of finite free probability in a systematic way from
eigenvalues to singular values, or said otherwise from square hermitian
matrices to rectangular matrices.

###  1.1 An introduction to rectangular free probability

In classical probability, if we are given two random variables $X_{1}$ and
$X_{2}$ in two probability spaces $(M_{1},\mu_{1})$ and $(M_{2},\mu_{2})$,
then one way to investigate the joint distribution $\mu_{X_{1},X_{2}}$ of
$X_{1}$ and $X_{2}$ is to have access to the expectation
$\mathbb{E}[p(X_{1},X_{2})]$ for all bivariate polynomials $p$. In particular,
one can calculate $\mu_{X_{1},X_{2}}$ knowing only $\mu_{1}$ and $\mu_{2}$ if
$X_{1}$ and $X_{2}$ are independent, in which case we have
$\mu_{X_{1},X_{2}}=\mu_{1}\otimes\mu_{2}$. Another way to see it is that we
know all product moments by the knowledge of the moments of $X_{1}$ and
$X_{2}$ separately.
In a similar spirit, we can extend these notions to noncommutative probability
spaces. A noncommutative probability space $(\mathcal{A},\phi)$ is a unital
algebra $\mathcal{A}$ over $\mathbb{C}$ and a unital linear functional (trace
or expectation) $\phi:\mathcal{A}\rightarrow\mathbb{C}$ with $\phi(1)=1$.
Elements of $\mathcal{A}$ are called non commutative random variables. All the
definitions and theorems that follow can be found in [12].

######  Definition 1.1 (distribution of a random variable).

If there exists a probability distribution $\mu_{a}$ on $\mathbb{R}$ (we
restrict to hermitian operators in the following) such that
$\phi(a^{k})=\int_{\mathbb{R}}z^{k}d\mu_{a}(z)$ for all $k\in\mathbb{N}$, then
we call $\mu_{a}$ the (spectral) distribution of the noncommutative random
variable $a$. In this situation, we call $a$ a spectral operator.

######  Remark 1.2.

One can wonder if it compatible with the spectrum of usual matrices when the
expectation function is a trace. Consider the noncommutative unital algebra
$M_{d}(\mathbb{C})$ of complex square matrices of size $d$, on which there is
the normalized trace defined by $\phi(a)=\frac{1}{d}\sum\lambda_{i}(a)$, where
the $\lambda_{i}$ are the eigenvalues of the matrix $a$.Then
$\phi(a^{k})=\frac{1}{d}\sum\lambda_{i}^{k}(a)$ and
$\mu_{a}=\frac{1}{d}\sum\delta_{\lambda_{i}(a)}$, which is the average of
delta masses. The measure $\mu_{a}$ is indeed the eigenvalue distribution of
$a$.

######  Definition 1.3 (joint distribution for hermitian spaces).

The joint distribution of spectral operators $a$ and $b$ consists of all
crossed moments:$\big{\\{}\phi(a_{1}a_{2}\dots a_{n})|n\geq
1,a_{i}\in\\{a,b\\}\big{\\}}$.

######  Definition 1.4 (freeness in spectral case).

We say that spectral operators $a$ and $b$ are free if for all
$n\in\mathbb{N}$ and all univariate polynomials $p_{1},p_{2},\dots p_{2n}$ the
following holds

$\phi[p_{1}(a)p_{2}(b)...p_{2n-1}(a)p_{2n}(b)]=0$

whenever $\phi[p_{2i-1}(a)]=\phi[p_{2i}(b)]=0$ for $i\leq n$. These equalities
give a way of computing recursively all crossed moments in terms of the
moments of $a$ and $b$.

######  Remark 1.5.

This notion of freeness is a generalization of the classical notion of
independence in probability theory. Independence of random variables $X$ and
$Y$ in the case where $\phi$ is the expectation is equivalent to

$\displaystyle\phi[p_{1}(X)p_{2}(Y)]=0$ $\displaystyle\text{ whenever
}\phi[p_{1}(X)]=\phi[p_{2}(Y)]=0$

Indeed, we have to notice that

$\phi[(p_{1}(X)-\phi(p_{1}(X)))(p_{2}(Y)-\phi(p_{2}(Y)))]=\phi[p_{1}(X)p_{2}(Y)]-\phi[p_{1}(X)]\phi[p_{2}(Y)]$

The main difference between free independence and classical independence is
that free independence respects the noncommutativity of the variables.

The most interesting use of free independence for us is the analysis of the
asymptotics of large random matrices.

######  Definition 1.6 (Orthogonal invariance).

Consider some hermitian random matrices $A_{d}$ (for $d\in\mathbb{N}$). We say
that they form an orthogonally invariant random ensemble if they are invariant
with respect to conjugation with a Haar orthogonal random matrix: that is the
entries of $A_{d}$ and $Q_{d}^{T}A_{d}Q_{d}$ have the same joint distribution
for a random $Q_{d}$ in the orthogonal group of size $d$ (for all
$d\in\mathbb{N}$) . Conjugation by a Haar orthogonal random matrix corresponds
to a random rotation of the eigenvectors.

The following theorem states that independence of the eigenvectors (obtained
by rotating randomly) in the classical sense leads to free asymptotic
independence of the spectral distributions.

######  Theorem 1.7 (Voiculescu).

Let $A_{d}$ and $B_{d}$ be $d\times d$ real independent hermitian random
orthogonally invariant matrix ensembles whose eigenvalue distributions
converge in probability to distributions $\mu$ and $\nu$. Then define
$a:=\lim_{d}A_{d}$ and $b:=\lim_{d}B_{d}$. These two spectral operators
operators are freely independent with respective spectral measures $\mu$ and
$\nu$.

######  Definition 1.8 (free convolution).

For freely independent spectral operators $a$ and $b$ with real distributions
$\mu_{a}$ and $\mu_{b}$ we denote by $\mu_{a}\boxplus\mu_{b}$ the distribution
of $a+b$.

######  Remark 1.9.

If we take any independent random matrices $A_{d}$ and $B_{d}$ whose
eigenvalue distributions converge in probability to distributions $\mu_{a}$
and $\mu_{b}$ without any assumption on orthogonal invariance, then
$Q_{d}^{T}A_{d}Q_{d}$ and $\tilde{Q_{d}}^{T}B_{d}\tilde{Q_{d}}$ are
orthogonally invariant (for $Q_{d}$ and $\tilde{Q_{d}}$ orthogonal Haar
random) and we get that they are asymptotically free. As the eigenvalue
distribution of $Q_{d}^{T}A_{d}Q_{d}+\tilde{Q_{d}}^{T}B_{d}\tilde{Q_{d}}$ is
the same as $A_{d}+Q_{d}^{T}B_{d}Q_{d}$ (conjugation doesn’t affect the
eigenvalue distribution), we can conclude that the eigenvalue distribution of
$A_{d}+Q_{d}^{T}B_{d}Q_{d}$ is converging to $\mu_{a}\boxplus\mu_{b}$.

In order to compute the free convolution, we associate power series to
distributions that behave well with respect to this operation.

######  Definition 1.10 (Cauchy and $R$-transforms).

Define the Cauchy transform of a Borel measure $\mu$ on $\mathbb{R}$ as

$\mathcal{G}_{\mu}(x)=\int_{t\in\mathbb{R}}\frac{d\mu(t)}{x-t},\ \ \text{ for
}Im(x)>0$

The $R$-transform is in turn defined as

$\mathcal{R}_{\mu}(x)=\mathcal{G}_{\mu}^{-1}(x)-\frac{1}{x}=\mathcal{G}_{\mu}^{-1}(x)-\mathcal{G}_{\mu_{0}}^{-1}(x)$

Where $\mu_{0}$ is the Dirac mass at zero. By inverse we mean compositional
inverse for power series (around $x=\infty$, that is $\mathcal{G}_{\mu}(x)$ is
a power series in $\frac{1}{x}$).

######  Remark 1.11.

If the measure $\mu$ is compactly supported, which is the case in this paper,
then we can also define $\mathcal{G}_{\mu}(x)$ for real $x>\max\textup{Supp}\
\mu$, where $\textup{Supp}\ \mu$ is the support of $\mu$.

The $R$-transform is useful because it fully characterizes the distribution
and linearizes the convolution.

######  Theorem 1.12 (linearization, see [12]).

$\mathcal{R}_{\mu_{a}\boxplus\mu_{b}}(x)=\mathcal{R}_{\mu_{a}}(x)+\mathcal{R}_{\mu_{b}}(x)$

As the $R$-transform characterizes the distribution, the previous theorem
allows us to compute the free sum distribution of two operators.

We can extend the definition of freeness for general (non spectral) operators
that admit an adjoint.The idea is that an operator $a$ that admits an adjoint
$a^{\star}$ is such that $a^{\star}a$ is spectral.

######  Definition 1.13 (joint distribution for $\star$-probability spaces).

The notion of joint distributions for hermitian operators can be generalized
to a similar one for all operators that admit some adjoint. The joint
distribution of $a$ and $b$ consists in all crossed moments:
$\big{\\{}\phi(a_{1},a_{2}\dots a_{n})|n\geq
1,a_{i}\in\\{a,b,a^{\star},b^{\star}\\}\big{\\}}$ where $a^{\star}$ represents
the adjoint.

######  Definition 1.14 (freeness for $\star$-probability spaces).

We say that $a$ and $b$ are free if for all $n$ and all bivariate polynomials
$p_{1},p_{2},\dots,p_{2n}$,

$\phi[p_{1}(a,a^{\star})p_{2}(b,b^{\star})...p_{2n-1}(a,a^{\star})p_{2n}(b,b^{\star})]=0$

whenever $\phi[p_{2i-1}(a,a^{\star})]=\phi[p_{2i}(b,b^{\star})]=0$ for all
$i\leq n$. These equalities give a way to compute recursively all crossed
moments in terms of the moments of $a$ and $b$.

We wish to connect this extended notion of freeness to random matrices. Let’s
recall that if a matrix $M$ of size $m\times d$ has singular decomposition
$M=UDV$, with $U$ and $V$ orthogonal matrices of size $m\times m$ and $d\times
d$, and $D$ is a nonnegative diagonal matrix of size $m\times d$, then the
singular values are the elements of the diagonal of $D$. They also correspond
to the square roots of the eigenvalues of $M^{T}M$. The uniform distribution
on the singular values will be called the singular law of $M$. A random matrix
is said to be bi-orthogonally invariant if its distribution is invariant under
the left and right actions of the orthogonal group. For $\mu$ probability
measure on $\mathbb{R}$, denote by $\tilde{\mu}$ the symmetrization of $\mu$,
which is the probability measure defined by
$\tilde{\mu}(B)=\frac{\mu(B)+\mu(-B)}{2}$ for all Borel sets $B$.

A first step was accomplished by Voiculescu [10] who proved the following.

######  Theorem 1.15 (free square singular addition).

The asymptotic singular law of the sum of two independent square random
matrices whose distributions are bi-orthogonally invariant and for which the
symmetrizations of the respective singular laws converge weakly to the
probability measures $\mu_{1}$ and $\mu_{2}$, respectively, only depends on
$\mu_{1}$ and $\mu_{2}$, and can be expressed easily from $\mu_{1}$ and
$\mu_{2}$: it is the probability measure on $[0,\infty)$, the symmetrization
of which is the free convolution of $\mu_{1}$ and $\mu_{2}$. Notice that if we
call the symmetrization of the limiting distribution of the sum
$\mu_{1}\boxplus^{1}\mu_{2}$, then
$\mu_{1}\boxplus^{1}\mu_{2}=\mu_{1}\boxplus\mu_{2}$. Furthermore, the random
matrices become free in the limit.

######  Remark 1.16.

if $A_{d}$ and $B_{d}$ are two independent square random matrix ensembles and
$Q_{d}$, $R_{d}$, $\tilde{Q_{d}}$, $\tilde{R_{d}}$ are Haar random orthogonal
matrices of size $d$, then $Q_{d}^{T}A_{d}R_{d}$ and
$\tilde{Q_{d}}^{T}B_{d}\tilde{R_{d}}$ are free in the limit $d$ going to
infinity, and we know the asymptotic singular law of
$Q_{d}^{T}A_{d}R_{d}+\tilde{Q_{d}}^{T}B_{d}\tilde{R_{d}}$. But as the singular
distribution is unchanged by left and right conjugation, then it is the same
as the singular law of $A_{d}+Q_{d}^{T}B_{d}R_{d}$.

It was then generalized by Benaych-Georges to rectangular matrices ([3]).
However it should be noted that in this case rectangular matrices don’t form
an algebra, therefore the notion of freeness has to be redefined to deal with
delicate undefined products. In this regard, a notion of freeness with
amalgamation is necessary.

######  Theorem 1.17 (free rectangular singular addition from [3]).

Let, for all $n\geq 1$,$M(1,n)$ and $M(2,n)$ be independent bi-orthogonally
invariant $q_{1}(n)\times q_{2}(n)$ random matrices with $q_{1}(n)\geq
q_{2}(n)$, and such that for all i= 1,2, the symmetrization of the singular
law of $M(i,n)$ converges in probability to $\mu_{i}$. Then the symmetrization
of the singular law of $M(1,n)+M(2,n)$ converges in probability to a symmetric
probability measure on the real line, denoted by
$\mu_{1}\boxplus^{\lambda}\mu_{2}$, which depends only on $\mu_{1},\mu_{2}$,
and $\lambda:=\lim_{n\rightarrow\infty}q_{2}(n)/q_{1}(n)$. Notice that
$\lambda\in[0,1]$.

The tool that plays the role of the Cauchy transform in the rectangular
setting is the following quadratic transform.

######  Definition 1.18 (from [3]).

The $\lambda$-rectangular Cauchy transform of a symmetric compact measure
$\mu$ (and $x$ in a nonnegative neighborhood of $0$) is given by

$H^{\lambda}_{\mu}(x)=\lambda\big{[}\mathcal{G}_{\mu}(\frac{1}{\sqrt{x}})\big{]}^{2}+(1-\lambda)\sqrt{x}\mathcal{G}_{\mu}(\frac{1}{\sqrt{x}})=x+\sum_{i=2}^{\infty}h_{i}^{(p)}x^{i}$

######  Lemma 1.19 (from [3]).

As the measure is compact, $H^{\lambda}_{\mu}(0)=0$ and
$\frac{dH^{\lambda}_{\mu}}{dx}(0)=1$ , the rectangular transform is analytic
in a neighborhood of $x=0$, and therefore admits a compositional inverse which
is also analytic in a neighborhood of zero.This inverse will be denoted by
$[{H_{\mu}^{\lambda}}]^{-1}(x)$.

######  Definition 1.20 (following [3]).

For $x$ small enough, let

$U^{\lambda}(x):=\frac{-\lambda-1+\big{[}(\lambda+1)^{2}+4\lambda
x\big{]}^{1/2}}{2\lambda}$

The rectangular $R$-transform is given by

$\mathcal{R}_{\mu}^{\lambda}(x):=U^{\lambda}\Big{(}\frac{x}{[{H_{\mu}^{\lambda}}]^{-1}(x)}-1\Big{)}$

######  Remark 1.21.

$\mathcal{R}_{\mu}^{\lambda}(x)$ is also analytic in a neighborhood of zero by
theorems of composition given that the square root is well-defined and
$\frac{x}{[{H_{\mu}^{\lambda}}]^{-1}(x)}$ can be expanded at zero (because it
doesn’t vanish).

######  Theorem 1.22 (from [3]).

The rectangular $R$-transform linearizes the rectangular additive convolution
for symmetric measures $\mu_{1}$ and $\mu_{2}$:

$\mathcal{R}_{\mu_{1}\boxplus^{\lambda}\mu_{2}}^{\lambda}(x)=\mathcal{R}_{\mu_{1}}^{\lambda}(x)+\mathcal{R}_{\mu_{2}}^{\lambda}(x)$

###  1.2 New results and organization of the paper

####  1.2.1 A new polynomial operation on singular values of rectangular
matrices

We start by introducing in Section 2 a new operation on polynomials with
nonnegative roots, or alternatively, on even realrooted polynomials. We define
the symmetrization operator: $\mathbb{S}p$ which denotes for a polynomial $p$
with all real nonnegative roots the polynomial $p(x^{2})$ (its roots are the
square-roots of $p$ plus the symmetric negative numbers). So being given the
polynomial $p$ with nonnegative roots or the even realrooted polynomial
$\mathbb{S}p$ is the same. In all this paper, we will consider a polynomial
and the uniform measure $\mu_{p}$ over its roots as giving the same
information. If $p:=\prod_{i}(x-\lambda_{i}(p))$, and $\lambda_{i}(p)\geq 0$
for all $i$, we associate to $p$ the measure

$\mu_{\mathbb{S}p}=\frac{1}{2d}\sum_{i=1}^{d}(\delta_{\sqrt{\lambda_{i}(p)}}+\delta_{-\sqrt{\lambda_{i}(p)}}).$

Consider two rectangular matrices $A$ and $B$ of size $m\times d$ where $m\geq
d$, with symmetrized singular distributions $\mu_{A}$ and $\mu_{B}$, and
orthogonal matrices $Q$ in $\mathcal{O}_{m}$ and $R$ in $\mathcal{O}_{d}$.
Free probability (see Theorem 1.17) tells us that the symmetrization of the
singular distribution of $A+Q^{T}BR$ is close to the Benaych-Georges’
rectangular free sum $\mu_{A}\boxplus^{\frac{d}{m}}\mu_{B}$ when $d$, $m$ are
large (for $\lambda=\frac{d}{m}$ fixed)— that is, the roots of
$\chi_{(A+QBR^{T})(A+QBR^{T})^{T}}$ can be predicted with a good accuracy. One
possible way to create a deterministic finite distribution on the model of
free probability that doesn’t depend on the instances $A$ and $B$ but only on
the distributions is to look at all possible characteristic polynomials
$\chi_{(A+QBR^{T})(A+QBR^{T})^{T}}$ and average them uniformly. In the limit,
as it converges to the same distribution for all random matrices, it will also
heuristically converge to this same distribution when we take the expectation.
This led us to define the following univariate convolution in [1]:

######  Definition 1.23 (Rectangular singular free sum).

For $m\times d$ rectangular matrices A and B, define

$\displaystyle\chi_{A^{T}A}\boxplus_{d}^{m-d}\chi_{B^{T}B}$
$\displaystyle:=\mathbb{E}_{R\in\mathcal{O}_{m},Q\in\mathcal{O}_{d}}\left\\{{\chi_{(A+QBR^{T})(A+QBR^{T})^{T}}}\right\\}$
$\displaystyle=\iint_{\mathcal{O}_{m}\times\mathcal{O}_{d}}\det\left[xI-(A+QBR^{T})^{T}(A+QBR^{T})\right]dRdQ$

where the measures are Haar on the respective orthogonal groups.

We derived a binomial formula that enabled us to extend the convolution to
polynomials of degree at most $d$:

######  Theorem 1.24 (From [1]).

Consider two polynomials $p$ and $q$ with only real nonnegative roots (they
can be written as $\chi_{A^{T}A}$ and $\chi_{B^{T}B}$ for some $m\times d$
matrices $A$ and $B$). If we write $p(x)=\sum_{i=0}^{d}(-1)^{i}a_{i}x^{d-i}$
and $q(x)=\sum_{i=0}^{d}(-1)^{i}b_{i}x^{d-i}$ the following holds

$\displaystyle
p\boxplus_{d}^{m-d}q=\sum_{k=0}^{d}x^{d-k}(-1)^{k}\sum_{i+j=k}\frac{(d-i)!(d-j)!}{d!(d-k)!}\frac{(m-i)!(m-j)!}{m!(m-k)!}a_{i}b_{j}$

We also proved that such a convolution is real-rooted with nonnegative roots:

######  Theorem 1.25 (Realrootedness of the convolution, from [1]).

For $p$,$q$ polynomials with nonnegative real roots of degrees at most than
$d$, the polynomial we get through the convolution is real rooted with
nonnegative roots (the operation $\boxplus_{d}^{m-d}$ is stable over
polynomials with nonnegative real roots). Furthermore, the operation
$\boxplus_{d}^{m-d}$ is associative and bilinear.

In this paper, we extend the definition to special bivariate polynomials.

######  Definition 1.26.

Fix $m$ and $d$. If $p$ is a realrooted polynomial with nonnegative real roots
of degree $d$, define the ”rectangular” polynomial extension of order $m-d$ as
$p(x,y):=y^{m-d}p(xy)$.

######  Remark 1.27.

If for a $m\times d$ matrix $A$, $p(x)=\chi_{A^{T}A}$ then

$p(x,y)=\det\begin{bmatrix}y\mathbb{I}_{m}&{A}\\\
{A}^{T}&x\mathbb{I}_{d}\end{bmatrix}$

Now given that if $p$ has only nonnegative roots, such a matrix $A$ can always
be exhibited whatever the value of $m\geq d$, we are led to the following
generalization:

######  Definition 2.1 (Rectangular bivariate convolution).

If

$p(x,y)=\det\begin{bmatrix}y\mathbb{I}_{m}&{A}\\\
{A}^{T}&x\mathbb{I}_{d}\end{bmatrix}\quad\text{and}\quad
q(x,y)=\det\begin{bmatrix}y\mathbb{I}_{m}&{B}\\\
{B}^{T}&x\mathbb{I}_{d}\end{bmatrix}$

Then for $\lambda:=\frac{d}{m}$ (for a given $d$, $m$ and $\lambda$ are in
bijection), define the $\lambda$ -rectangular additive convolution of $p$ and
$q$ as

$\displaystyle[p\boxplus_{d,\lambda}q](x,y)$
$\displaystyle:=\iint\det\left[\begin{bmatrix}y\mathbb{I}_{m}&{A}\\\
{A}^{T}&x\mathbb{I}_{d}\end{bmatrix}+\begin{bmatrix}\hat{Q}&0\\\
0&\hat{R}\end{bmatrix}\begin{bmatrix}y\mathbb{I}_{m}&{B}\\\
{B}^{T}&x\mathbb{I}_{d}\end{bmatrix}\begin{bmatrix}\hat{Q}&0\\\
0&\hat{R}\end{bmatrix}^{T}\right]d\hat{Q}d\hat{R}$
$\displaystyle=\iint\det\begin{bmatrix}y\mathbb{I}_{m}&{(A+\hat{Q}B\hat{R})}\\\
{(A+\hat{Q}B\hat{R})}^{T}&x\mathbb{I}_{d}\end{bmatrix}d\hat{Q}d\hat{R}$

where $d\hat{Q},d\hat{R}$ can be taken to be the Haar measure over the
orthogonal group (it could be made more general but we won’t extend it for the
sake of simplicity).

######  Remark 2.2.

We get the univariate rectangular convolution by plugging $y=1$:

$[p\boxplus_{d,\lambda}q](x,1)=p\boxplus_{d}^{m-d}q(x)=\iint\det\left[x\mathbb{I}-(A+\hat{Q}B\hat{R})^{T}(A+\hat{Q}B\hat{R}))\right]d\hat{Q}d\hat{R}$

In this paper, we will denote by $\boxplus_{d,\lambda}$ the convolution
$\boxplus_{d}^{m-d}$ for $\lambda=\frac{d}{m}$ even when restricting to the
univariate case. This will be convenient as the parameter that is constant
will be $\lambda$ and not $m-d$.

####  1.2.2 Defining a transform fit for this polynomial convolution

In Section 3, we associate bijectively to every polynomial of degree $d$ with
nonnegative real roots $p$ its finite $R$-transform
$\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)$ as another polynomial of degree
$d$. We prove the following crucial linearization emulating property:

######  Theorem 3.9.

For two polynomials $p$ and $q$ with nonnegative real roots of degree $d$, we
have

$\mathcal{R}^{d,\lambda}_{\mathbb{S}[p\boxplus_{d,\lambda}q]}(s)=\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)+\mathcal{R}^{d,\lambda}_{\mathbb{S}q}(s)$

It is the direct analogue of the free probability additivity property that
defines the free $R$-rectangular transform:

$\displaystyle\mathcal{R}_{\mu_{\mathbb{S}p}\boxplus_{\lambda}\mu_{\mathbb{S}q}}^{\lambda}(s)$
$\displaystyle=\mathcal{R}_{\mu_{\mathbb{S}p}}^{\lambda}(s)+\mathcal{R}_{\mu_{\mathbb{S}q}}^{\lambda}(s)$

####  1.2.3 Adapting free probability notions to connect them to polynomials

We derive in Section 4 more simple formulas for discrete measures. In the
following definition we define the $\lambda$-rectangular Cauchy transform of a
polynomial $p$ that is not exactly the $\lambda$-rectangular Cauchy transform
of a measure $\mu$ from the Definition 1.18.

######  Definition 4.1.

For a polynomial $p$, the $\lambda$-rectangular Cauchy transform of its
symmetrized polynomial $\mathbb{S}p$ with all roots nonnegative is given by

$\mathcal{H}^{\lambda}_{\mathbb{S}p}(x):=\mathcal{G}_{\mathbb{S}p}(x)\Big{(}\lambda\mathcal{G}_{\mathbb{S}p}(x)+(1-\lambda){\mathcal{G}}_{0}(x)\Big{)}\textit{\
\ \ for \ }x>\sqrt{\mathrm{maxroot}\left\\{p\right\\}}$

Recall that ${\mathcal{G}}_{0}(x)=\frac{1}{x}$. It is easy to check that

$\mathcal{H}^{\lambda}_{\mathbb{S}p}(x)=H^{\lambda}_{\mu_{\mathbb{S}p}}(\frac{1}{x^{2}})$

This definition is a slightly modified version of the Benaych Georges’
transform fit for polynomials. Our definition also incorporates the
symmetrization.

######  Lemma 4.3.

$\mathcal{H}^{\lambda}_{\mathbb{S}p}$ is a bijection from
$[\sqrt{\mathrm{maxroot}\left\\{p\right\\}},+\infty]$ to $[0,+\infty]$. We
denote the inverse by $\mathcal{J}^{\lambda}_{\mathbb{S}p}$. In particular,
for $x>0$

$\mathcal{H}^{\lambda}_{\mathbb{S}p}\circ\mathcal{J}^{\lambda}_{\mathbb{S}p}[x]=x$

and for $x>\sqrt{\mathrm{maxroot}\left\\{p\right\\}}$

$\mathcal{J}^{\lambda}_{\mathbb{S}p}\circ\mathcal{H}^{\lambda}_{\mathbb{S}p}[x]=x$

We can rewrite the $R$-transform for polynomials:

######  Claim 4.5.

$\mathcal{R}_{\mathbb{S}p}^{\lambda}(s^{2}):=\mathcal{R}_{\mu_{\mathbb{S}p}}^{\lambda}(s^{2})=\frac{-\lambda-1}{2\lambda}+\sqrt{\frac{(\lambda-1)^{2}}{4\lambda^{2}}+\frac{s^{2}[\mathcal{J}^{\lambda}_{\mathbb{S}p}(s^{2})]^{2}}{\lambda}}$

####  1.2.4 Defining a modified finite $R$-transform

We define in Section 5 an alternative power series
$\widetilde{\mathcal{R}}^{d,\lambda}_{\mathbb{S}p}(s)$ that converges by
design to $\mathcal{R}_{\mathbb{S}p}^{\lambda}(s^{2})$ using approximations of
$\mathcal{J}^{\lambda}_{\mathbb{S}p}(s^{2})$.

######  Theorem 5.3 (Convergence of the modified finite $R$-transform to the
free $R$-transform).

For $s$ small enough,

$\widetilde{\mathcal{R}}^{d,\lambda}_{\mathbb{S}p}(s)\xrightarrow{d\to\infty}\mathcal{R}_{\mathbb{S}p}^{\lambda}(s^{2})$

Secondly, we show it that has the same first $d/2$ coefficients as
$\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)$.

######  Theorem 5.12.

$\widetilde{\mathcal{R}}^{d,\lambda}_{\mathbb{S}p}(s)\equiv\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s^{2})\mod[s^{d+1}]$

####  1.2.5 Convergence: from finite free probability to free probability

In Section 6, we gather all the pieces to prove the main theorem of this
paper.

######  Theorem 6.1.

Coefficientwise or for $s$ small enough, the following power series
convergence holds:

$\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)\xrightarrow{d\to\infty}\mathcal{R}_{\mathbb{S}p}^{\lambda}(s)$

####  1.2.6 Polynomials as random variables: limit theorems

Finally, in Section 7, we show that the connexion between probability and
polynomials goes actually further. We consider polynomials of the form
$\mathbb{S}p$ as random variables in our rectangular polynomial framework.

######  Definition 7.1.

For a realrooted polynomial $r$ with roots $r_{i}$, define its expectation and
variance as

$\displaystyle\mathbb{E}(r)$ $\displaystyle:=\frac{1}{d}\sum_{i}r_{i}$
$\displaystyle Var(r)$
$\displaystyle:=\frac{1}{d}\sum_{i}(r_{i}-\mathbb{E}(r))^{2}$

######  Lemma 7.2.

Exactly like for a sum of independent random variables, we get

  * •

$\mathbb{E}\big{[}\mathbb{S}[p\boxplus_{d,\lambda}q]\big{]}=\mathbb{E}[\mathbb{S}p]+\mathbb{E}[\mathbb{S}q]=0$
,

  * •

$Var\big{[}\mathbb{S}[p\boxplus_{d,\lambda}q]\big{]}=Var[\mathbb{S}p]+Var[\mathbb{S}q]$.

######  Definition 7.3 (Normalizing operator).

For $p$ of degree $d$, and $\alpha>0$, define
$\mathit{R_{\alpha}}(p):=\alpha^{-d}p(\alpha x)$.

$\mathbb{S}q=x^{2d}$ is the zero polynomial in terms of rectangular addition
in the sense that $\mathbb{S}[p\boxplus_{d,\lambda}q]=\mathbb{S}p$. For two
polynomials $p$ and $q$, we will use $p\approx q$ to express that they have
the same roots but not the same leading coefficient.

######  Proposition 7.7 (Law of large numbers).

Let $p_{1},p_{2}$,… be a sequence of degree $d$ polynomials with real
nonnegative roots and means uniformly bounded by $\sigma^{2}$,

$\displaystyle p_{i}:=\prod_{j}(x-r_{i,j}^{2})$
$\displaystyle\frac{1}{d}\sum_{j}r_{i,j}^{2}\leq\sigma^{2}$

then

$\lim_{N\to\infty}\mathit{R_{1/{N}}}(\mathbb{S}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}])(x)\approx
x^{2d}$

######  Lemma 7.8 (Laguerre polynomials).

For p of degree $d$ with nonnegative real roots, and $\sigma^{2}>0$, the
following are equivalent:

  1. 1.

$\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)=m\sigma^{2}s$, so only the first
nontrivial cumulant is nonzero.

  2. 2.

$p\approx L_{d}^{(m-d)}(\frac{x}{\sigma^{2}})$, that is $p$ is up to scaling a
generalized Laguerre polynomial of parameter $m-d$.

Therefore Laguerre polynomials play the role of the Gaussians in our
framework, which is made clearer by the following.

######  Proposition 7.10 (Central limit theorem).

Let $p_{1},p_{2}$,… be a sequence of degree $d$ with real nonnegative roots
and same mean $\sigma^{2}$, with

$\displaystyle p_{i}=\prod_{j}(x-r_{i,j}^{2})$
$\displaystyle\frac{1}{d}\sum_{j}r_{i,j}^{2}=\sigma^{2}$

then

$\lim_{N\to\infty}\mathit{R_{1/\sqrt{N}}}(\mathbb{S}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}])(x)\approx
L_{d}^{(m-d)}(\frac{x^{2}m}{\sigma^{2}})$

##  2 Bivariate definition of the convolution

We first need to generalize the usual characteristic polynomial in order to
store information not only about the singular values but also about the
dimensions. The main issue is that there really are two different
characteristic polynomials associated with a given set of singular values,
that is, if $A$ is an $m\times d$ matrix with $m\geq d$,

$\psi_{A}(x,1):=\det[x\mathbb{I}_{d}-A^{T}A]$

and

$\psi_{A}(1,x):=\det[x\mathbb{I}_{m}-AA^{T}]$

The first one contains all the singular values but fails to give information
about the additional rectangular dimension, $m$. The second gives $m$ but not
$d$, and adds a bunch of zeros to the singular values. We need to keep track
of both dimensions and in this respect, considering a bivariate polynomial
works well. We consider

$\displaystyle\psi_{A}(x,y)$
$\displaystyle=\det\begin{bmatrix}y\mathbb{I}_{m}&{A}\\\
{A}^{T}&x\mathbb{I}_{d}\end{bmatrix}$
$\displaystyle=y^{m-d}\det\left[xy\mathbb{I}_{d}-A^{T}A\right]$
$\displaystyle=\left(\frac{1}{x}\right)^{m-d}\det\left[xy\mathbb{I}_{m}-AA^{T}\right]$

Note that although this is “technically” a bivariate polynomial, it is in some
sense a univariate polynomial with an extra dimension term added on. Now, let
$A,B$ be $m\times d$ matrices and let their generalized (singular)
characteristic polynomials be:

$p(x,y)=\det\begin{bmatrix}y\mathbb{I}_{m}&{A}\\\
{A}^{T}&x\mathbb{I}_{d}\end{bmatrix}\quad\text{and}\quad
q(x,y)=\det\begin{bmatrix}y\mathbb{I}_{m}&{B}\\\
{B}^{T}&x\mathbb{I}_{d}\end{bmatrix}.$

######  Definition 2.1.

For $\lambda:=\frac{d}{m}$ (for a given $d$, $m$ and $\lambda$ are in
bijection), define the $\lambda$ -rectangular additive convolution of $p$ and
$q$ as

$\displaystyle[p\boxplus_{d,\lambda}q](x,y)$
$\displaystyle:=\iint\det\left[\begin{bmatrix}y\mathbb{I}_{m}&{A}\\\
{A}^{T}&x\mathbb{I}_{d}\end{bmatrix}+\begin{bmatrix}\hat{Q}&0\\\
0&\hat{R}\end{bmatrix}\begin{bmatrix}y\mathbb{I}_{m}&{B}\\\
{B}^{T}&x\mathbb{I}_{d}\end{bmatrix}\begin{bmatrix}\hat{Q}&0\\\
0&\hat{R}\end{bmatrix}^{T}\right]d\hat{Q}d\hat{R}$
$\displaystyle=\iint\det\begin{bmatrix}y\mathbb{I}_{m}&{(A+\hat{Q}B\hat{R})}\\\
{(A+\hat{Q}B\hat{R})}^{T}&x\mathbb{I}_{d}\end{bmatrix}d\hat{Q}d\hat{R}$

where $d\hat{Q},d\hat{R}$ can be taken to be the Haar measure over the
orthogonal group (it could be made more general but we won’t extend it for the
sake of simplicity).

######  Remark 2.2.

We get the univariate rectangular convolution by plugging $y=1$:

$[p\boxplus_{d,\lambda}q](x,1)=p\boxplus_{d}^{m-d}q=\iint\det\left[x\mathbb{I}-(A+\hat{Q}B\hat{R})^{T}(A+\hat{Q}B\hat{R}))\right]d\hat{Q}d\hat{R}$

######  Proposition 2.3.

Let $A,B$ be $m\times d$ matrices and let

$p(x,y)=\det\begin{bmatrix}y\mathbb{I}_{m}&{A}\\\
{A}^{T}&x\mathbb{I}_{d}\end{bmatrix}\quad\text{and}\quad
q(x,y)=\det\begin{bmatrix}y\mathbb{I}_{m}&{B}\\\
{B}^{T}&x\mathbb{I}_{d}\end{bmatrix}.$

There exists an $m\times d$ matrix $C$ such that

$[p\boxplus_{d,\lambda}q](x,y)=\det\begin{bmatrix}y\mathbb{I}_{m}&{C}\\\
{C}^{T}&x\mathbb{I}_{d}\end{bmatrix}.$

###### Proof.

Using theorem 1.25, we get that $[p\boxplus_{d,\lambda}q](x,1)$ has
nonnegative roots. Therefore we can take square-roots of such values, and we
choose $C$ to be a matrix such that it’s singular values are these square
roots. We get

$[p\boxplus_{d,\lambda}q](x,1)=\det(x\mathbb{I}_{d}-C^{T}C)$

It follows that

$[p\boxplus_{d,\lambda}q](x,y)=y^{m-d}\det(xy\mathbb{I}_{d}-C^{T}C)$

∎

It defines a “rotation invariant” operation on the algebra of singular values.
To push the analogy with the hermitian case (see [6]), we can also exhibit a
derivative expression.

######  Lemma 2.4.

Consider polynomials $p$ and $q$ with nonnegative real roots and
$p(x,y)=y^{m-d}p(xy),\\\ q(x,y)=y^{m-d}q(xy)$ their bivariate extensions, then

$[p\boxplus_{d,\lambda}q](x,y)=\frac{(m-d)!}{d!m!}\sum_{k=0}^{d}(\partial_{x}\partial_{y})^{d-k}p(x,y)(\partial_{x}\partial_{y})^{k}q(0,1)$
$[p\boxplus_{d}^{m-d}q](x)=\frac{(m-d)!}{d!m!}\sum_{k=0}^{d}(\partial_{x}\partial_{y})^{d-k}p(x,1)(\partial_{x}\partial_{y})^{k}q(0,1)$

###### Proof.

$[p\boxplus_{d,\lambda}q](x,y)=y^{m-d}[p\boxplus_{d,\lambda}q](xy,1)=y^{m-d}[p\boxplus_{d}^{m-d}q](xy)$

So it is sufficient to check the equality for $p(x,y)=y^{m-d}(xy)^{d-i}$ and
$q(x,y)=y^{m-d}(xy)^{d-j}$ and extend it by bilinearity.

$y^{m-d}(xy)^{d-i}\boxplus_{d,\lambda}y^{m-d}(xy)^{d-j}=y^{m-d}(xy)^{d-i-j}\frac{(d-i)!(d-j)!}{d!(d-i-j)!}\frac{(m-i)!(m-j)!}{m!(m-i-j)!}$

On the other hand, $(\partial_{x}\partial_{y})^{k}q(0,1)\neq 0\iff k=d-j$ and
therefore

$\frac{(m-d)!}{d!m!}\sum_{k=0}^{d}(\partial_{x}\partial_{y})^{d-k}p(x,y)(\partial_{x}\partial_{y})^{k}q(0,1)=\frac{(m-d)!}{d!m!}(\partial_{x}\partial_{y})^{j}p(x,y)(\partial_{x}\partial_{y})^{d-j}q(0,1)$
$\displaystyle(\partial_{x}\partial_{y})^{j}p(x,y)$
$\displaystyle=\frac{(m-i)!}{(m-i-j)!}y^{m-i-j}\frac{(d-i)!}{(d-i-j)!}x^{d-i-j}$
$\displaystyle(\partial_{x}\partial_{y})^{d-j}q(0,1)$
$\displaystyle=(d-j)!\frac{(m-j)!}{(m-d)!}$

We conclude by gathering terms. ∎

######  Lemma 2.5.

Let $\mathcal{L}=\sum_{i}c_{i}(\partial_{x}\partial_{y})^{i}$ be a linear
differential operator. Then we can pull out the differential operator from the
convolution operation:

$\mathcal{L}\left\\{p\boxplus_{d,\lambda}q\right\\}=\mathcal{L}\\{p\\}\boxplus_{d,\lambda}q=p\boxplus_{d,\lambda}\mathcal{L}\\{q\\}.$

So if $\mathcal{P}$ and $\mathcal{Q}$ are linear differential operators such
that

$p(x,y)=\mathcal{P}\\{y^{m}x^{d}\\}\quad\text{and}\quad
q(x,y)=\mathcal{Q}\\{y^{m}x^{d}\\}$

then

$\displaystyle[p\boxplus_{d,\lambda}q]$
$\displaystyle=[\mathcal{P}\\{y^{m}x^{d}\\}\boxplus_{d,\lambda}\mathcal{Q}\\{y^{m}x^{d}\\}]$
$\displaystyle=\mathcal{P}\\{\mathcal{Q}[y^{m}x^{d}]\boxplus_{d,\lambda}y^{m}x^{d}\\}$
$\displaystyle=\mathcal{P}\mathcal{Q}\\{y^{m}x^{d}\\}.$

###### Proof.

Using Lemma 2.4, we see that the operation $\boxplus_{d,\lambda}$ commutes
with $\partial_{x}\partial_{y}$ and the result follows. ∎

##  3 Definition of the rectangular finite $R$-transform for bivariate
polynomials

###  3.1 From polynomials to random variables: the rectangular $T$-Transform

Consider a symmetric discrete finite probability measure $\mu_{A}$ such that
$\mu_{A}=\mu_{\mathbb{S}p_{A}}$, for a polynomial $p_{A}$ with nonnegative
roots. If such a polynomial $p_{A}$ of degree $d$ exists, then it is the
unique polynomial of degree $d$ such that this holds. Now fix
$\lambda\in]0,1]$. For each $\lambda$, and each $d$ such that $d=m\lambda$
(for an integer $m\geq d$), we can uniquely associate a random variable to the
measure $\mu_{A}$. We call it the $T^{(m,d)}$-transform or also the
$T^{(d,\lambda)}$-transform. Alternatively, for a given polynomial of degree
$d$ with nonnegative real roots $p_{A}$, we associate uniquely a
$T^{(m,d)}$-transform. In the following, the expectation symbol will stand for
a uniform mean over all the values taken by the random variable
($T$-transform).

######  Definition 3.1.

For $d$, $m$, $\lambda$, $\mu_{A}$ and $p_{A}$ as above, we define the random
variable $T^{(m,d)}_{\mathbb{S}p_{A}}$ or $T^{(m,d)}_{\mu_{A}}$ , associated
to the measure $\mu_{A}$ and vice versa by

$\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mu_{A}}\partial_{x}\partial_{y}}\big{]}\\{y^{m}x^{d}\\}=\frac{1}{d}\sum_{i=1}^{d}e^{-t_{i}^{(m,d)}(A)[\partial_{x}\partial_{y}]}\\{y^{m}x^{d}\\}:=y^{m-d}p_{A}(xy)$

where $T^{(m,d)}_{\mathbb{S}p_{A}}$, the rectangular $T$-transform associated
to $p_{A}$, is a random variable taking with same probability any value from
the multiset $({t_{1}^{(m,d)}(A),....,t_{d}^{(m,d)}(A))}$.

######  Claim 3.2.

$T^{(m,d)}_{\mathbb{S}\mu_{A}}$ is well-defined and uniquely determines
$\mu_{A}$ (or $p_{A}$).

###### Proof.

Write:

$y^{m-d}p_{A}(xy)=y^{m-d}\sum_{i=1}^{d}(xy)^{d-i}(-1)^{i}p_{i}.$

$\displaystyle\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mu_{A}}\partial_{x}\partial_{y}}\big{]}\\{y^{m}x^{d}\\}$
$\displaystyle=\sum_{i=0}^{d}\frac{\mathbb{E}\big{[}(-T^{(m,d)}_{\mu_{A}})^{i}\big{]}}{i!}(\partial_{x}\partial_{y})^{i}\\{y^{m}x^{d}\\}$
(1)
$\displaystyle=\sum_{i=0}^{d}\frac{\mathbb{E}\big{[}(-T^{(m,d)}_{\mu_{A}})^{i}\big{]}}{i!}\frac{m!}{(m-i)!}\frac{d!}{(d-i)!}y^{m-i}x^{d-i}$
(2)
$\displaystyle=y^{m-d}\sum_{i=0}^{d}(xy)^{d-i}(-1)^{i}\frac{\mathbb{E}\big{[}(T^{(m,d)}_{\mu_{A}})^{i}\big{]}}{i!}\frac{m!}{(m-i)!}\frac{d!}{(d-i)!}$
(3)

So we have to prove the existence and the unicity of $T^{(m,d)}_{\mu_{A}}$
such that for all $i\in[|1,d|]$:

$\mathbb{E}\big{[}(T^{(m,d)}_{\mu_{A}})^{i}\big{]}=\frac{i!(m-i)!}{m!}\frac{(d-i)!}{d!}p_{i}$
$\sum_{j=1}^{d}[t_{j}^{(m,d)}(A)]^{i}=d\frac{i!(m-i)!}{m!}\frac{(d-i)!}{d!}p_{i}$

It amounts to finding $d$ complex numbers for which the $d$ first power sums
are given. The existence and the unicity follows by using Newton’s identities.
∎

######  Remark 3.3.

Notice that the numbers $t_{j}^{(m,d)}(A)$ are a priori complex. But the
expectations $\mathbb{E}\big{[}(-T^{(m,d)}_{\mu_{A}})^{i}]$ are real.

Now, let’s see why this representation turns out to be useful in the framework
of finite free probability.

######  Proposition 3.4.

For any multisets $T^{(m,d)}_{\mathbb{S}p}$ and
$T^{(m,d)}_{\mathbb{S}q}$associated to polynomials $p$ and $q$ with
nonnegative roots and degree $d$, considered by assumption independent (we
choose the joint distribution to be the tensor product of distributions) ,

$T^{(m,d)}_{\mathbb{S}[p\boxplus_{d,\lambda}q]}=_{law}T^{(m,d)}_{\mathbb{S}p}+T^{(m,d)}_{\mathbb{S}q}$

that is, the random variables have the same distribution.

###### Proof.

$\displaystyle\big{[}\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}[p\boxplus_{d,\lambda}q]}\partial_{x}\partial_{y}}\big{]}\\{y^{m}x^{d}\\}=y^{m-d}[p\boxplus_{d,\lambda}q](xy)$
(4)
$\displaystyle=y^{m-d}\sum_{k=0}^{d}[xy]^{d-k}(-1)^{k}\sum_{i+j=k}p_{i}q_{j}\frac{(d-i)!(d-j)!}{d!(d-i-j)!}\frac{(m-i)!(m-j)!}{m!(m-i-j)!}$
(5)
$\displaystyle=y^{m-d}\sum_{k=0}^{d}[xy]^{d-k}(-1)^{k}\sum_{i+j=k}\frac{\mathbb{E}\big{[}{T^{(m,d)}_{\mathbb{S}p}}^{i}\big{]}m!d!}{i!(m-i)!(d-i)!}\frac{\mathbb{E}\big{[}{T^{(m,d)}_{\mathbb{S}q}}^{j}\big{]}m!d!}{j!(m-j)!(d-j)!}\frac{(d-i)!(d-j)!}{d!(d-i-j)!}\frac{(m-i)!(m-j)!}{m!(m-i-j)!}$
(6)
$\displaystyle=y^{m-d}\sum_{k=0}^{d}[xy]^{d-k}(-1)^{k}\sum_{i+j=k}\frac{\mathbb{E}\big{[}{T^{(m,d)}_{\mathbb{S}p}}^{i}\big{]}\mathbb{E}\big{[}{T^{(m,d)}_{\mathbb{S}q}}\big{]}^{j}}{i!j!}\frac{m!d!}{(d-i-j)!(m-i-j)!}$
(7)
$\displaystyle=y^{m-d}\sum_{k=0}^{d}[(xy]^{d-k}(-1)^{k}\sum_{i+j=k}\frac{\mathbb{E}\big{[}{(T^{(m,d)}_{\mathbb{S}p}})^{i}({T^{(m,d)}_{\mathbb{S}q}})^{j}\big{]}\binom{k}{i}}{k!}\frac{m!d!}{(m-k)!(d-k)!}~{}(\textit{by
independence})$ (8)
$\displaystyle=\mathbb{S}\big{[}\sum_{k=0}^{d}\sum_{i+j=k}\frac{\mathbb{E}\big{[}(-T^{(m,d)}_{\mathbb{S}p})^{i}(-T_{\mathbb{S}q})^{j}\big{]}\binom{k}{i}}{k!}(\partial_{x}\partial_{y})^{k}\\{y^{m}x^{d}\\}\big{]}$
(9)
$\displaystyle=\mathbb{S}\big{[}\sum_{k=0}^{d}\frac{\mathbb{E}\big{[}(-1)^{k}(T^{(m,d)}_{\mathbb{S}p}+T^{(m,d)}_{\mathbb{S}q})^{k}\big{]}(\partial_{x}\partial_{y})^{k}}{k!}\\{y^{m}x^{d}\\}\big{]}$
(10)
$\displaystyle=\mathbb{E}\big{[}e^{-(T^{(m,d)}_{\mathbb{S}p}+T^{(m,d)}_{\mathbb{S}q})\partial_{x}\partial_{y}}\big{]}\\{y^{m}x^{d}\\}$
(11)

∎

The $T$-transform therefore linearizes the rectangular convolution. We will
use this property to define a polynomial of degree $d$ that also shares this
linearization property (the rectangular finite $R$-transform.

###  3.2 The $R$-transform as a polynomial in partial derivatives

We start by the definition of the ($m$,$d$)-rectangular finite $R$ transform
of a measure $\mu_{\mathbb{S}p_{A}}$. First let’s introduce a notation.

######  Definition 3.5 (Modulo truncation).

For two formal power series $f(s)=\sum_{i}f_{i}s^{i}$ and
$g(s)=\sum_{i}g_{i}s^{i}$ and an integer $d$ we write

$f(s)\equiv g(s)\mod[s^{d}]$

if

$\sum_{i=0}^{d-1}f_{i}s^{i}=\sum_{i=0}^{d-1}g_{i}s^{i}$

We will also denote by $f(s)\mod[s^{d}]$ the polynomial
$\sum_{i=0}^{d-1}f_{i}s^{i}$.

######  Remark 3.6.

It was proven in [8] that for any power series $h$, if $f(s)=g(s)\mod[s^{d}]$
then

$h[f(s)]\equiv h[g(s)]\mod[s^{d}]$

and we also have

$\displaystyle\frac{df}{ds}(s)$
$\displaystyle\equiv\frac{dg}{ds}(s)\mod[s^{d}]$ $\displaystyle sf(s)$
$\displaystyle\equiv sg(s)\mod[s^{d+1}]$ (12)

######  Definition 3.7 (Rectangular finite $R$-transform).

We define $\mathcal{R}^{d,\lambda}_{\mathbb{S}p_{A}}(s)$ as the unique
polynomial of degree $d$ verifying

$\mathcal{R}^{d,\lambda}_{\mathbb{S}p_{A}}(s)\equiv\frac{-1}{d}s\frac{d}{ds}\log\big{(}\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}p_{A}}smd}\big{]}\big{)}\mod[s^{d+1}]$

######  Remark 3.8.

It will become clear later on why we choose such constants in the definition
(it is related to convergence). Notice that
$\mathcal{R}^{d,\lambda}_{\mathbb{S}p_{A}}(s)$ has zero constant term so that
$\frac{\mathcal{R}^{d,\lambda}_{\mathbb{S}p_{A}}(s)}{s}$ is a polynomial of
degree $d-1$.

We can deduce the fundamental additivity property for the rectangular finite
$R$-transform:

######  Theorem 3.9 (Finite additivity).

For two polynomials $p$ and $q$ of degree $d$ with nonnegative real roots, we
have

$\mathcal{R}^{d,\lambda}_{\mathbb{S}[p\boxplus_{d,\lambda}q]}(s)=\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)+\mathcal{R}^{d,\lambda}_{\mathbb{S}q}(s)$

It is the direct analogue of the free probability additivity property that
defines the free $R$-rectangular transform:

$\displaystyle\mathcal{R}_{\mu_{\mathbb{S}p}\boxplus_{\lambda}\mu_{\mathbb{S}q}}^{\lambda}(s)$
$\displaystyle=\mathcal{R}_{\mu_{\mathbb{S}p}}^{\lambda}(s)+\mathcal{R}_{\mu_{\mathbb{S}q}}^{\lambda}(s)$
$\displaystyle=\mathcal{R}_{\mathbb{S}p}^{\lambda}(s)+\mathcal{R}_{\mathbb{S}q}^{\lambda}(s)$

###### Proof.

$\displaystyle\mathcal{R}^{d,\lambda}_{\mathbb{S}[p\boxplus_{d,\lambda}q]}(s)$
$\displaystyle\equiv\frac{-s}{d}\frac{d}{ds}\Big{(}\log\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}[p\boxplus_{d,\lambda}q]}smd}\big{]}\Big{)}\mod[s^{d+1}]$
$\displaystyle\equiv\frac{-s}{d}\frac{d}{ds}\Big{(}\log\mathbb{E}\big{[}e^{(-T^{(m,d)}_{\mathbb{S}p}-T^{(m,d)}_{\mathbb{S}q})smd}\big{]}\Big{)}\mod[s^{d+1}]\text{
using Corollary \ref{corT}}$
$\displaystyle\equiv\frac{-s}{d}\frac{d}{ds}\Big{(}\log\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}p}smd}\big{]}+\log\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}q}smd}\big{]}\Big{)}\mod[s^{d+1}]$
$\displaystyle=\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)+\mathcal{R}^{d,\lambda}_{\mathbb{S}q}(s)$

∎

We can also deduce the following meaningful representation:

######  Proposition 3.10 (Inversion formula).

For any monic polynomial $p$ with nonnegative roots of degree $d$,

$y^{m-d}p(xy)=e^{-[\widehat{\mathcal{R}}^{d,\lambda}_{\mathbb{S}p}(\partial_{x}\partial_{y})]}\\{y^{m}x^{d}\\}$

where
$\widehat{\mathcal{R}}^{d,\lambda}_{\mathbb{S}p}(s):=d\int_{s}\frac{\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(\frac{s}{md})}{(\frac{s}{md})}$
(the primitive which is zero at zero). It follows that
$\mathcal{R}^{d,\lambda}_{\mathbb{S}p}$ characterizes uniquely $p$ in the
sense that we can recover $p$ if we know
$\mathcal{R}^{d,\lambda}_{\mathbb{S}p}$. It should be noted that as the
transform is a function of the roots only, we need to require $p$ to be monic
to be able to recover the polynomial.

###### Proof.

Consider the polynomial $Q_{p}$ of degree $d$ such that

$Q_{p}[\frac{{\partial_{x}\partial_{y}}}{md}]\\{y^{m}x^{d}\\}=y^{m-d}p(xy)=\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mu_{\mathbb{S}p}}\partial_{x}\partial_{y}}\big{]}\\{y^{m}x^{d}\\}$
(13)

Its coefficients can be found explicitly as the reverse of the coefficients of
$p$ renormalized. Furthermore it is clear that such a polynomial is defined
uniquely. In particular, we have that $Q_{p}$ is the following truncated
polynomial:

$Q_{p}[\frac{s}{md}]\equiv\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mu_{\mathbb{S}p}}s}\big{]}\mod[s^{d+1}]\\\
$ (14)

Indeed, if a polynomial $Q$ verifies 14 (there is only one such possible
polynomial of degree $d$), then it verifies 13 by plugging
$s=\partial_{x}\partial_{y}$ into 14. And we have necessarily $Q=Q_{p}$. We
then get using 3.6 and the fact that $\log(1+s)$ expands as a power series:

$\displaystyle-\log[Q_{p}(s)]$
$\displaystyle\equiv-\log\big{[}\mathbb{E}\big{(}e^{-T^{(m,d)}_{\mu_{\mathbb{S}p}}smd}\big{)}\big{]}\mod[s^{d+1}]$
$\displaystyle-s\frac{d\log[Q_{p}(s)]}{ds}$
$\displaystyle\equiv-s\frac{d}{ds}\log\big{[}\mathbb{E}\big{(}e^{-T^{(m,d)}_{\mu_{\mathbb{S}p}}smd}\big{)}\big{]}\mod[s^{d+1}]$
$\displaystyle-s\frac{d\log[Q_{p}(s)]}{ds}$ $\displaystyle\equiv
d\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)\mod[s^{d+1}]$
$\displaystyle\log[Q_{p}(s)]$
$\displaystyle\equiv-d\int_{s}\frac{\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)}{s}\mod[s^{d+1}]$
$\displaystyle Q_{p}(s)$ $\displaystyle\equiv
e^{-d\int_{s}\frac{\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)}{s}}\mod[s^{d+1}]$
$\displaystyle Q_{p}(\frac{s}{md})$ $\displaystyle\equiv
e^{-d\int_{s}\frac{\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(\frac{s}{md})}{\frac{s}{md}}}\mod[s^{d+1}]$

We plug in $s=\partial_{x}\partial_{y}$ so that

$\displaystyle
y^{m-d}p(xy)=Q_{p}(\frac{\partial_{x}\partial_{y}}{md})\\{y^{m}x^{d}\\}$
$\displaystyle=\Big{(}e^{-\widehat{\mathcal{R}}^{d,\lambda}_{\mathbb{S}p}(\partial_{x}\partial_{y})}\mod[(\partial_{x}\partial_{y})^{d+1}]\Big{)}\\{y^{m}x^{d}\\}$
$\displaystyle=e^{-\widehat{\mathcal{R}}^{d,\lambda}_{\mathbb{S}p}(\partial_{x}\partial_{y})}\\{y^{m}x^{d}\\}$

as
$h(\partial_{x}\partial_{y})\\{y^{m}x^{d}\\}=\big{[}h(\partial_{x}\partial_{y})\mod[(\partial_{x}\partial_{y})^{d+1}]\big{]}\\{y^{m}x^{d}\\}$,
because $(\partial_{x}\partial_{y})^{k}\\{y^{m}x^{d}\\}=0$ for $k\geq d+1$. ∎

##  4 Adapting free probability notions to polynomials

###  4.1 Redefining transforms for compact discrete symmetric measures

######  Definition 4.1.

The $\lambda$-rectangular Cauchy transform of a polynomial $p$ with all roots
nonnegative or equivalently of the symmetric polynomial $\mathbb{S}p$ is given
by

$\mathcal{H}^{\lambda}_{\mathbb{S}p}(x):=\mathcal{G}_{\mathbb{S}p}(x)\Big{(}\lambda\mathcal{G}_{\mathbb{S}p}(x)+(1-\lambda){\mathcal{G}}_{0}(x)\Big{)}\textit{\
\ \ for \ }x>\sqrt{\mathrm{maxroot}\left\\{p\right\\}}$

Recall that ${\mathcal{G}}_{0}(x)=\frac{1}{x}$. It is easy to check that

$\mathcal{H}^{\lambda}_{\mathbb{S}p}(x)=H^{\lambda}_{\mu_{\mathbb{S}p}}(\frac{1}{x})$

This definition is a slightly modified version of the Benaych Georges’
transform fit for polynomials. Our definition also incorporates the
symmetrization. We can rewrite it more explicitly if $p$ is of degree $d$ as:

$\mathcal{H}^{\lambda}_{\mathbb{S}p}(x)=\frac{x}{d}\frac{p^{\prime}(x^{2})}{p(x^{2})}\Big{(}\lambda\frac{1}{d}\frac{xp^{\prime}(x^{2})}{p(x^{2})}+\frac{(1-\lambda)}{x}\Big{)}$

Note that we get some expansion around $x=\infty$:

$\mathcal{H}^{\lambda}_{\mathbb{S}p}(x)=\sum_{i=1}^{\infty}h_{i}^{(p)}\frac{1}{x^{2i}}$

Also, after the last root, the $H$-transform will be monotonous decreasing. It
leads to the following.

######  Lemma 4.2.

$\mathcal{H}^{\lambda}_{\mathbb{S}p}$ is a bijection from
$[\sqrt{\mathrm{maxroot}\left\\{p\right\\}},+\infty]$ to $[0,+\infty]$. We
denote the inverse by $\mathcal{J}^{\lambda}_{\mathbb{S}p}$. In particular,
for $x>0$

$\mathcal{H}^{\lambda}_{\mathbb{S}p}\circ\mathcal{J}^{\lambda}_{\mathbb{S}p}[x]=x$

and for $x>\sqrt{\mathrm{maxroot}\left\\{p\right\\}}$

$\mathcal{J}^{\lambda}_{\mathbb{S}p}\circ\mathcal{H}^{\lambda}_{\mathbb{S}p}[x]=x$

######  Lemma 4.3.

We can get an expression for $\mathcal{J}^{\lambda}_{\mathbb{S}p}$ in terms of
the free probability power series inverse
$[H^{\lambda}_{\mu_{\mathbb{S}p}}]^{-1}$. Given the quadratic nature in terms
of the inverse variable, it makes sense for symmetry to replace $x$ by
$s^{2}$, which is relevant as all the quantities we consider are positive. It
will avoid us to restrict to positive variables. For s in a neighbohood of
$0$,

$\big{[}\mathcal{J}^{\lambda}_{\mathbb{S}p}(s^{2})\big{]}^{2}=\frac{1}{[H^{\lambda}_{\mu_{\mathbb{S}p}}]^{-1}(s^{2})}$

###### Proof.

Use the formula
$\mathcal{H}^{\lambda}_{\mathbb{S}p}(y)=H^{\lambda}_{\mu_{\mathbb{S}p}}(\frac{1}{y^{2}})$.
∎

######  Corollary 4.4.

$s^{2}\big{[}\mathcal{J}^{\lambda}_{\mathbb{S}p}(s^{2})\big{]}^{2}$ is
analytic for $s$ in a neighborhood of $0$, even though
$\mathcal{J}^{\lambda}_{\mathbb{S}p}(s^{2})$ is not. It is important to
emphasize that $\mathcal{J}^{\lambda}_{\mathbb{S}p}(s^{2})$ is not a power
series in $s^{2}$.

###### Proof.

$\frac{s^{2}}{s^{2}+\sum_{i=2}^{\infty}\beta_{i}s^{2i}}=\frac{1}{1+\sum_{i=1}^{\infty}\beta_{i+1}s^{2i}}$

Which is analytic (multiplying by $s^{2}$ gets rid of the singularity at $0$).
∎

We are now able to get a more simple formula for the Benaych-Georges’
rectangular $R$-transform applied to a symmetric discrete measure.

######  Claim 4.5.

$\mathcal{R}_{\mu_{\mathbb{S}p}}^{\lambda}(s^{2})=\frac{-\lambda-1}{2\lambda}+\sqrt{\frac{(\lambda-1)^{2}}{4\lambda^{2}}+\frac{s^{2}[\mathcal{J}^{\lambda}_{\mathbb{S}p}(s^{2})]^{2}}{\lambda}}$

###### Proof.

$\displaystyle\mathcal{R}_{\mu_{\mathbb{S}p}}^{\lambda}(s^{2})=U\Big{(}\frac{s^{2}}{[{H_{\mu_{\mathbb{S}p}}^{\lambda}}]^{-1}(s^{2})}-1\Big{)}$
$\displaystyle=\frac{-\lambda-1}{2\lambda}+\frac{\sqrt{(\lambda+1)^{2}+4\lambda\big{(}\frac{s^{2}}{{[H^{\lambda}_{\mu_{\mathbb{S}p}}]^{-1}(s^{2})}}-1\big{)}}}{2\lambda}$
$\displaystyle=\frac{-\lambda-1}{2\lambda}+\frac{\sqrt{(\lambda-1)^{2}+\frac{4\lambda
s^{2}}{{[H^{\lambda}_{\mu_{\mathbb{S}p}}]^{-1}(s^{2})}}}}{2\lambda}$
$\displaystyle=\frac{-\lambda-1}{2\lambda}+\sqrt{\frac{(\lambda-1)^{2}}{4\lambda^{2}}+\frac{s^{2}[\mathcal{J}^{\lambda}_{\mathbb{S}p}(s^{2})]^{2}}{\lambda}}$

∎

######  Definition 4.6.

For $p$ with all roots nonnegative, we define:

$\mathcal{R}_{\mathbb{S}p}^{\lambda}(s^{2}):=\mathcal{R}_{\mu_{\mathbb{S}p}}^{\lambda}(s^{2})$

It is important to stress that here
$\mathcal{R}_{\mathbb{S}p}^{\lambda}(s^{2})$ is indeed a power series in
$s^{2}$ and not just a function.

###  4.2 Truncating bivariate power series

We start by generalizing the notion of truncation to bivariate power series.
For $f(r,t)=\sum_{i,j}a_{i,j}r^{i}t^{j}$, we denote by

$f\mod[<r,t>^{d}]:=\sum_{i+j\leq d-1}a_{i,j}r^{i}t^{j}$

We will write:

$f\equiv g\mod[<r,t>^{d}]$

when $f-g\mod[<r,t>^{d}]=0$. The following properties follow easily. If

$\displaystyle a(r,t)$ $\displaystyle\equiv b(r,t)\mod[<r,t>^{d}]$
$\displaystyle c(r,t)$ $\displaystyle\equiv d(r,t)\mod[<r,t>^{d}]$

then, there is additivity and multiplicativity:

$\displaystyle a(r,t)+c(r,t)$ $\displaystyle\equiv
b(r,t)+d(r,t)\mod[<r,t>^{d}]$ $\displaystyle a(r,t)c(r,t)$
$\displaystyle\equiv b(r,t)d(r,t)\mod[<r,t>^{d}]$

Therefore for any power series $h$,

$h[a(r,t)]\equiv h[b(r,t)]\mod[<r,t>^{d}]$

We can also differentiate the modulo equalities, up to losing one degree:

$\displaystyle\frac{\partial}{\partial r}a(r,t)$
$\displaystyle\equiv\frac{\partial}{\partial r}b(r,t)\mod[<r,t>^{d-1}]$
$\displaystyle\frac{\partial}{\partial t}a(r,t)$
$\displaystyle\equiv\frac{\partial}{\partial t}b(r,t)\mod[<r,t>^{d-1}]$

###  4.3 $L_{q}$-norms on intervals depending on the largest root

######  Definition 4.7.

Take a two-variable continuous function $f(x,y)$ that is positive on a
bidimensional interval $X$. Define the $L_{q}$-norm associated with this
interval as

$\|f(x,y)\|_{q}:=(\int\int_{(x,y)\in X}f(x,y)^{q}dxdy)^{\frac{1}{q}}$

######  Definition 4.8.

We will denote, similarly, by

$\|f(x,y)\|_{\infty}:=\sup_{(x,y)\in X}|f(x,y)|$

######  Lemma 4.9.

If $f$ is such that its $L_{q}$ and $L_{\infty}$ norms are finite, then

$\lim_{q\to\infty}\|f(x,y)\|_{q}=\|f(x,y)\|_{\infty}$

In all the following, the function $f(x,y)$ that we will consider will contain
some problematic polynomial factor that doesn’t have constant sign:
$p_{A}(xy)$. We will therefore choose an interval X that depends on the
polynomial (or the operator), more specifically if we denote by $R$ the
largest root of $p$, we take

$X:=[\sqrt{\mathrm{maxroot}\left\\{p\right\\}},+\infty]\times[\sqrt{\mathrm{maxroot}\left\\{p\right\\}},+\infty]=[\sqrt{R},+\infty]\times[\sqrt{R},+\infty]$

This ensures that for $(x,y)$ in this interval, the product $xy$ is above the
largest root and the polynomial is positive.

###  4.4 Rectangular Fuglede-Kadison determinant and potential

We first recall the usual simple definition of the Fuglede-Kadison
determinant. For a finite dimensional positive definite $d\times d$ matrix, it
corresponds to the following normalized determinant:

$\Delta^{+}(A):=\det(A)^{1/d}$

But the main object needed will be the bivariate following object, defined for
a general rectangular matrix $A$.

######  Definition 4.10.

For $\lambda\in]0,1]$, and $m$, $d$ integers such that $\frac{d}{m}=\lambda$
and for $(x,y)\in X$, introduce the bivariate $\lambda$-rectangular Fuglede-
Kadison determinant of an $m\times d$ matrix $A$:

$\Delta_{\lambda}^{+}(x,y,A):=y^{\frac{1-\lambda}{\sqrt{\lambda}}}\big{[}\Delta^{+}\big{(}xyI_{d}-A^{T}A\big{)}\big{]}^{\sqrt{\lambda}}$

It is well defined because for $(x,y)\in X$, the polynomial inside takes
positive values. This definition shows clearly that it doesn’t depend on the
dimensions, only on the root distribution and the parameter $\lambda$. We can
however expand and we get

$\Delta_{\lambda}^{+}(x,y,A)=y^{\frac{m-d}{\sqrt{md}}}[p_{A}(xy)]^{\frac{1}{\sqrt{md}}}$

for $(x,y)\in X$.

######  Remark 4.11.

We normalize the characteristic polynomial so that the polynomials associated
to two matrices of different sizes but with the same underlying measure on the
roots don’t differ. Take $A_{1}$, $m_{1}\times d_{1}$ and $A_{2}$,
$m_{2}\times d_{2}$, two rectangular matrices such that
$\frac{d1}{m1}=\frac{d2}{m2}=\lambda$ and $\mu_{p_{A_{1}}}=\mu_{p_{A_{2}}}$.
Then

$\Delta_{\lambda}^{+}(x,y,A_{1})=\Delta_{\lambda}^{+}(x,y,A_{2})$

Therefore we can increase the dimensions and the normalized polynomials stay
the same. In all the following we will denote by

$\Delta_{\lambda}^{+}(x,y,\mu_{A}):=\Delta_{\lambda}^{+}(x,y,A)$

if $\mu_{\mathbb{S}p_{A}}=\mu_{A}$.

######  Definition 4.12 (Fuglede-Kadison Potential).

For a given symmetric compact discrete measure $\mu_{A}$, we consider the
following $\lambda$-Fuglede Kadison potential defined on $X$:

$F^{\lambda}_{\mu_{A}}(x,y):=-\log\Delta_{\lambda}^{+}(x,y,\mu_{A})$

Rewrite $F^{\lambda}_{\mu_{A}}$ as

$F^{\lambda}_{\mu_{A}}(x,y)=-\frac{1}{\sqrt{md}}\log[y^{m-d}{p_{A}}(xy)]$

######  Lemma 4.13.

$F^{\lambda}_{\mu_{A}}$ is convex and
$\partial_{x}{F^{\lambda}_{\mu_{A}}}(z,z)\partial_{y}{F^{\lambda}_{\mu_{A}}}(z,z)=\mathcal{H}^{\lambda}_{\mu_{A}}(z)$
for $z>0$.

###### Proof.

$\displaystyle\partial_{x}{F^{\lambda}_{\mu_{A}}}(z,z)\partial_{y}{F^{\lambda}_{\mu_{A}}}(z,z)$
$\displaystyle=\frac{1}{md}\Big{(}\frac{zp_{A}^{\prime}(z^{2})}{p_{A}(z^{2})}\Big{)}\Big{(}\frac{zp_{A}^{\prime}(z^{2})}{p_{A}(z^{2})}+\frac{(m-d)}{z}\Big{)}=\Big{(}\frac{1}{d}\frac{zp_{A}^{\prime}(z^{2})}{p_{A}(z^{2})}\Big{)}\Big{(}\frac{d}{m}\frac{1}{d}\frac{zp_{A}^{\prime}(z^{2})}{p_{A}(z^{2})}+\frac{(m-d)}{m}\frac{1}{z}\Big{)}$
$\displaystyle=\mathcal{H}^{\lambda}_{\mathbb{S}p_{A}}(z)$

As for the convexity it suffices to compute the second derivatives and show
that $\partial_{xx}{F^{\lambda}_{\mu_{A}}}\geq 0$,
$\partial_{yy}{F^{\lambda}_{\mu_{A}}}\geq 0$ and
$\partial_{xx}{F^{\lambda}_{\mu_{A}}}\partial_{yy}{F^{\lambda}_{\mu_{A}}}-(\partial_{xy}{F^{\lambda}_{\mu_{A}}})^{2}\geq
0$. But we have

$\displaystyle\partial_{xx}{F^{\lambda}_{\mu_{A}}}(x,y)$
$\displaystyle=\frac{1}{\sqrt{md}}\frac{y^{2}}{p_{A}^{2}(xy)}\Big{(}p_{A}^{\prime
2}(xy)-p_{A}^{\prime\prime}(xy)p_{A}(xy)\Big{)}$
$\displaystyle\partial_{yy}{F^{\lambda}_{\mu_{A}}}(x,y)$
$\displaystyle=\frac{1}{\sqrt{md}}\frac{x^{2}}{p_{A}^{2}(xy)}\Big{(}p_{A}^{\prime
2}(xy)-p_{A}^{\prime\prime}(xy)p_{A}(xy)\Big{)}+\frac{m-d}{\sqrt{md}}\frac{1}{y^{2}}$
$\displaystyle\partial_{xy}{F^{\lambda}_{\mu_{A}}}(x,y)$
$\displaystyle=\frac{1}{\sqrt{md}}\frac{xy}{p_{A}^{2}(xy)}\Big{(}p_{A}^{\prime
2}(xy)-p_{A}^{\prime\prime}(xy)p_{A}(xy)\Big{)}$

We notice that as $p_{A}$ is real-rooted,

$p_{A}^{\prime 2}(xy)-p_{A}^{\prime\prime}(xy)p_{A}(xy)\geq 0$

by Laguerre’s inequality, and therefore we have
$\partial_{xx}{F^{\lambda}_{\mu_{A}}}\geq 0$,
$\partial_{yy}{F^{\lambda}_{\mu_{A}}}\geq 0$. Lastly,

$\partial_{xx}{F^{\lambda}_{\mu_{A}}}\partial_{yy}{F^{\lambda}_{\mu_{A}}}-(\partial_{xy}{F^{\lambda}_{\mu_{A}}})^{2}=\frac{m-d}{md}\frac{1}{{p_{A}}^{2}(xy)}\Big{(}p_{A}^{\prime
2}(xy)-p_{A}^{\prime\prime}(xy)p_{A}(xy)\Big{)}\geq 0$

∎

###  4.5 Bidimensional Legendre transform

The main tool to inverse gradients of convex functions is the Legendre
transform. We first recall the definition the bidimensional Legendre transform
and basic porperties. Denote by $X^{*}=\nabla f(X)$, a convex subset of
$\mathbb{R}^{2}$, the image of $X$ by the gradient of $f$.

######  Definition 4.14 (bidimensional Legendre transform).

If $f(x,y)$ is a convex function on $X$ we define for $(r,t)\in X^{*}$ :

$\displaystyle f^{*}(r,t)$ $\displaystyle:=\|xr+yt-f(x,y)\|_{\infty}$
$\displaystyle=\sup_{(x,y)\in X}\\{xr+yt-f(x,y)\\}$

######  Lemma 4.15.

If $f$ is strictly convex on $X$, then we have for all $(x,y)\in X,(r,t)\in
X^{*}$

$[\nabla f^{*}\circ\nabla f](x,y)=(x,y)$

$[\nabla f\circ\nabla f^{*}](r,t)=(r,t)$

######  Lemma 4.16.

For all $(r,t)\in X^{*}$,

$f^{*}(r,t)=\log\|e^{xr+yt-f(x,y)}\|_{\infty}$

###### Proof.

It follows from the fact that
$\|e^{g(x,y)}\|_{\infty}=e^{\|g(x,y)\|_{\infty}}$. ∎

###  4.6 Approximation of the inverse transform

The goal in this section is to derive an explicit expression for
$\mathcal{J}^{\lambda}_{\mathbb{S}p_{A}}(s^{2})$ using the bivariate tools
introduced in the last section.

######  Lemma 4.17 (Inverting the rectangular Cauchy transform).

For $s\neq 0$, then we can exhibit $r_{p_{A}}(s)$ and $t_{p_{A}}(s)$ functions
of $s$, such that $r_{p_{A}}(s)t_{p_{A}}(s)=s^{2}$, and

${\mathcal{J}^{\lambda}_{\mu_{A}}}(s^{2})^{2}={\mathcal{J}^{\lambda}_{\mathbb{S}p_{A}}}(s^{2})^{2}={\frac{\partial}{\partial
r}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\infty}\frac{\partial}{\partial
t}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\infty}}_{\big{|}_{r=r_{p_{A}}(s),t=t_{p_{A}}(s)}}$

###### Proof.

Recall that

$F^{\lambda}_{\mu_{A}}(x,y)=-\log\Delta_{\lambda}^{+}(x,y,\mu_{A})$

Take $s\neq 0$.We have by definition
$\mathcal{H}^{\lambda}_{\mathbb{S}p_{A}}\circ\mathcal{J}^{\lambda}_{\mathbb{S}p_{A}}(s^{2})=s^{2}$.
Call $z_{p_{A}}(s):=\mathcal{J}^{\lambda}_{\mathbb{S}p_{A}}(s^{2})$ to
alleviate the computations. Ideally, we would want some explicit power series
of $z_{p_{A}}$ in $s$. This being too complicated, we look for an expression
that would explicitly depend on $r_{p_{A}}$ and $t_{p_{A}}$ such that
$r_{p_{A}}t_{p_{A}}=s^{2}=\mathcal{H}^{\lambda}_{\mathbb{S}p_{A}}(z_{p_{A}})$.

Define:

$\displaystyle r_{p_{A}}(s):=$
$\displaystyle-\partial_{x}F^{\lambda}_{\mu_{A}}(z_{p_{A}},z_{p_{A}})=\frac{1}{\sqrt{md}}\frac{z_{p_{A}}p_{A}^{\prime}(z_{p_{A}}^{2})}{p_{A}(z_{p_{A}}^{2})}$
$\displaystyle t_{p_{A}}(s):=$
$\displaystyle-\partial_{y}F^{\lambda}_{\mu_{A}}(z_{p_{A}},z_{p_{A}})=\frac{1}{\sqrt{md}}\frac{z_{p_{A}}p_{A}^{\prime}(z_{p_{A}}^{2})}{p_{A}(z_{p_{A}}^{2})}+\frac{1}{\sqrt{md}}\frac{m-d}{z_{p_{A}}}$

Given that $z_{p_{A}}>\sqrt{R}$ as mentioned, we get that $r_{p_{A}}>0$,
$t_{p_{A}}>0$. Furthermore, using the fact that the gradient of the Legendre
transform is the inverse of the gradient of $F$, like stated in 4.15, we
obtain, applying the equality at $(z_{p_{A}},z_{p_{A}})$

$\begin{bmatrix}\partial_{r}{F^{\lambda}_{\mu_{A}}}^{*}[\partial_{x}F^{\lambda}_{\mu_{A}}(z_{p_{A}},z_{p_{A}}),\partial_{y}F^{\lambda}_{\mu_{A}}(z_{p_{A}},z_{p_{A}})]\\\
\partial_{t}{F^{\lambda}_{\mu_{A}}}^{*}[\partial_{x}F^{\lambda}_{\mu_{A}}(z_{p_{A}},z_{p_{A}}),\partial_{y}F^{\lambda}_{\mu_{A}}(z_{p_{A}},z_{p_{A}})]\end{bmatrix}=\begin{bmatrix}z_{p_{A}}\\\
z_{p_{A}}\end{bmatrix}$

Making the product of the two lines we get

$\partial_{r}{F^{\lambda}_{\mu_{A}}}^{*}(-r_{p_{A}},-t_{p_{A}})\partial_{t}{F^{\lambda}_{\mu_{A}}}^{*}(-r_{p_{A}},-t_{p_{A}})=z_{p_{A}}^{2}$

Now let’s rewrite the left hand side quantity and conclude:

$\displaystyle\partial_{r}{F^{\lambda}_{\mu_{A}}}^{*}(-r_{p_{A}},-t_{p_{A}})\partial_{t}{F^{\lambda}_{\mu_{A}}}^{*}(-r_{p_{A}},-t_{p_{A}})=\frac{\partial}{\partial
r}\log\|e^{-xr-
yt-F^{\lambda}_{\mu_{A}}(x,y)}\|_{\infty}\frac{\partial}{\partial
t}\log\|e^{-xr-
yt-F^{\lambda}_{\mu_{A}}(x,y)}\|_{\infty}\big{|}_{r=r_{p_{A}},t=t_{p_{A}}}$
(15) $\displaystyle=\frac{\partial}{\partial r}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\infty}\frac{\partial}{\partial
t}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\infty}\big{|}_{r=r_{p_{A}},t=t_{p_{A}}}$
(16)

∎

######  Corollary 4.18.

$r_{p_{A}}(s)$ and $t_{p_{A}}(s)$ are analytic functions of $s$ in the
neighborhood of zero.

###### Proof.

The proofs are similar. We start by the decomposition:

$r_{p_{A}}(s)=\sum_{i=1}^{\infty}\rho_{i}\frac{1}{z_{p_{A}}^{i}}$

where $z_{p_{A}}=\mathcal{J}^{\lambda}_{\mu_{A}}(s^{2})$. Now remember that

$\frac{1}{z_{p_{A}}}=\sqrt{[H^{\lambda}_{\mu_{A}}]^{-1}(s^{2})}=\sum_{i=1}^{\infty}z_{i}s^{i}$

The result follows by composition of power series around the origin. ∎

######  Remark 4.19.

Notice that in the square case, $r_{p_{A}}=s$ and $t_{p_{A}}=s$. In the
rectangular case, they are approximations of the identity, and in all the
following we will show that in this case, the higher order terms in the power
expansions in $s$ can still be neglected (but it is not trivial).

Computing the infinity norm is not easy, and we try to reach it through
$L_{q}$-norms and taking limits as $q$ goes to infinity.

######  Definition 4.20.

For an $m\times d$ matrix A with singular distribution $\mu_{A}$, define the
finite inverse rectangular transform as

$\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s):={\frac{\partial}{\partial
r}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\big{)}\frac{\partial}{\partial
t}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\big{)}}_{\big{|}_{r=r(s),t=t(s)}}$

######  Lemma 4.21.

For $s$ small enough such that ${\mathcal{J}^{\lambda}_{\mu_{A}}}(s^{2})$ is
well defined and finite,

$\lim_{\begin{subarray}{c}d\to\infty\\\
\end{subarray}}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)={\mathcal{J}^{\lambda}_{\mu_{A}}}(s^{2})^{2}$

######  Remark 4.22.

As $s^{2}{\mathcal{J}^{\lambda}_{\mu_{A}}}(s^{2})^{2}$ is analytic, we will
show in the next section that similarly
$s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)$ is analytic too.

We will need to use the following results of uniform convergence to prove
Lemma 4.21.

######  Lemma 4.23 (Second theorem of Dini).

Assume that are given a sequence of functions
$\big{(}f_{n}(x)\big{)}_{n}$defined on a segment of the real line and such
that

  * •

f is continuous,

  * •

there is monotonicity in $x$: for $x\leq y,f_{n}(x)\leq f_{n}(y)$,

  * •

there is pointwise convergence: $f_{n}(x)$ converges to $f(x)$,

then the convergence is uniform.

######  Lemma 4.24 (Derivative uniform convergence).

If $f_{n}$ converges pointwise to $f,f_{n}^{\prime}$ converges uniformly to
$f^{\prime}$, then $f_{n}$ converges uniformly to $f$and
$f^{\prime}=\lim_{n\to\infty}f_{n}^{\prime}$.

We will also need to use convexity of Laplace transforms.

######  Lemma 4.25 (Convexity of Laplace transform).

Consider a function $g$ that is positive and sufficiently smooth so that it’s
Laplace tranform is well defined on a domain $X$,

$L(s):=\log\big{(}\int_{X}e^{-su}g(u)du\big{)}$

Then $L$ is convex, therefore $L^{\prime}$ is monotonous.

###### Proof.

$L^{\prime\prime}(s)=\frac{(\int_{X}e^{-su}g(u)du)(\int_{X}e^{-su}g(u)u^{2}du)-(\int_{X}e^{-su}g(u)udu)^{2}}{\big{(}\int_{X}e^{-su}g(u)du\big{)}^{2}}$

The numerator is positive using the inequality of Cauchy Schwarz. ∎

###### Proof of Lemma 4.21.

$\lim_{\begin{subarray}{c}d\to\infty\\\
d=m\lambda\end{subarray}}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)=\lim_{\begin{subarray}{c}d\to\infty\\\
d=m\lambda\end{subarray}}\frac{\partial}{\partial r}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\frac{\partial}{\partial
t}\log\|e^{-xr-yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}$

The goal is to show that we can permute the limit and the derivative signs. We
would get the result as

$\displaystyle\lim_{\begin{subarray}{c}d\to\infty\\\
d=m\lambda\end{subarray}}\frac{\partial}{\partial r}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\frac{\partial}{\partial
t}\log\|e^{-xr-yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}$ (17)
$\displaystyle=\frac{\partial}{\partial r}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\infty}\frac{\partial}{\partial
t}\log\|e^{-xr-yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\infty}$ (18)

By Lemma 4.9. Now let’s do only the first term (derivative in $r$), the other
one is done the same way. Define:

$f_{d}(r):=\log\|e^{-xr-yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}$

Now we can use Lemma 4.25 to get that $f_{d}^{\prime}(r)$ is monotonous in $r$
therefore we can apply Lemma 4.23 to get uniform convergence of the
derivatives and eventually we can apply Lemma 4.24 to conclude that
$\big{[}\lim_{d\to\infty}f_{d}(r)\big{]}^{\prime}=\lim_{d\to\infty}f_{d}^{\prime}(r)$

∎

##  5 The modified rectangular finite $R$-transform and convergence

We define in this section a modified finite $R$-transform as a quantity that
converges by design to the analytical rectangular free probability
$R$-transform defined by Benaych Georges. And we then operate the shift from
the convergence of the modified rectangular finite $R$-transform to the actual
rectangular finite $R$-transform.

###  5.1 The modified finite $R$-transform definition and convergence

Denote by

$\mathcal{K}_{\mu_{A}}^{d,\lambda}(s):=\sqrt{\frac{1}{\lambda}s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)+\frac{(1-\lambda)^{2}}{4\lambda^{2}}}$

Using the expression of $\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)$ computed in the
next section, we easily get

$\mathcal{K}_{\mu_{0}}^{d,\lambda}=\Big{(}\frac{(\lambda+1)}{2\lambda}+\frac{1}{d}\Big{)}$

######  Definition 5.1 (Modified rectangular finite $R$-transform).

For all $s>0$, the modified rectangular finite $R$-transform is defined as

$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s):=\mathcal{K}_{\mu_{A}}^{d,\lambda}(s)-\mathcal{K}_{\mu_{0}}^{d,\lambda}$

######  Remark 5.2.

It will follow from 5.6 below that
$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)$ can be expressed as a power
series in $s$ like $\mathcal{R}_{\mu_{A}}^{\lambda}(s^{2})$ (which makes the
definition not totally arbitrary). Also note that by definition
$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)$ has no constant term.

We can explicitly see the convergence to the free probability transform as

$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)=\sqrt{\frac{1}{\lambda}s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)+\frac{(1-\lambda)^{2}}{4\lambda^{2}}}-\Big{(}\frac{(\lambda+1)}{2\lambda}+\frac{1}{d}\Big{)}$

and

$\mathcal{R}_{\mu_{A}}^{\lambda}(s^{2})=\sqrt{\frac{s^{2}[\mathcal{J}^{\lambda}_{\mu_{A}}(s^{2})]^{2}}{\lambda}+\frac{(\lambda-1)^{2}}{4\lambda^{2}}}-\frac{\lambda+1}{2\lambda}$

As on the other hand, using the convergence Lemma 4.17, for s small enough
such that ${\mathcal{J}^{\lambda}_{\mu_{A}}}(s^{2})$ is well defined:

$\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)\xrightarrow{d\to\infty}{\mathcal{J}^{\lambda}_{\mu_{A}}}(s^{2})^{2}$

we get

######  Theorem 5.3 (Convergence of the modified finite $R$-transform to the
free $R$-transform).

For $s$ small enough, such that $\mathcal{R}_{\mu_{A}}^{\lambda}(s^{2})$ is
absolutely convergent,

$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)\xrightarrow{d\to\infty}\mathcal{R}_{\mu_{A}}^{\lambda}(s^{2})$

######  Remark 5.4.

$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)$ is a power series in $s$
but in the limit ${\begin{subarray}{c}d\to\infty\end{subarray}}$ becomes a
power series in $s^{2}$ like $\mathcal{R}_{\mu_{A}}^{\lambda}(s^{2})$.

###  5.2 Relating the modified finite $R$-transform to the finite
$R$-transform

The goal is now to get some explicit expression for
$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)$ , or at least the first
coefficients in the power series expansion, so that we can relate it to the
actual $R$-trasnform $\mathcal{R}^{d,\lambda}_{\mu_{A}}(s)$.

####  5.2.1 From integrals to power series

In this section, we explictly integrate and group summation terms.

######  Lemma 5.5.

For $r,t>0$, we get

$\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{m\sqrt{\lambda}}^{m\sqrt{\lambda}}=\frac{m!d!}{(rm\sqrt{\lambda
R})^{d+1}(tm\sqrt{\lambda
R})^{m+1}}\sum_{i=0}^{d}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)$

where

$\displaystyle L_{i}^{d,\lambda}(r):=e^{-rm\sqrt{\lambda
R}}\sum_{l=0}^{d-i}\frac{(rm\sqrt{\lambda R})^{l}}{l!}$ $\displaystyle
M_{i}^{d,\lambda}(t):=e^{-tm\sqrt{\lambda
R}}\sum_{j=0}^{m-i}\frac{(tm\sqrt{\lambda R})^{j}}{j!}$

###### Proof.

$\displaystyle\|e^{-xr-yt}\Delta_{\lambda}^{+}(x,y,A)\|_{m\sqrt{\lambda
R}}^{m\sqrt{\lambda R}}$
$\displaystyle=\int_{x=\sqrt{R}}^{\infty}\int_{y=\sqrt{R}}^{\infty}e^{-xrm\sqrt{\lambda}-ytm\sqrt{\lambda}}y^{m-d}p_{A}(xy)dxdy$
$\displaystyle=\int_{x=\sqrt{R}}^{\infty}\int_{y=\sqrt{R}}^{\infty}e^{-xrm\sqrt{\lambda}-ytm\sqrt{\lambda}}\mathbb{E}(e^{-T^{(m,d)}_{\mathbb{S}p_{A}}\partial{x}\partial_{y}}\\{y^{m}x^{d}\\})dxdy$
$\displaystyle=\sum_{i=0}^{d}\int_{x=\sqrt{R}}^{\infty}\int_{y=\sqrt{R}}^{\infty}\frac{\mathbb{E}[(-T^{(m,d)}_{\mathbb{S}p_{A}})^{i}]}{i!}e^{-xrm\sqrt{\lambda}-ytm\sqrt{\lambda}}\frac{m!d!}{(m-i)!(d-i)!}y^{m-i}x^{d-i}dxdy$

Now we use the fact that for $n$ integer and $a,b>0$,

$\int_{b}^{\infty}x^{n}e^{-ax}dx=\frac{n!}{a^{n+1}}e^{-ab}\sum_{i=0}^{n}\frac{(ab)^{i}}{i!}.$

We therefore get that

$\displaystyle\int_{x=\sqrt{R}}^{\infty}\int_{y=\sqrt{R}}^{\infty}\frac{\mathbb{E}[(-T^{(m,d)}_{\mathbb{S}p_{A}})^{i}]}{i!}e^{-xrm\sqrt{\lambda}-ytm\sqrt{\lambda}}\frac{m!d!}{(m-i)!(d-i)!}y^{m-i}x^{d-i}dxdy$
$\displaystyle=\frac{\mathbb{E}[(-T^{(m,d)}_{\mathbb{S}p_{A}})^{i}]}{i!}\frac{m!d!L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)}{(r\sqrt{\lambda
R}m)^{d-i+1}(t\sqrt{\lambda R}m)^{m-i+1}}$

∎

######  Corollary 5.6 (Analycity of finite transforms ).

$s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)$ is analytic. As a consequence,
$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)$ is also analytic.

###### Proof.

Let’s rewrite it as a product of two terms and show that any of the two is a
convergent power series in $s$:

$s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)=r{\frac{\partial}{\partial
r}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\big{)}\times
t\frac{\partial}{\partial t}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\big{)}}_{\big{|}_{r=r(s),t=t(s)}}.$

Using Lemma 5.5, we obtain

$\displaystyle\frac{\partial}{\partial r}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{m\sqrt{\lambda}}\big{)}$
$\displaystyle=\frac{1}{m\sqrt{\lambda}}\frac{\partial}{\partial
r}\log\Bigg{(}\frac{m!d!}{(rm\sqrt{\lambda})^{d+1}(tm\sqrt{\lambda})^{m+1}}\sum_{i=0}^{d}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)\Bigg{)}$
$\displaystyle=\frac{1}{m\sqrt{\lambda}}\frac{\partial}{\partial
r}\log\Big{(}\frac{m!d!}{(rm\sqrt{\lambda})^{d+1}(tm\sqrt{\lambda})^{m+1}}\Big{)}+\frac{1}{m\sqrt{\lambda}}\frac{\partial}{\partial
r}\log\Big{(}\sum_{i=0}^{d}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)\Big{)}$
$\displaystyle=\frac{1}{m\sqrt{\lambda}}\frac{-(d+1)}{r}+\frac{1}{m\sqrt{\lambda}}\frac{\partial}{\partial
r}\log\Big{(}1+\sum_{i=1}^{d}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)\Big{)}.$

Therefore,

$r\frac{\partial}{\partial r}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{m\sqrt{\lambda}}\big{)}=-\frac{1}{m\sqrt{\lambda}}(d+1)+r\frac{\partial}{\partial
r}\log\Big{(}1+\sum_{i=1}^{d}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)\Big{)}.$

If we call

$\sum_{i,j=1}^{\infty}a_{i,j}r^{i}t^{j}:=\sum_{i=1}^{d}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)$

which is analytic in two variables (see [9] for an introduction to
multivariate complex analysis), we have that
$\log\big{(}1+\sum_{i,j=1}^{\infty}a_{i,j}r^{i}t^{j}\big{)}=\sum_{i,j=1}^{\infty}b_{i,j}r^{i}t^{j}$
for $r$ and $t$ small enough and some coefficients $b_{i,j}$, by composition
of analytic functions, and also $r\frac{\partial}{\partial
r}\log\big{(}1+\sum_{i,j=1}^{\infty}a_{i,j}r^{i}t^{j}\big{)}=\sum_{i,j=1}^{\infty}ib_{i,j}r^{i}t^{j}$.
We conclude by plugging in $r=r_{p_{A}}(s)$ and $t=t_{p_{A}}(s)$ which are
both analytic which leads us to the analyticity of $r\frac{\partial}{\partial
r}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{m\sqrt{\lambda}}\big{)}_{\big{|}_{r=r(s),t=t(s)}}$,
which is what was required. ∎

####  5.2.2 Truncating the superfluous terms

The expression we get for the $L_{m\sqrt{\lambda}}$-norms depend on $r$ and
$t$ independently. We want to get rid of this bi-dependence and keep only a
dependence in $rt=s^{2}$.

######  Lemma 5.7.

$\sum_{i=0}^{d}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)\equiv\sum_{i=0}^{d}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}\mod[<r,t>^{d+1}]$

###### Proof.

We need to prove that we can get rid of the
$L_{i}^{d,\lambda}(r)$,$M_{i}^{d,\lambda}(t)$ factors.

$e^{-rm\sqrt{\lambda}}\sum_{l=0}^{d-i}\frac{(rm\sqrt{\lambda})^{l}}{l!}=e^{-rm\sqrt{\lambda}}(e^{rm\sqrt{\lambda}}-\sum_{l=d-i+1}^{\infty}\frac{(rm\sqrt{\lambda})^{l}}{l!})\equiv
1\mod[<r,t>^{d-i+1}]$

Similarly,

$e^{-tm\sqrt{\lambda}}\sum_{l=0}^{m-i}\frac{(tm\sqrt{\lambda})^{l}}{l!}\equiv
1\mod[<r,t>^{m-i+1}]\equiv 1\mod[<r,t>^{d-i+1}].$

It follows that

$\displaystyle L_{i}^{d,\lambda}(r)$ $\displaystyle=M_{i}^{d,\lambda}(t)\equiv
1\mod[<r,t>^{d-i+1}]\text{,}$ $\displaystyle
L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)$ $\displaystyle\equiv
1\mod[<r,t>^{d-i+1}]$

And finally,

$\displaystyle\sum_{i=0}^{d}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)$
$\displaystyle\equiv\sum_{i=0}^{d}\Big{[}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}.\big{(}1\mod[<r,t>^{d-i+1}]\big{)}\Big{]}$
$\displaystyle\equiv\sum_{i=0}^{d}\frac{E[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}\mod[<r,t>^{d+1}]$

Where we used that $(rt)^{i}\big{(}0\mod[<r,t>^{d-i+1}]\big{)}\equiv
0\mod[<r,t>^{d+1}]$ ∎

######  Lemma 5.8.

$r\frac{\partial}{\partial r}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\big{)}\equiv\frac{1}{\sqrt{md}}\Big{(}-(d+1)+rt\frac{d}{dx}\mathcal{E}^{d,\lambda}_{\mu_{A}}(x)\Big{|}_{x=rt}\Big{)}\mod[<r,t>^{d+1}]$
$t\frac{\partial}{\partial t}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\big{)}\equiv\frac{1}{\sqrt{md}}\Big{(}-(m+1)+rt\frac{d}{dx}\mathcal{E}^{d,\lambda}_{\mu_{A}}(x)\Big{|}_{x=rt}\Big{)}\mod[<r,t>^{d+1}]$

where

$\mathcal{E}^{d,\lambda}_{\mu_{A}}(x)=\log\big{(}\mathbb{E}\sum_{i=0}^{d}\frac{(-T^{(m,d)}_{\mathbb{S}p_{A}}xmd)^{i}}{i!}\big{)}$

In the following we will denote by :
$[\mathcal{E}{{}^{d,\lambda}_{\mu_{A}}}]^{\prime}(u):=\frac{d}{dx}\mathcal{E}^{d,\lambda}_{\mu_{A}}(x)\Big{|}_{x=u}$

###### Proof.

$\displaystyle\frac{\partial}{\partial r}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\big{)}=\frac{1}{m\sqrt{\lambda}}\frac{\partial}{\partial
r}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{m\sqrt{\lambda}}^{m\sqrt{\lambda}}\big{)}$
(19)
$\displaystyle=\frac{1}{m\sqrt{\lambda}}\Big{(}-\frac{d+1}{r}+\frac{\partial}{\partial
r}\log\big{(}\sum_{i=0}^{d}\frac{\mathbb{E}[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)\big{)}\Big{)}$
(20)

Then,

$\displaystyle r\frac{\partial}{\partial r}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\big{)}=\frac{1}{m\sqrt{\lambda}}\Big{(}-(d+1)+r\frac{\partial}{\partial
r}\log\big{[}\sum_{i=0}^{d}\frac{\mathbb{E}[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)\big{]}\Big{)}$
(21)
$\displaystyle\equiv\frac{1}{m\sqrt{\lambda}}\Big{(}-(d+1)+r\frac{\partial}{\partial
r}\log\Big{\\{}\sum_{i=0}^{d}\frac{\mathbb{E}[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}\mod[<r,t>^{d+1}]\Big{\\}}\Big{)}\text{Using
Lemma \ref{trunc}}$ (22)
$\displaystyle\equiv\frac{1}{m\sqrt{\lambda}}\Big{(}-(d+1)+r\frac{\partial}{\partial
r}\log\Big{\\{}\sum_{i=0}^{d}\frac{\mathbb{E}[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}\Big{\\}}\mod[<r,t>^{d+1}]\Big{)}$
(23)

where we used that applying a power series ($\log$) or the operator
$r\partial_{r}$ doesn’t affect a modulo equality, which follows from 4.2. Now
we conclude by noticing that

$\frac{\partial}{\partial
r}\mathcal{E}^{d,\lambda}_{\mu_{A}}(rt)=t[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(rt)$

######  Remark 5.9.

One could wonder if the logarithm defined above is well defined, as we are not
sure whether the quantity inside is positive. It follows from the fact that
the norm is always positive and that for $r$ and $t$ small enough (using the
asymptotic expansion in $t$ and $r$),

$\sum_{i=0}^{d}\frac{\mathbb{E}[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}L_{i}^{d,\lambda}(r)M_{i}^{d,\lambda}(t)>0\implies\sum_{i=0}^{d}\frac{\mathbb{E}[(-T^{(m,d)}_{\mathbb{S}p_{A}}rtmd)^{i}]}{i!}>0$

∎

####  5.2.3 Back to univariate power series

######  Corollary 5.10.

$s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)\equiv\frac{1}{md}\big{[}(m+1)-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})\big{]}\big{[}(d+1)-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})\big{]}\mod[s^{d+1}]$

###### Proof.

$\displaystyle r\frac{\partial}{\partial r}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\big{)}\times
t\frac{\partial}{\partial t}\big{(}\log\|e^{-xr-
yt}\Delta_{\lambda}^{+}(x,y,\mu_{A})\|_{\sqrt{md}}\big{)}\equiv$
$\displaystyle\frac{1}{\sqrt{md}}\Big{(}-(d+1)+rt\frac{d}{dx}\mathcal{E}^{d,\lambda}_{\mu_{A}}(x)\Big{|}_{x=rt}\Big{)}\times\frac{1}{\sqrt{md}}\Big{(}-(m+1)+rt\frac{d}{dx}\mathcal{E}^{d,\lambda}_{\mu_{A}}(x)\Big{|}_{x=rt}\Big{)}\mod[<r,t>^{d+1}]$

Now plug in $r=r_{p_{A}}(s)$ and $t=t_{p_{A}}(s)$ and notice that as
$r_{p_{A}}(s)=s+\sum_{i=2}^{\infty}r_{i}s^{i}$ and
$t_{p_{A}}(s)=s+\sum_{i=2}^{\infty}t_{i}s^{i}$,

$f\equiv g\mod[<r_{p_{A}}(s),t_{p_{A}}(s)>^{d+1}]\implies f\equiv
g\mod[s^{d+1}]$

∎

######  Corollary 5.11.

$\sqrt{\frac{1}{\lambda}s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)+\frac{(1-\lambda)^{2}}{4\lambda^{2}}}\equiv\frac{1}{d}\Big{(}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})+\big{[}m\frac{(\lambda+1)}{2}+1\big{]}\Big{)}\mod[s^{d+1}]$

###### Proof.

$\displaystyle
s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)\equiv\frac{1}{m^{2}\lambda}\Big{(}\big{[}s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})\big{]}^{2}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})\big{[}m+1+m\lambda+1\big{]}\Big{)}+\frac{(m+1)(m\lambda+1)}{m^{2}\lambda}\mod[s^{d+1}]$
(24)
$\displaystyle\equiv\frac{1}{m^{2}\lambda}\Big{(}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})+\big{[}m\frac{(\lambda+1)}{2}+1\big{]}\Big{)}^{2}+\frac{-\Big{(}m\frac{(\lambda+1)}{2}+1\Big{)}^{2}+(m+1)(m\lambda+1)}{m^{2}\lambda}\mod[s^{d+1}]$
(25)
$\displaystyle\equiv\frac{1}{m^{2}\lambda}\Big{(}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})+\big{[}m\frac{(\lambda+1)}{2}+1\big{]}\Big{)}^{2}-\frac{(\lambda-1)^{2}}{4\lambda}\mod[s^{d+1}]$
(26)

Therefore, multiplying both sides of the equation by $\frac{1}{\lambda}$ and
adding $\frac{(\lambda-1)^{2}}{4\lambda^{2}}$:

$\frac{1}{\lambda}s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)+\frac{(\lambda-1)^{2}}{4\lambda^{2}}\equiv\frac{1}{m^{2}\lambda^{2}}\Big{(}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})+\big{[}m\frac{(\lambda+1)}{2}+1\big{]}\Big{)}^{2}\mod[s^{d+1}]$

As
$\Big{(}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})+\big{[}m\frac{(\lambda+1)}{2}+1\big{]}\Big{)}\geq
0$, for $s$ small enough, we can take the square root of our modulo equality
(using the stability of the modulo equalities with respect to composition with
the analytic function $\sqrt{1+s}$ near $0$) :

$\displaystyle\sqrt{\frac{1}{\lambda}s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)+\frac{(\lambda-1)^{2}}{4\lambda^{2}}}$
$\displaystyle\equiv\sqrt{\frac{1}{m^{2}\lambda^{2}}\Big{(}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})+\big{[}m\frac{(\lambda+1)}{2}+1\big{]}\Big{)}^{2}}\mod[s^{d+1}]$
$\displaystyle\equiv\frac{1}{d}\Big{(}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})+\big{[}m\frac{(\lambda+1)}{2}+1\big{]}\Big{)}\mod[s^{d+1}]$

We finally obtain

$\sqrt{\frac{1}{\lambda}s^{2}\mathcal{Q}_{\mu_{A}}^{d,\lambda}(s)+\frac{(\lambda-1)^{2}}{4\lambda^{2}}}-\Big{(}\frac{[\lambda+1]}{2\lambda}+\frac{1}{d}\Big{)}\equiv\frac{1}{d}\Big{(}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})\Big{)}\mod[s^{d+1}]$

∎

######  Theorem 5.12.

$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)\equiv\mathcal{R}^{d,\lambda}_{\mu_{A}}(s^{2})\mod[s^{d+1}]$

###### Proof.

The previous equality gives us

$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)\equiv\frac{1}{d}\Big{(}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})\Big{)}\mod[s^{d+1}]$

On the other hand,

$\displaystyle\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}p_{A}}s^{2}md}\big{]}=\sum_{i=0}^{\infty}\frac{\mathbb{E}\big{[}-T^{(m,d)}_{\mathbb{S}p_{A}}s^{2}md\big{]}^{i}}{i!}\equiv\sum_{i=0}^{d}\frac{\mathbb{E}\big{[}-T^{(m,d)}_{\mathbb{S}p_{A}}s^{2}md\big{]}^{i}}{i!}\mod[s^{d+1}]$
$\displaystyle\log\big{(}\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}p_{A}}s^{2}md}\big{]}\big{)}\equiv\log\big{(}\sum_{i=0}^{d}\frac{\mathbb{E}\big{[}-T^{(m,d)}_{\mathbb{S}p_{A}}s^{2}md\big{]}^{i}}{i!}\big{)}\mod[s^{d+1}]$
$\displaystyle\frac{d}{ds}\log\big{(}\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}p_{A}}s^{2}md}\big{]}\big{)}\equiv\frac{d}{ds}\log\big{(}\sum_{i=0}^{d}\frac{\mathbb{E}\big{[}-T^{(m,d)}_{\mathbb{S}p_{A}}s^{2}md\big{]}^{i}}{i!}\big{)}\mod[s^{d}]$
$\displaystyle\frac{-s}{d}\frac{d}{ds}\log\big{(}\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}p_{A}}s^{2}md}\big{]}\big{)}\equiv\frac{-s}{d}\frac{d}{ds}\log\big{(}\sum_{i=0}^{d}\frac{\mathbb{E}\big{[}-T^{(m,d)}_{\mathbb{S}p_{A}}s^{2}md\big{]}^{i}}{i!}\big{)}\mod[s^{d+1}]$
$\displaystyle\mathcal{R}^{d,\lambda}_{\mu_{A}}(s^{2})\equiv\frac{1}{d}\Big{(}-s^{2}[\mathcal{E}^{d,\lambda}_{\mu_{A}}]^{\prime}(s^{2})\Big{)}\mod[s^{d+1}]$

∎

######  Remark 5.13.

We notice that the degree $d$ truncation of
$\tilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)$ is in fact a polynomial in
$s^{2}$. Only high order odd terms can be nonzero.

##  6 From finite free probability to free probability: a bridge between
algebra and analysis

######  Theorem 6.1.

(Convergence of the finite free rectangular transform to the free transform)
The following convergence holds coefficientwise and also pointwise for $s$
small enough:

$\mathcal{R}^{d,\lambda}_{\mu_{A}}(s)\xrightarrow{d\to\infty}\mathcal{R}^{\lambda}_{\mu_{A}}(s)$

###### Proof.

Let’s write

$\displaystyle\mathcal{R}^{\lambda}_{\mu_{A}}(s)$
$\displaystyle:=\sum_{k}a_{k}s^{k}$
$\displaystyle\mathcal{R}^{d,\lambda}_{\mu_{A}}(s)$
$\displaystyle:=\sum_{k}a^{d}_{k}s^{k}$
$\displaystyle\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)$
$\displaystyle:=\sum_{k}\tilde{a}^{d}_{k}s^{k}$

By Theorem 5.3, we get the convergence for all $s$ small enough of
$\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)$ to
$\mathcal{R}^{\lambda}_{\mu_{A}}(s^{2})$. In particular we get (as convergence
over an interval gives us the convergence of all coefficients separately) that

$\displaystyle\tilde{a}^{d}_{2k+1}$ $\displaystyle\xrightarrow{d\to\infty}0$
$\displaystyle\tilde{a}^{d}_{2k}$ $\displaystyle\xrightarrow{d\to\infty}a_{k}$

We also get from Corollary 5.12 that $\tilde{a}^{d}_{2k}=a^{d}_{k}$ for $k\leq
d/2$, which in particular gives us that
$a^{d}_{k}\to_{d\rightarrow\infty}a_{k}$ for all $k$. This doesn’t mean that
$\mathcal{R}^{d,\lambda}_{\mu_{A}}(s)$ converges to
$\mathcal{R}^{\lambda}_{\mu_{A}}(s)$ or does have a limit when $d$ goes to
infinity.

######  Lemma 6.2.

For $s$ small enough, $\mathcal{R}^{d,\lambda}_{\mu_{A}}(s)$ has a tail of
order $d/2$ that goes to zero when $d$ goes to infinity.

###### Proof.

Let’s recall that $p_{A}(x)=\sum_{i}(-1)^{i}x^{d-i}p_{i}$ and that

$\frac{\mathbb{E}\Big{(}\big{[}T^{(m,d)}_{\mathbb{S}p_{A}}md\big{]}^{i}\Big{)}}{i!}=\frac{m^{i}(m-i)!}{m!}\frac{d^{i}(d-i)!}{d!}p_{i}$

As the roots are uniformly bounded by $R$, we get that $|p_{i}|\leq
R^{i}\binom{d}{i}\leq R^{i}\binom{d}{d/2}$. On the other hand,

$\displaystyle\binom{d}{d/2}$ $\displaystyle\leq 2^{d}$
$\displaystyle\frac{d^{i}(d-i)!}{d!}$
$\displaystyle\leq\frac{d^{i}}{(d/2)^{i}}\leq 2^{d}$
$\displaystyle\frac{m^{i}(m-i)!}{m!}$ $\displaystyle\leq
2^{\frac{d}{\lambda}}$

Grouping the upper bounds we get for $\alpha>0$ and all $i\leq d$,

$\Big{|}\frac{\mathbb{E}\Big{(}\big{[}-T^{(m,d)}_{\mathbb{S}p_{A}}md\big{]}^{i}\Big{)}}{i!}\Big{|}\leq
e^{d\alpha}$

We want to investigate the expansion of

$\mathcal{R}^{\lambda}_{\mu_{A}}(s)=\frac{-1}{d}s\frac{d}{ds}\log\big{(}\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}p_{A}}smd}\big{]}\big{)}\mod[s^{d+1}]$

Now when we look at the coefficients in the expansion in $s$ of
$\log\big{(}\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}p_{A}}smd}\big{]}\big{)}$.
We get $p(k)$ terms of order $k$ expanding the series where $p(k)$ is the
number of partitions of $k$. Overall this number can be upper bounded by
$e^{\beta k}$ for some $\beta>0$ ( see [13]), so that the $k^{th}$ coefficient
in the expansion will be upper bounded by $e^{(\alpha+\beta)d}$, for $k\geq
d/2$. We conclude as

$\sum_{k=d/2}^{d}|a_{k}^{d}s^{k}|\leq\sum_{k=d/2}^{d}k|s|^{k}e^{(\alpha+\beta)k}$

And the last sum goes to zero for $s$ small enough.

∎

Back to the convergence. We readily get from Corollary 5.12 that

$\mathcal{R}^{d,\lambda}_{\mu_{A}}(s)=(\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)\mod[s^{d/2}])+\sum_{k=d/2}^{d}a_{k}^{d}s^{k}\
$ (27)

As $\widetilde{\mathcal{R}}^{d,\lambda}_{\mu_{A}}(s)$ converges to
$\mathcal{R}^{\lambda}_{\mu_{A}}(s)$ for $s$ small enough, then its partial
sum of order $d/2$, the first term in 27, also converges to this same limit.
Using Lemma 6.2, the second term in 27 goes to zero and finally we get that
$\mathcal{R}^{d,\lambda}_{\mu_{A}}(s)$ converges to
$\mathcal{R}^{\lambda}_{\mu_{A}}(s)$ . ∎

.

######  Lemma 6.3.

A good way to exhibit actual matrices/polynomials that have constant
underlying measure and whose dimensions go to infinity is to stack up (by
block) $n$ identical matrices $A$ of fixed size $m\times d$(associated to a
polynomial $p$) along the diagonal, and their characteristic polynomials will
be $p^{n}$. We can make $n$ go to infinity and we get

$\mathcal{R}^{dn,\lambda}_{\mathbb{S}p^{n}}(s)\xrightarrow{n\to\infty}\mathcal{R}_{\mathbb{S}p}^{\lambda}(s)$

###### Proof.

Consider $p^{n}$, a polynomial with roots repeated $n$ times. It is a way to
keep the measure on the roots constant and increase the size of the matrix. We
have

$\mu_{\mathbb{S}p}=\mu_{\mathbb{S}p^{n}}$

And using Theorem 6.1 we get that

$\mathcal{R}^{dn,\lambda}_{\mathbb{S}p^{n}}(s)\xrightarrow{n\to\infty}\mathcal{R}_{\mathbb{S}p}^{\lambda}(s)$

∎

######  Corollary 6.4.

$\mathcal{R}^{dn,\lambda}_{\mathbb{S}[p^{n}\boxplus_{d,\lambda}q^{n}]}(s)\xrightarrow{n\to\infty}\mathcal{R}_{\mu_{\mathbb{S}p}\boxplus_{\lambda}\mu_{\mathbb{S}q}}^{\lambda}(s)$

###### Proof.

We have on the one hand

$\mathcal{R}^{dn,\lambda}_{\mathbb{S}[p^{n}\boxplus_{d,\lambda}q^{n}]}(s)=\mathcal{R}^{dn,\lambda}_{\mathbb{S}p^{n}}(s)+\mathcal{R}^{dn,\lambda}_{\mathbb{S}q^{n}}(s)$

so that

$\mathcal{R}^{dn,\lambda}_{\mathbb{S}[p^{n}\boxplus_{d,\lambda}q^{n}]}(s)\xrightarrow{n\to\infty}\mathcal{R}_{\mathbb{S}p}^{\lambda}(s)+\mathcal{R}_{q}^{\lambda}(s)=\mathcal{R}_{\mu_{\mathbb{S}p}\boxplus_{\lambda}\mu_{\mathbb{S}q}}^{\lambda}(s)$

∎

We recall the main theorem that we proved in [1], which reads

######  Theorem 6.5.

For degree $d$ polynomials $p$ and $q$ having only nonnegative real roots and
$s>0$:

$\mathcal{R}_{\mathbb{S}[p\boxplus_{d,\lambda}q]}^{\lambda}(s)\leq\mathcal{R}_{\mathbb{S}p}^{\lambda}(s)+\mathcal{R}_{\mathbb{S}q}^{\lambda}(s)$
(28)

######  Corollary 6.6 (tightness of the maxroot inequality).

The inequality 28 is tight as the dimension $d$ goes to infinity.

###### Proof.

Combine additivity:

$\mathcal{R}^{d,\lambda}_{\mathbb{S}[p\boxplus_{d,\lambda}q]}(s)=\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)+\mathcal{R}^{d,\lambda}_{\mathbb{S}q}(s)$

and convergence:

$\displaystyle\mathcal{R}^{d,\lambda}_{\mathbb{S}[p\boxplus_{d,\lambda}q]}(s)$
$\displaystyle\xrightarrow{d\to\infty}\mathcal{R}_{\mu_{\mathbb{S}p}\boxplus_{\lambda}\mu_{\mathbb{S}q}}^{\lambda}(s)$
$\displaystyle\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)$
$\displaystyle\xrightarrow{d\to\infty}\mathcal{R}_{\mathbb{S}p}^{\lambda}(s)$
$\displaystyle\mathcal{R}^{d,\lambda}_{\mathbb{S}q}(s)$
$\displaystyle\xrightarrow{d\to\infty}\mathcal{R}_{\mathbb{S}q}^{\lambda}(s)$

$d$ going to infinity, we get the tightness. ∎

##  7 Limit theorems and special polynomials

In this section, we try to show how we can actually consider even realrooted
polynomials as equivalent of independent random variables in our framework.

######  Definition 7.1.

For a realrooted polynomial $r$ with roots $r_{i}$, define its expectation and
variance as

$\displaystyle\mathbb{E}(r)$ $\displaystyle:=\frac{1}{d}\sum_{i}r_{i}$
$\displaystyle Var(r)$
$\displaystyle:=\frac{1}{d}\sum_{i}(r_{i}-\mathbb{E}(r))^{2}$

######  Lemma 7.2.

Exactly like for a sum of independent random variables, we get

  * •

$\mathbb{E}\big{[}\mathbb{S}[p\boxplus_{d,\lambda}q]\big{]}=\mathbb{E}[\mathbb{S}p]+\mathbb{E}[\mathbb{S}q]=0$
,

  * •

$Var\big{[}\mathbb{S}[p\boxplus_{d,\lambda}q]\big{]}=Var[\mathbb{S}p]+Var[\mathbb{S}q]$

###### Proof.

it follows by inspection of the two first coefficients (associated to $1$ and
$x$) in the finite $R$-transform. ∎

We study in this section aymptotics related to the convolution. The only
single-rooted symmetric polynomial is the zero polynomial $x^{d}$. The law of
large numbers in probability states that for random variables that have the
same mean and are independent, then the average converges to this mean, so the
constant random variable. We adapt it in this context: we add polynomials
”freely” and renormalize the symmetrization of their sum. The result is that
it converges to the zero polynomial. Let’s first introduce a renormalization
operator and establish a few basic properties verified by $R$\- transforms.

######  Definition 7.3 (Normalization operator on roots).

For $p$ of degree $d$, and $\alpha>0$, define
$\mathit{R_{\alpha}}(p):=\alpha^{-d}p(\alpha x)$.

######  Lemma 7.4.

For all $\alpha>0$,
$\mathit{R}_{\alpha}\mathbb{S}p=\mathbb{S}[\mathit{R}_{\alpha^{2}}p]$. Also,
$\mathit{R}_{\alpha}[p\boxplus_{d,\lambda}q]=\mathit{R}_{\alpha}p\boxplus_{d,\lambda}\mathit{R}_{\alpha}q$.

###### Proof.

Straightforward using the definitions. ∎

######  Lemma 7.5.

If $\mathcal{R}^{d,\lambda}_{\mathbb{S}p}=\sum_{k=1}^{d}a^{d}_{k}s^{k}$, then
$\mathcal{R}^{d,\lambda}_{\mathbb{S}\mathit{R}_{\alpha}p}=\sum_{k=1}^{d}a^{d}_{k}\alpha^{k}s^{k}$.

###### Proof.

Using the fact that $\mathit{R}_{\alpha}$ multiplies the roots by $\alpha$ and
therefore rescales the coefficients of $p$, $p_{i}$, by $\alpha^{i}$, we get

$\mathbb{E}\Big{(}[T^{(m,d)}_{\mathbb{S}\mathit{R}_{\alpha}p}]^{i}\Big{)}=\mathbb{E}\Big{(}[T^{(m,d)}_{\mathbb{S}p}]^{i}\Big{)}\alpha^{i}$

We then get expanding and using linearity,

$\mathcal{R}^{d,\lambda}_{\mathbb{S}R_{\alpha}p}=\sum_{k=1}^{d}a^{d}_{k}\alpha^{k}s^{k}$

∎

######  Lemma 7.6.

Fix $d$. Then if $p_{1},p_{2}$,… are a sequence of degree $d$ polynomials with
real nonnegative roots and means uniformly bounded, then all the coefficients
are uniformly bounded too, and the coefficients of the
$\mathcal{R}^{d,\lambda}_{\mathbb{S}p_{i}}$ as well.

###### Proof.

The first part follows from the $k$-norms of roots being all equivalent and
therefore bounded by the $1$-norm (the parameter depending only on the
dimensions), when the size $d$ is fix (and $k\in[|1,d|]$), and the
coefficients of the polynomials being combinations of the $k$-norms of the
roots. As for the second part, notice that if $d$ and $m$ are fix, then the
coefficients of $\mathcal{R}^{d,\lambda}_{\mathbb{S}p_{i}}$ are polynomial in
the coefficients of the polynomial $p_{i}$, that are uniformly bounded as we
just saw, so the uniform boundedness of the coefficients of
$\mathcal{R}^{d,\lambda}_{\mathbb{S}p_{i}}$ follows. ∎

######  Proposition 7.7 (Law of large numbers).

Let $p_{1},p_{2}$,… be a sequence of degree $d$ polynomials with real
nonnegative roots and means uniformly bounded by $\sigma^{2}$,

$\displaystyle p_{i}:=\prod_{j}(x-r_{i,j}^{2})$
$\displaystyle\frac{1}{d}\sum_{j}r_{i,j}^{2}\leq\sigma^{2}$

then

$\lim_{N\to\infty}\mathit{R_{1/{N}}}(\mathbb{S}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}])(x)\approx
x^{d}$

###### Proof.

Let’s write for all $i$,

$\mathcal{R}^{d,\lambda}_{\mathbb{S}p_{i}}:=\sum_{k=1}^{d}a^{(d,i)}_{k}s^{k}$

Using successively Lemma 7.5 and Lemma 7.4,

$\displaystyle\mathcal{R}^{d,\lambda}_{\mathit{R}_{1/{N}}(\mathbb{S}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}])}$
$\displaystyle=\mathcal{R}^{d,\lambda}_{\mathbb{S}\mathit{R}_{1/{N^{2}}}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}]}$
$\displaystyle=\mathcal{R}^{d,\lambda}_{\mathbb{S}\big{[}(\mathit{R}_{1/{N^{2}}}p_{1})\boxplus_{d,\lambda}...\boxplus_{d,\lambda}(\mathit{R}_{1/{N^{2}}}p_{N})\big{]}}$
$\displaystyle=\sum_{i=1}^{N}\mathcal{R}^{d,\lambda}_{\mathbb{S}\mathit{R}_{1/{N^{2}}}p_{i}}$
$\displaystyle=\sum_{i=1}^{N}\sum_{k=1}^{d}a^{(d,i)}_{k}\frac{1}{N^{2k}}s^{k}$

The uniform boundedness of the $\big{(}a^{(d,i)}_{k}\big{)}_{k,i}$ gives us
$|a^{(d,i)}_{k}|\leq K$ for all $i\leq N$,$k\leq d$ and $K>0$, and
$\frac{|\sum_{i=1}^{N}a^{(d,i)}_{k}|}{N^{2}}\leq\frac{K}{N}$. Therefore

$\lim_{N\to\infty}\mathcal{R}^{d,\lambda}_{\mathit{R}_{1/{N}}(\mathbb{S}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}])}=0$

Using the inversion formula through which one recovers the polynomial from its
$R$-transform, it means, as the polynomial with $0$ $R$-transform is trivially
$x^{d}$, that

$\lim_{N\to\infty}\mathit{R}_{1/{N}}(\mathbb{S}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}])=x^{d}$

∎

For two polynomials $p$ and $q$, we will use $p\approx q$ to express that they
have the same roots but not the same leading coefficient.

######  Lemma 7.8 (Laguerre $R$ and $T$-transforms).

For p of degree $d$ with nonnegative real roots, and $\sigma^{2}>0$, the
following are equivalent:

  1. 1.

$T^{(m,d)}_{\mathbb{S}p}=\sigma^{2}$, so the $T$-transform is constant (it has
to be positive).

  2. 2.

$\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)=m\sigma^{2}s$, so only the first
nontrivial cumulant is nonzero.

  3. 3.

$p\approx L_{d}^{(m-d)}(\frac{x}{\sigma^{2}})$, that is $p$ is up to scaling a
generalized Laguerre polynomial of parameter $m-d$.

###### Proof.

Assume that $T^{(m,d)}_{\mathbb{S}p}=\sigma^{2}$. Then

$\displaystyle\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)$
$\displaystyle\equiv\frac{-1}{d}s\frac{d}{ds}\log\big{(}\mathbb{E}\big{[}e^{-T^{(m,d)}_{\mathbb{S}p}smd}\big{]}\big{)}\mod[s^{d+1}]$
$\displaystyle\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)$
$\displaystyle\equiv\frac{-1}{d}s\frac{d}{ds}\log\big{(}\big{[}e^{-\sigma^{2}smd}\big{]}\big{)}\mod[s^{d+1}]$
$\displaystyle\mathcal{R}^{d,\lambda}_{\mathbb{S}p}(s)$
$\displaystyle=\sigma^{2}ms$

We can use the bijection between transforms to conclude that the two first
points are equivalent. We get the equivalence with the third point as

$\displaystyle e^{-\sigma^{2}\partial{x}\partial_{y}}\\{y^{m}x^{d}\\}$
$\displaystyle=\sigma^{2d}y^{m-d}\sum_{i=0}^{d}\big{(}\frac{-xy}{\sigma^{2}}\big{)}^{d-i}\frac{m!d!}{i!(m-i)!(d-i)!}$
(29)
$\displaystyle=d!\sigma^{2d}y^{m-d}\sum_{i=0}^{d}\big{(}\frac{-xy}{\sigma^{2}}\big{)}^{i}\frac{m!}{i!(m-d+i)!(d-i)!}$
(30)
$\displaystyle=d!\sigma^{2d}y^{m-d}\sum_{i=0}^{d}\big{(}\frac{-xy}{\sigma^{2}}\big{)}^{i}\frac{1}{i!}{m\choose
d-i}$ (31)
$\displaystyle=d!\sigma^{2d}y^{m-d}L_{d}^{(m-d)}(\frac{xy}{\sigma^{2}})$ (32)

∎

######  Corollary 7.9 (Divisibility of the Laguerre polynomials).

$L_{d}^{(m-d)}(\frac{x}{\sigma^{2}+\tau^{2}})\approx
L_{d}^{(m-d)}(\frac{x}{\sigma^{2}})\boxplus_{d,\lambda}L_{d}^{(m-d)}(\frac{x}{\tau^{2}})$

The central limit theorem in probability states that for independent random
variables that have zero mean and constant variance, then the square-root
average converges to a Gaussian random variable that has the same variance. We
adapt it in this context: we add polynomials freely, and renormalize the roots
by $\sqrt{N}$ . The result is that it converges to a generalized Laguerre
polynomial, the equivalent of a Gaussian random variable.

######  Proposition 7.10 (Central limit theorem).

Let $p_{1},p_{2}$,… be a sequence of degree $d$ with real nonnegative roots
and same mean $\sigma^{2}$, with

$\displaystyle p_{i}=\prod_{j}(x-r_{i,j}^{2})$
$\displaystyle\frac{1}{d}\sum_{j}r_{i,j}^{2}=\sigma^{2}$

then

$\lim_{N\to\infty}\mathit{R_{1/\sqrt{N}}}(\mathbb{S}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}])(x)\approx
L_{d}^{(m-d)}(\frac{x^{2}m}{\sigma^{2}})$

###### Proof.

Let’s write again for all $i$,

$\mathcal{R}^{d,\lambda}_{\mathbb{S}p_{i}}:=\sum_{k=1}^{d}a^{(d,i)}_{k}s^{k}$

Using successively Lemma 7.5 and Lemma 7.4,

$\displaystyle\mathcal{R}^{d,\lambda}_{\mathit{R}_{1/{\sqrt{N}}}(\mathbb{S}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}])}$
$\displaystyle=\mathcal{R}^{d,\lambda}_{\mathbb{S}\mathit{R}_{1/{N}}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}]}$
$\displaystyle=\mathcal{R}^{d,\lambda}_{\mathbb{S}\big{[}(\mathit{R}_{1/{N}}p_{1})\boxplus_{d,\lambda}...\boxplus_{d,\lambda}(\mathit{R}_{1/{N}}p_{N})\big{]}}$
$\displaystyle=\sum_{i=1}^{N}\mathcal{R}^{d,\lambda}_{\mathbb{S}\mathit{R}_{1/{N}}p_{i}}$
$\displaystyle=\sum_{i=1}^{N}\sum_{k=1}^{d}a^{(d,i)}_{k}\frac{1}{N^{k}}s^{k}$

The uniform boundedness of the $\big{(}a^{(d,i)}_{k}\big{)}_{k,i}$ gives us
$|a^{(d,i)}_{k}|\leq K$ for all $i\leq N$, $k\leq d$ and $K>0$. We have for
$k\geq 2$,

$\frac{|\sum_{i=1}^{N}a^{(d,i)}_{k}|}{N^{k}}\leq\frac{K}{N}$

Therefore

$\lim_{N\to\infty}\sum_{i=1}^{N}\sum_{k=2}^{d}a^{(d,i)}_{k}\frac{1}{N^{k}}s^{k}=0$

Notice that $a^{(d,i)}_{1}=m\sigma^{2}$ and as a consequence,

$\sum_{i=1}^{N}a^{(d,i)}_{1}\frac{1}{N}s=m\sigma^{2}s$

We therefore get

$\lim_{N\to\infty}\mathcal{R}^{d,\lambda}_{\mathit{R}_{1/{N}}(\mathbb{S}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}])}=m\sigma^{2}s$

Using the inversion formula through which one recovers the polynomial from its
$R$-transform, it means

$\lim_{N\to\infty}\mathit{R}_{1/{N}}(\mathbb{S}[p_{1}\boxplus_{d,\lambda}...\boxplus_{d,\lambda}p_{N}])\approx
L_{d}^{(m-d)}(\frac{x^{2}m}{\sigma^{2}})$

∎

## Acknowledgements

I would like to thank my advisor Adam Marcus for suggesting this problem to me
and taking time to discuss the details with me.

## References

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