Rank One Non-Hermitian Perturbations of Hermitian $\beta$-Ensembles of Random Matrices

Abstract

We provide a tridiagonal matrix model and compute the joint eigenvalue density of a rank one non-Hermitian perturbation of a random matrix from the Gaussian or Laguerre $\beta$-ensemble.

1 Introduction

The energy Hamiltonian of a closed quantum system is usually modelled by a Hermitian random matrix H. The Hamiltonian of this system after coupling it to the outer world via s open channels is modelled by the so-called effective Hamiltonian¹

$$\begin{aligned} H_{eff}=H + i \Gamma , \end{aligned}$$

(1.1)

where $\Gamma \ge {\varvec{0}}$ is a rank s positive semi-definite Hermitian matrix that is independent of H. The eigenvalues of $H_{eff}$ are the mathematical model for the resonances, which are the long-lived decaying states of our open quantum system.

In this paper we are concerned with the exact joint distribution of these eigenvalues when there is one open channel $(s=1)$, and H is a Gaussian or Laguerre (Wishart) orthogonal/unitary/symplectic random matrix. $\Gamma$ may be deterministic or random with a given distribution function. We obtain tridiagonal models (in the spirit of Dumitriu–Edelman [2]) and compute the joint eigenvalue distribution for any $\beta >0$, not merely $\beta =1,2,4$ (Theorems 3 and 4).

The joint eigenvalue law for non-Hermitian perturbations of Laguerre ensembles has not been addressed in the literature before (however, see [11] for a related topic), while the joint eigenvalue law for non-Hermitian perturbations of Gaussian ensembles has been studied in the physics literature by numerous authors: Ullah [19] (for the case $\beta =1$), Sokolov–Zelevinsky [15] ($\beta =1$), Stöckmann–Šeba [17] ($\beta =1,2$), Fyodorov–Khoruzhenko [5] ($\beta =2$). The present paper provides a rigorous derivation of this law which works for any $\beta >0$ and for any choice of $\Gamma$—deterministic or random. More importantly, our approach can be applied to other models, e.g., perturbations of Laguerre $\beta$-ensembles (done in this paper); of chiral Gaussian $\beta$-ensembles; multiplicative perturbations of Gaussian and Laguerre $\beta$-ensembles (to be explored in a forthcoming paper). We also expect that the tridiagonal matrix models proposed here will be useful for establishing asymptotic properties of these “weakly non-Hermitian” ensembles. Finally, we note that our methods can provide matrix models (namely, block Jacobi matrices with independent (matrix-valued) Jacobi coefficients) for higher order perturbations $s\ge 2$ as well, which could prove to be useful for computing their eigenvalue density (for the case $\beta =2$, $s\ge 2$, Fyodorov–Khoruzhenko [5] provide another approach). The solution to this matrix-valued eigenvalue problem is currently beyond our reach. We leave this as a challenging open problem.

The asymptotic analysis of the weakly non-Hermitian ensembles are of high interest in the mathematics and physics literature and have been studied in [3, 4, 6, 14], see also [11, 12]. The numerous physical applications of such random matrices can be found in the review papers [6, 7, 10].

The important cornerstones of our proofs are the Dumitriu–Edelman Hermitian matrix models [2], and the Arlinskiĭ–Tsekanovskiĭ result [1] on the spectral analysis of (deterministic) Jacobi matrices.

2 Preliminaries

2.1 Gaussian and Laguerre Ensembles

Definition 1

Denote by $N (0,\sigma )$, $N (0,\sigma \mathbf {I}_2)$, and $N(0,\sigma \mathbf {I}_4)$ the real, complex, and quaternionic normal random variables (r.v.) with variance $\beta \sigma ^2$ $(\beta =1,2,4$, respectively).

Denote by $\chi ^2_k$ $(k>0)$ a real r.v. with p.d.f. $\tfrac{1}{2^{k/2}\Gamma (k/2)} x^{k/2-1}e^{-x/2}$. Denote by $\chi _k$ $(k>0)$ a square root of a $\chi ^2_k$ r.v., and $\tilde{\chi }_k$ $(k>0)$ to be $\tfrac{1}{\sqrt{2}} \chi _k$.

Definition 2

Let Y be an $n\times n$ matrix with independent identically distributed (i.i.d.) entries chosen from N(0, 1), $N(0, \mathbf {I}_2)$, or $N(0, \mathbf {I}_4)$. Then we say that $X=\tfrac{1}{2} (Y + Y^*)$ belongs to the Gaussian orthogonal/unitary/symplectic ensemble, respectively. We denote it by $GOE_n$, $GUE_n$, $GSE_n$, respectively.

Definition 3

Let Y be an $m\times n$ matrix with i.i.d. entries chosen from N(0, 1), $N(0, \mathbf {I}_2)$, or $N(0, \mathbf {I}_4)$. Then we say that the $n\times n$ matrix $X=Y^* Y$ belongs to the Laguerre (Wishart) orthogonal/unitary/symplectic ensemble, respectively. We denote it by $LOE_{(m,n)}$, $LUE_{(m,n)}$, $LSE_{(m,n)}$, respectively.

2.2 Tridiagonalization of Hermitian Matrices

Let H be an $n \times n$ Hermitian matrix. Denote $\mathbf {e}_j$ to be the j-th standard vector in ${\mathbb {C}}^n$, that is, having 1 in its j-th entry and 0 everywhere else. Let $\langle \mathbf {x}, \mathbf {y} \rangle := \mathbf {x}^* \mathbf {y}$, the usual inner product in ${\mathbb {C}}^n$. Let us apply the Gram–Schmidt orthogonalization procedure in ${\mathbb {C}}^n$ to the sequence of vectors $\mathbf {e}_1, H \mathbf {e}_1, H^2 \mathbf {e}_1,\ldots , H^{k-1} \mathbf {e}_1$, where $k= \dim \mathrm{span}\{ H^j \mathbf {e}_1: j\ge 0 \}$. Note that $1\le k\le n$. After normalization we obtain an orthonormal sequence of vectors $\mathbf {v}_1,\ldots ,\mathbf {v}_k$ in ${\mathbb {C}}^n$. If $k< n$, then we choose an arbitrary unit vector $\mathbf {v}_{k+1}$ in ${\mathbb {C}}^n \ominus \mathrm{span}\{ \mathbf {v}_1,\ldots ,\mathbf {v}_k \}$ and repeat the procedure but with $\mathbf {v}_{k+1}$ instead of $\mathbf {e}_1$. By repeating this procedure finitely many times more if necessary and combining all the resulting vectors together, we obtain an orthonormal basis $\{\mathbf {v}_j\}_{j=1}^n$ of ${\mathbb {C}}^n$.

Standard arguments (see, e.g., [16, Sect 1.3]) show that the matrix of H in the basis $\{\mathbf {v}_j\}_{j=1}^n$ is tridiagonal. In other words, if we form unitary matrix S with $\{\mathbf {v}_j\}_{j=1}^n$ as its columns, then $S^* H S = {\mathcal J}$, where

$$\begin{aligned} {\mathcal J}= S^* H S = \left( \begin{array}{ccccc} b_1&{}a_1&{}0&{} &{}\\ a_1&{}b_2&{}a_2&{}\ddots &{}\\ 0&{}a_2&{}b_3&{}\ddots &{} 0 \\ &{}\ddots &{}\ddots &{}\ddots &{} a_{n-1} \\ &{} &{} 0 &{} a_{n-1} &{} b_n \end{array}\right) , \quad a_j \ge 0, b_n\in {\mathbb {R}}. \end{aligned}$$

(2.1)

We call matrices of the form (2.1) Jacobi, and the coefficients $\{a_j,b_j\}$—their Jacobi coefficients. For a future reference, observe that

$$\begin{aligned} S \mathbf {e}_1 = S^* \mathbf {e}_1 = \mathbf {e}_1 \end{aligned}$$

(2.2)

since $\mathbf {v}_1 = \mathbf {e}_1$ in the Gram–Schmidt procedure. Note that in the tridiagonalization procedure above, if $\dim \mathrm{span}\{ H^j \mathbf {e}_1: j\ge 0 \} = k < n$, then $a_j > 0$ for $1\le j\le k-1$, and $a_k = 0$, i.e., ${\mathcal J}$ becomes a direct sum of Jacobi matrices of smaller sizes.

2.3 Matrix Models for Gaussian and Laguerre Ensembles

Now let us apply the tridiagonalization procedure from the previous section to a random matrix from a Gaussian or a Laguerre ensemble. This is the idea of Dumitriu–Edelman [2] (see also Trotter’s [18]).

If H is from $GOE_n$, $GUE_n$, or $GSE_n$, then $\mathbf {e}_1$ is a cyclic vector for H with probability 1. Therefore we obtain (2.1) with $a_j>0$ for all $1\le j\le n-1$.

The same is true for a random matrix H from $LOE_{(m,n)}$, $LUE_{(m,n)}$, or $LSE_{(m,n)}$, but only if $m \ge n$. If $m<n$, then with probability 1, $\dim \mathrm{span}\{ H^j \mathbf {e}_1: j\ge 0 \} = m +1 \le n$, and ${\mathbb {C}}^n \ominus \mathrm{span}\{ H^j \mathbf {e}_1: j\ge 0 \} \subseteq \ker H$, so that the resulting Jacobi matrix (2.1) that we obtain has $a_{m+1} = \cdots = a_{n-1} = 0, b_{m+2} = \cdots = b_n=0$. In other words, we have that ${\mathcal J}$ is the direct sum of an $(m+1)\times (m+1)$ Jacobi matrix and the $(n-m-1)\times (n-m-1)$ zero matrix. The proof of this case can be done by following the Dumitriu–Edelman [2] arguments.

Lemma 1

(Dumitriu–Edelman [2]) Let H be a random matrix taken from $GOE_n$, $GUE_n$, or $GSE_n$ ensemble. There exists a (random) unitary matrix S satisfying (2.2) such that $S H S^* = {\mathcal J}$ is tridiagonal (2.1), where

$$\begin{aligned} a_j&\sim \tilde{\chi }_{\beta (n-j)},&1\le j\le n-1, \\ b_j&\sim N(0,1),&1\le j\le n, \end{aligned}$$

where $\beta =1,2,4$ for $GOE_n$, $GUE_n$, $GSE_n$, respectively.

Lemma 2

(Dumitriu–Edelman [2]) Let H be a random matrix taken from $LOE_{(m,n)}$, $LUE_{(m,n)}$, or $LSE_{(m,n)}$ ensemble. There exists a (random) unitary matrix S satisfying (2.2) such that $S H S^* = {\mathcal J}= B^* B$ is tridiagonal (2.1), where

$$\begin{aligned} B = \left( \begin{array}{ccccc} x_1&{}y_1&{}0&{} &{}\\ 0&{}x_2&{}y_2&{}\ddots &{}\\ 0&{}0&{}x_3&{}\ddots &{} 0 \\ &{}\ddots &{}\ddots &{}\ddots &{} y_{n-1} \\ &{} &{} 0 &{} 0 &{} x_n \end{array}\right) , \quad \text{ with } \end{aligned}$$

(2.3)

(i) If $m \ge n$:

$$\begin{aligned} x_j&\sim \chi _{\beta (m-j+1)},&1\le j\le n, \\ y_j&\sim \chi _{\beta (n-j)},&1\le j\le n-1; \end{aligned}$$

(ii) If $m \le n-1$:

$$\begin{aligned} x_j&\sim {\left\{ \begin{array}{ll} \chi _{\beta (m-j+1)}, &{} \text{ if } 1\le j\le m, \\ 0, &{} \text{ if } m+1\le j\le n, \end{array}\right. } \\ y_j&\sim {\left\{ \begin{array}{ll} \chi _{\beta (n-j)}, &{} \text{ if } 1\le j\le m, \\ 0, &{} \text{ if } m+1\le j\le n-1 ; \end{array}\right. } \end{aligned}$$

where $\beta =1,2,4$ for $LOE_{(m,n)}$, $LUE_{(m,n)}$, $LSE_{(m,n)}$, respectively.

Remarks

• 1. For $GSE_n$ and $LSE_{(m,n)}$ every entry is quaternionic, so all the instances of ${\mathbb {C}}$ in the arguments above should be replaced with the algebra of quaternions. The resulting coefficients $a_j$, $b_j$, $x_j$, $y_j$ in Lemmas 1 and 2 are quaternionic too, but with the $\mathsf {i}$, $\mathsf {j}$, and $\mathsf {k}$ parts equal to zero.

• 2. It is worth reminding the reader that the random matrix S in Lemmas 1 and 2 is statistically independent of ${\mathcal J}$.

2.4 $\beta$-Ensembles

The tridiagonal matrix ensembles from Lemmas 1 and 2 make sense for any $\beta >0$, not merely for $\beta =1,2,4$. They are called the Gaussian $\beta$-ensemble $G\beta E_n$ and the Laguerre $\beta$-ensemble $L\beta E_{(m,n)}$, respectively.

2.5 Spectral Measures of Gaussian and Laguerre $\beta$-Ensembles

By the Riesz representation theorem, for any Hermitian matrix H there exists a probability measure $\mu$ (called the spectral measure) satisfying

$$\begin{aligned} \langle \mathbf {e}_1, H^k \mathbf {e}_1 \rangle = \int _{\mathbb {R}}x^k d\mu (x), \quad \text{ for } \text{ all } \quad k\ge 0. \end{aligned}$$

(2.4)

In fact, any Hermitian can be unitarily diagonalized, so that we can write $H = U D U^*$, where D is the diagonal matrix with eigenvalues $\lambda _1,\ldots ,\lambda _n$ of H on the diagonal, and the columns $\mathbf u _1,\ldots ,\mathbf u _n$ of U are the corresponding orthonormal eigenvectors of H. This easily implies (2.4) with

$$\begin{aligned} \mu (x) = \sum _{j=1}^n w_j \delta _{\lambda _j}, \quad \text{ where } \quad w_j = |\langle \mathbf {e}_1, \mathbf {u}_j \rangle |^2. \end{aligned}$$

(2.5)

Here $\delta _\lambda$ is the Dirac measure at $\lambda$. The support of $\mu$ consists of $\le n$ points.

As our matrix H is random, its spectral measure is random too. The joint law of $w_j$’s and $\lambda _j$’s in (2.5) will be referred to as the law of the spectral measure of H.

Because of (2.2), the laws of the spectral measures of H and of its Jacobi form ${\mathcal J}$ coincide, that is, H and ${\mathcal J}$ have identically distributed eigenvalues $\lambda _j$’s and eigenweights $w_j$’s. In particular, laws of the spectral measures of $GOE_n$ and $G \beta E_n$ with $\beta =1$ coincide; laws of the spectral measures of $GUE_n$ and $G \beta E_n$ with $\beta =2$ coincide; laws of the quaternion-valued spectral measures of $GSE_n$ and $G \beta E_n$ with $\beta =4$ (viewed as a matrix with purely-real quaternion entries) coincide. The analogous statements hold true for the Laguerre case.

Laws of the spectral measures for $G\beta E_n$ and $L\beta E_{(m,n)}$ with $m \ge n$ have been computed in [2], see Lemmas 3 and 4 below. We also need the spectral measure of $L\beta E_{(m,n)}$ when $m < n$, which we compute in Proposition 1 below.

Lemma 3

(Dumitriu–Edelman [2]) For any $\beta >0$, the spectral measure of a random matrix from the $G \beta E_n$-ensemble is (2.5) with the joint distribution

$$\begin{aligned} \tfrac{1}{g_{\beta ,n}} \prod _{j=1}^n e^{-\lambda _j^2/2} \prod _{1\le j<k \le n} |\lambda _j - \lambda _k|^\beta d\lambda _1 \ldots d\lambda _n \times \tfrac{1}{c_{\beta ,n}} \prod _{j=1}^n w_j^{\beta /2 -1 }dw_1 \ldots dw_{n-1}, \end{aligned}$$

(2.6)

where

$$\begin{aligned}&\sum _{j=1}^n w_j =1; \quad w_j > 0, \quad 1 \le j \le n; \quad \lambda _j \in {\mathbb {R}}, \end{aligned}$$

(2.7)

$$\begin{aligned}&g_{\beta ,n} = (2\pi )^{n/2} \prod _{j=1}^{n} \frac{\Gamma (1+\beta j/2)}{\Gamma (1+\beta /2)}, \quad c_{\beta ,n} = \frac{\Gamma (\beta /2)^n}{\Gamma (\beta n/2)}. \end{aligned}$$

(2.8)

Lemma 4

(Dumitriu–Edelman [2]) For any $m \ge n$ and $\beta >0$, the spectral measure of a random matrix from the $L \beta E_{(m,n)}$-ensemble is (2.5) with the joint distribution

$$\begin{aligned}&\tfrac{1}{h_{\beta ,n,a}} \prod _{j=1}^n \lambda _j^{\beta a/2} e^{-\lambda _j/2} \prod _{1\le j<k \le n} |\lambda _j - \lambda _k|^\beta d\lambda _1 \ldots d\lambda _n \nonumber \\&\quad \times \, \Gamma (\beta n/2) \prod _{j=1}^n \frac{w_j^{\beta /2 -1 }}{\Gamma (\beta /2)} dw_1 \ldots dw_{n-1}, \end{aligned}$$

(2.9)

where $a=m-n+1-2/ \beta$ and

$$\begin{aligned}&\sum _{j=1}^n w_j =1; \quad w_j> 0, \quad 1 \le j \le n; \quad \lambda _j >0, \end{aligned}$$

(2.10)

$$\begin{aligned}&h_{\beta ,n,a} = 2^{n(a\beta /2 +1+(n-1)\beta /2)} \prod _{j=1}^{n} \frac{\Gamma (1+\beta j /2) \Gamma (1+\beta a/2+\beta (j-1)/2)}{\Gamma (1+\beta /2)}, \end{aligned}$$

(2.11)

Proposition 1

If $m \le n-1$ and $\beta >0$, the spectral measure of a random matrix from the $L \beta E_{(m,n)}$ is

$$\begin{aligned} \mu (x) = w_0 \delta _0 + \sum _{j=1}^m w_j \delta _{\lambda _j}, \end{aligned}$$

(2.12)

with the joint distribution

$$\begin{aligned}&\tfrac{1}{h_{\beta ,m,a}} \prod _{j=1}^m \lambda _j^{\beta a/2} e^{-\lambda _j/2} \prod _{1\le j<k \le m} |\lambda _j - \lambda _k|^\beta d\lambda _1 \ldots d\lambda _m \nonumber \\&\quad \times \, \frac{w_0^{\beta (n-m) /2 -1}}{\Gamma (\beta (n-m)/2)}\times \Gamma (\beta n/2) \prod _{j=1}^m \frac{w_j^{\beta /2 -1 }}{\Gamma (\beta /2)} dw_1 \ldots dw_{m}, \end{aligned}$$

(2.13)

where $a=n-m+1-2/ \beta$; $h_{\beta ,m,a}$ is as in (2.11); and

$$\begin{aligned} \sum _{j=0}^m w_j =1; \quad w_j> 0, \quad 0 \le j \le m; \quad \lambda _j >0. \end{aligned}$$

(2.14)

Let us denote the normalization constant for $w_j$’s as

$$\begin{aligned} d_{\beta ,m,n} = \frac{\Gamma (\beta (n-m)/2) \Gamma (\beta /2)^m}{\Gamma (\beta n/2)}. \end{aligned}$$

(2.15)

Proof

Let us first deal with $\beta =1$ case. The distribution of the eigenvalues of a matrix H from $LOE_{(m,n)}$ is well-known. Let its eigenvalues be $\lambda _1>\cdots>\lambda _m>0=0=\cdots =0$ ($n-m$ zeros). Now choose an orthonormal system of (real) eigenvectors $\mathbf {u}_1,\ldots ,\mathbf {u}_n$ of H corresponding to these eigenvalues, respectively. We pick each $\mathbf {u}_j$ at random uniformly from the set of all possible choices. Since for any $n\times n$ orthogonal matrix O, the matrix $O^T H O$ also belongs to $LOE_{(m,n)}$, we can see that: $\mathbf {u}_1$ is uniformly distributed on the unit sphere $\{\mathbf {u}\in {\mathbb {R}}^n: ||\mathbf {u}|| = 1 \}$; and for any $1\le j \le n$, the vector $\mathbf {u}_j$ conditionally on $\mathbf {u}_1, \ldots , \mathbf {u}_{j-1}$ is uniformly distributed on the subset of this unit sphere orthogonal to $\mathbf {u}_1, \ldots , \mathbf {u}_{j-1}$. So the matrix consisting of $\mathbf {u}_1,\ldots ,\mathbf {u}_n$ as its columns is a Haar distributed orthogonal matrix (see, e.g., [9, Prop. 2.2(a)]). Then its first row $(v_1,\ldots ,v_n)$ is distributed uniformly on the unit sphere $\{\mathbf {u}\in {\mathbb {R}}^n: ||\mathbf {u}|| = 1 \}$. Now recalling (2.5), we obtain that $w_j = v_j^2$, $1\le j \le m$, and $w_0 = v_{m+1}^2+\cdots + v_n^2$. Now one can apply arguments from the proof of [8, Cor. A.2] (note that $dw_j = 2w_j^{1/2} dv_j$) to see that the joint distribution of $w_1,\ldots ,w_m$ is proportional to $w_0^{(n-m-2)/2} \prod _{j=1}^m w_j^{-1/2} dw_1 \ldots dw_m$.

This allows us to compute the Jacobian for the change of variables from $\{x_j,y_j\}_{j=1}^m$ in (2.3) to $\{\lambda _j,w_j\}_{j=1}^m$. Why is this change of variables bijective? By Favard’s theorem (see, e.g., [16, Thms. 1.3.2–1.3.3]), there is 1-to-1 correspondence between all $(m+1) \times (m+1)$ Jacobi matrices (2.1) with $a_j>0$ ($1\le j\le m$) and all probability measures supported on $m+1$ distinct points. This means there is 1-to-1 correspondence between all positive semi-definite $(m+1) \times (m+1)$ Jacobi matrices ${\mathcal J}$ with $a_j>0$ ($1\le j\le m$), $\det {\mathcal J}=0$ and all probability measures supported on $m+1$ points of the form (2.12), (2.14). By semi-definiteness, any such ${\mathcal J}$ can be Cholesky factorized ${\mathcal J}= B^* B$ with B upper-triangular with non-negative entries on the diagonal. Since ${\mathcal J}$ is tridiagonal, this $(m+1)\times (m+1)$ matrix B must be two-diagonal as in (2.3) with $x_j\ge 0$, $1\le j \le m+1$. Since $\det {\mathcal J}=0$, we must have that $x_j=0$ for at least one $1\le j\le m+1$. But since all $a_j>0$, we obtain that $x_{m+1}=0$, $x_j> 0$ for $1\le j\le m$, and $y_j>0$, $1\le j\le m$. Conversely, any $(m+1)\times (m+1)$ matrix B (2.3) with $x_j>0, y_j>0$ for $1\le j\le m$ and $x_{m+1}=0$ leads to a positive semi-definite $(m+1) \times (m+1)$ Jacobi matrix ${\mathcal J}$ with $\det {\mathcal J}=0$ and $a_j>0$ ($1\le j\le m$).

Using the matrix model in Lemma 2 (case $m<n$) and the distribution (2.13) that we proved for $\beta =1$, we obtain that the Jacobian is proportional (let us ignore the normalizing constants for now) to

$$\begin{aligned}&\det \frac{\partial (x_1,\ldots ,x_m,y_1,\ldots ,y_m)}{\partial (\lambda _1,\ldots ,\lambda _m,w_1,\ldots ,w_m)} \propto \prod _{j=1}^m x_j^{-m+j} e^{x_j^2/2} \prod _{j=1}^m y_j^{-n+j+1} e^{y_j^2/2} \\&\quad \times \, w_0^{\tfrac{n-m}{2} -1} \prod _{j=1}^m w_j^{-\tfrac{1}{2} } \prod _{j=1}^m \lambda _j^{\tfrac{n-m-1}{2}} e^{-\tfrac{\lambda _j}{2}} \prod _{1\le j<k\le m} |\lambda _j - \lambda _k|. \end{aligned}$$

Now taking the specified in Lemma 2(ii) joint distribution of $\{x_1,\ldots ,x_m,y_1,\ldots ,y_m\}$ for $L \beta E_{(m,n)}$, $m<n$, applying the the above Jacobian, and using the identities from Lemma 5 below, one obtains (2.13), up to a normalization. Finally, note that $h_{\beta ,m,a}$ is the right normalization constant for the eigenvalues in (2.13) by Lemma 4. The normalization constant $d_{\beta ,m,n}$ can be computed by evaluating the Dirichlet integral, see, e.g., [8, Cor. A.4]. $\square$

Lemma 5

The following identities hold:

$$\begin{aligned} \prod _{j=1}^m x_j^{m-j+1} y_j^{m-j+1}&= \prod _{j=0}^m w^{1/2}_j \prod _{1\le j<k\le m} |\lambda _j-\lambda _k| \prod _{j=1}^m \lambda _j , \end{aligned}$$

(2.16)

$$\begin{aligned} \prod _{j=1}^m y_j^2&= w_0 \prod _{j=1}^m \lambda _j. \end{aligned}$$

(2.17)

Proof

(2.16) follows immediately by noting that $x_j y_j = a_j$, $1\le j \le m$, and then applying [2, Lemma 2.7]. Note the clash of notations: their n is our $m+1$, their $\{b_1,\ldots ,b_{n-1}\}$, $\{\lambda _1,\ldots ,\lambda _n\}$, and $\{q_1^2,\ldots ,q_n^2\}$ are ours $\{a_m,\ldots ,a_1\}$, $\{\lambda _1,\ldots ,\lambda _m,0\}$, and $\{w_1,\ldots ,w_m,w_0\}$, respectively. To prove (2.17), we use theory of orthogonal polynomials, see, e.g., [16]. By combining [16, Prop. 3.2.8] and [16, Prop. 2.3.12] we get

$$\begin{aligned} w_0 = -\lim _{z\rightarrow 0} \langle \mathbf {e}_1, z ({\mathcal J}-z)^{-1} \mathbf {e}_1 \rangle = \lim _{z\rightarrow 0} \tfrac{z q_{m+1}(z)}{p_{m+1}(z)} = \tfrac{q_{m+1}(0)}{p'_{m+1}(0)}, \end{aligned}$$

where $p_j$’s and $q_j$’s are the orthonormal polynomials associated to ${\mathcal J}$ of the first and second kind, respectively (in order to define $p_{m+1}$ and $q_{m+1}$ we need $a_{m+1}$ which we take to be an arbitrary positive number). By [16, Thm. 1.2.4], $p_{m+1}(z) = \left( \prod _{j=1}^{m+1} a_j^{-1} \right) \det (z-{\mathcal J})$, so $p'_{m+1}(0) = (-1)^m \prod _{j=1}^{m+1} a_j^{-1} \prod _{j=1}^m \lambda _j$. Using the Wronskian relation [16, Prop. 3.2.3] and $p_{m+1}(0) = 0$ (since 0 is an eigenvalue of ${\mathcal J}$), we obtain $q_{m+1}(0)= 1/(a_{m+1} p_{m}(0))$. Finally, $p_m(z) = \left( \prod _{j=1}^{m} a_j^{-1} \right) \det (z-{\mathcal J}_{m\times m})$, where ${\mathcal J}_{m\times m}$ is the $m\times m$ top left corner of ${\mathcal J}$. Recall that ${\mathcal J}= B^* B$. It is easy to see that ${\mathcal J}_{m\times m} = B_{m\times m} ^* B_{m\times m}$, where $B_{m\times m}$ is the $m\times m$ top left corner of B. Therefore $p_m(0)=(\prod _{j=1}^{m} a_j^{-1}) \det (-B_{m\times m} ^* B_{m\times m}) = (-1)^m (\prod _{j=1}^{m} a_j^{-1}) \prod _{j=1}^m x_j^2$. Combining this all together with $a_j = x_j y_j$, $1\le j \le m$, we obtain (2.17). $\square$

3 Rank One Perturbations: Location of the Eigenvalues

Let us discuss all attainable configurations of eigenvalues of rank one perturbations of (deterministic) Hermitian matrices. Part (i) of the following statement is certainly well-known (see, e.g., [1, 11]), but (ii) and (iii) seem to be new.

For the rest of the paper let ${\mathbb {C}}_+ := \{z\in {\mathbb {C}}: {{\mathrm{Im}}}z>0\}$.

Theorem 1

Let $H_{eff}$ be as in (1.1), where $H=H^*$, $\Gamma \ge \mathbf {0}$, $\mathrm{rank}\,\Gamma = 1$. Choose any $\mathbf {w} \in \mathrm{Ran}\, \Gamma$, $\mathbf {w} \ne 0$, and let $k = \dim {\text {span}}\{H^j \mathbf {w}: j\ge 0\}$. Then:

• (i) $H_{eff}$ has k complex eigenvalues in ${\mathbb {C}}_+$ and $n-k$ real eigenvalues (counted with their algebraic multiplicities).

• (ii) If $H > \mathbf {0}$, then the k complex eigenvalues $\{z_j\}_{j=1}^k$ of $H_{eff}$ belong to the set $\{ (z_j)_{j=1}^k \in ({\mathbb {C}}_+)^k : \sum _{j=1}^k \mathrm{Arg}\,z_j < \tfrac{\pi }{2} \},$ and every such a configuration may occur.

• (iii) If $H \ge \mathbf {0}$ and $\det H = 0$, then the k complex eigenvalues $\{z_j\}_{j=1}^k$ of $H_{eff}$ belong to the set $\{ (z_j)_{j=1}^k \in ({\mathbb {C}}_+)^k : \sum _{j=1}^k \mathrm{Arg}\,z_j \le \tfrac{\pi }{2} \},$ and every such a configuration may occur.

Remark

Using similar ideas one can prove the analogue for the case when H is not positive semi-definite, but has s negative eigenvalues. The k complex eigenvalues (the other $n-k$ being real) of $H_{eff}$ then belong to $\big \{ (z_j)_{j=1}^k \in ({\mathbb {C}}_+)^k : \tfrac{\pi }{2}+\pi (s-1) < \sum _{j=1}^k \mathrm{Arg}\,z_j \le \tfrac{\pi }{2}+\pi s \big \}$, and every such a configuration may occur.

The proof relies on the following uniqueness+existence result for Jacobi matrices. We use n in (i) and $m+1$ in (ii) as the size of our matrices in order to be consistent with what follows later.

Proposition 2

For $l>0$, let

$$\begin{aligned} {\mathcal J}_l = {\mathcal J}+ il I_{1\times 1}, \end{aligned}$$

(3.1)

where $I_{1\times 1}$ is the matrix with (1, 1)-entry equal to 1 and 0 everywhere else.

• (i) Let ${\mathcal J}$ be an $n\times n$ positive definite (real) Jacobi matrix (2.1) with $a_j>0$, $j=1,\ldots ,n-1$. Eigenvalues of ${\mathcal J}_l$, counting algebraic multiplicities, belong to

  $$\begin{aligned} \left\{ (z_j)_{j=1}^n \in ({\mathbb {C}}_+)^n : \sum _{j=1}^n \mathrm{Arg}\,z_j < \tfrac{\pi }{2} \right\} . \end{aligned}$$

  (3.2)

  Moreover, for every configuration of n points from (3.2) there exists a unique matrix ${\mathcal J}_l$ of the form above with such a system of eigenvalues.

• (ii) Let ${\mathcal J}$ be an $(m+1)\times (m+1)$ positive semi-definite (real) Jacobi matrix (2.1) with $a_j>0$, $j=1,\ldots ,m$, satisfying $\det {\mathcal J}= 0$. Eigenvalues of ${\mathcal J}_l$, counting with their algebraic multiplicities, belong to

  $$\begin{aligned} \left\{ (z_j)_{j=1}^{m+1} \in ({\mathbb {C}}_+)^{m+1} : \sum _{j=1}^{m+1} \mathrm{Arg}\,z_j = \tfrac{\pi }{2} \right\} . \end{aligned}$$

  (3.3)

  Moreover, for every configuration of $m+1$ points from (3.3) there exists a unique matrix ${\mathcal J}_l$ of the form above with such a system of eigenvalues.

We will prove Proposition 2 in Sect. 5.2.

Proof of Theorem 1

Since $\Gamma \ge \mathbf {0}$, we can diagonalize $\Gamma = U (l I_{1\times 1}) U^*$, where $l>0$ and U is unitary. We may assume $\mathbf {w} = U e_1$. Then $H_{eff}=U( U^*HU + i l I_{1\times 1} ) U^*.$ Applying the tridiagonalization procedure from Sect. 2.2, we can reduce $U^*HU$ to the Jacobi form (2.1): $U^*HU = S {\mathcal J}S^*$ with S unitary. Note that $k = \dim {\text {span}}\{H^j \mathbf {w}: j\ge 0\} = \dim {\text {span}}\{ (U^* H U)^j \mathbf {e}_1: j\ge 0\}$, so $a_j > 0$ for $1\le j\le k-1$, $a_k =0$ (see Sect. 2.2). Therefore ${\mathcal J}$ is a direct sum of a $k\times k$ Jacobi matrix with positive $a_j$’s and some other $(n-k)\times (n-k)$ Jacobi matrix. Because of (2.2), $S^* I_{1\times 1} S=I_{1\times 1}$ and therefore

$$\begin{aligned} H_{eff}=U S ( {\mathcal J}+ i l I_{1\times 1} ) S^* U^*. \end{aligned}$$

(3.4)

Part (i) now follows from [1]. Part (ii) follows from Proposition 2(i). For the case (iii), $\det {\mathcal J}=0$, but it might happen that the zero eigenvalue of ${\mathcal J}$ is an eigenvalue either of the $k\times k$ or $(n-k)\times (n-k)$ submatrix of ${\mathcal J}$. Thus either Proposition 2(i) or (ii) applies and finishes the proof. $\square$

4 Rank One Perturbations: Tridiagonal Matrix Models

Let H be an $n \times n$ matrix from one of the six ensembles $GOE_n$, $LOE_{m\times n}$ ($\beta =1$); $GUE_n$, $LUE_{m\times n}$ ($\beta =2$); $GSE_n$, $LSE_{m\times n}$ ($\beta =4$). Let $H_{eff}$ be as in (1.1), where $\Gamma = (\Gamma _{jk})_{j,k=1}^n$ is an $n\times n$ positive definite (deterministic or random) matrix with real (if $\beta =1$), complex (if $\beta =2$), or quaternionic (if $\beta =4$) entries. We assume that $\Gamma$ is independent of H and has rank 1 (for the case $\beta =4$, the (right) rank is viewed over quaternions, see, e.g., [13]).

Since $\Gamma \ge \mathbf {0}$, we can write $\Gamma = U (l I_{1\times 1}) U^*$, where U is orthogonal, unitary, or unitary symplectic for $\beta =1,2,4$, respectively (for quaternion diagonalization, see, e.g., [13, Thm. 5.3.6]). Since the Hilbert–Schmidt norm should be preserved, we see that $l=||\Gamma ||_{HS} = \left( \sum _{j,k=1}^n |\Gamma _{jk}|^2 \right) ^{1/2}$.

Then $H_{eff}=U( U^*HU + i l I_{1\times 1} ) U^*,$ where U is independent of H. From Definitions 2 and 3, it is clear that $U^*HU$ belongs to the same ensemble as H. Therefore we can apply the tridiagonalization procedure from Sect. 2.2 to reduce $U^*HU$ to the Dumitriu–Edelman form: $U^*HU = S {\mathcal J}S^*$ with ${\mathcal J}$ as in Lemma 1 or 2, and S unitary satisfying $S^* I_{1\times 1} S=I_{1\times 1}$ (by 2.2), so (3.4) holds. We proved

Theorem 2

(Matrix model for rank one non-Hermitian perturbations of Gaussian and Laguerre ensembles) Let H be taken from one of the six ensembles $GOE_n$, $GUE_n$, $GSE_n$, $LOE_{m\times n}$, $LUE_{m\times n}$, $LSE_{m\times n}$. Suppose the (deterministic or random) matrix $\Gamma$ is independent of H and $\Gamma \ge \mathbf {0}$, $\mathrm{rank}\,\Gamma = 1$. Then $H_{eff} = H+ i \Gamma$ is unitarily equivalent to

$$\begin{aligned} {\mathcal J}+ i l I_{1\times 1} \end{aligned}$$

(4.1)

where ${\mathcal J}$ is as in Lemma 1 or 2, respectively, and $l=||\Gamma ||_{HS} = (\sum _{j,k=1}^n |\Gamma _{jk}|^2)^{1/2}$ is independent of ${\mathcal J}$.

Remark

This tridiagonal matrix ensemble (4.1) makes sense for any $\beta >0$.

5 Rank One Perturbations: Joint Eigenvalue Distribution

5.1 Perturbations of Gaussian $\beta$-Ensembles

Theorem 3

Fix a deterministic $l>0$, and for any $\beta >0$ let ${\mathcal J}$ be from $G\beta E_n$ ensemble. Then the eigenvalues of ${\mathcal J}_l$, (4.1), are distributed on $\{(z_j)\in ({\mathbb {C}}_+)^n: \sum _{j=1}^n {{\mathrm{Im}}}z_j =l \}$ according to

$$\begin{aligned}&\tfrac{1}{h_{\beta ,n}} \, e^{-\frac{1}{2} \sum _{j=1}^n {{\mathrm{Re}}}(z_j^2) } \times \prod _{j,k=1}^n |z_j-\bar{z}_k |^{\tfrac{\beta }{2} -1} \prod _{j<k} |z_j-z_k|^2 \nonumber \\&\quad \times \, l^{-\tfrac{\beta n}{2}+1} e^{-\frac{l^2}{2} } d^2 z_1\ldots d^2 z_{n-1} d({{\mathrm{Re}}}z_n), \end{aligned}$$

(5.1)

where $d^2 z$ stands for the 2-dimensional Lebesgue measure on ${\mathbb {C}}$; and

$$\begin{aligned} h_{\beta ,n} = 2^{n(\beta /2-1)} g_{\beta ,n} c_{\beta ,n}, \end{aligned}$$

(5.2)

where $g_{\beta ,n}$ and $c_{\beta ,n}$ are as in (2.8).

Remarks

• 1. In view of Theorem 2, distribution (5.1) with $\beta =1,2,4$ is the eigenvalue distribution of rank one perturbations of $GOE_n$, $GUE_n$, $GSE_n$, respectively.

• 2. If we suppose that $l>0$ is random (independent of ${\mathcal J}_l$) with a distribution $\gamma$, then the expression in (5.1) should be viewed as the conditional distribution of $z_j$’s given l. The joint distribution of $z_j$’s and l is therefore equal to the product of (5.1) and $d\gamma (l)$. In the special case when $\gamma$ is absolutely continuous $d\gamma (l) = F(l) dl$, we get that the eigenvalues of ${\mathcal J}_l$ are distributed on $\{(z_j)\in ({\mathbb {C}}_+)^n: \sum _{j=1}^n {{\mathrm{Im}}}z_j \in \mathrm{supp}(F) \}$ according to

  $$\begin{aligned} \tfrac{1}{h_{\beta ,n}} \, e^{-\frac{1}{2} \sum _{j=1}^n {{\mathrm{Re}}}(z_j^2) }\times & {} \prod _{j,k=1}^n |z_j-\bar{z}_k |^{\tfrac{\beta }{2} -1} \prod _{j<k} |z_j-z_k|^2 \nonumber \\\times & {} l^{-\tfrac{\beta n}{2}+1} e^{-\frac{l^2}{2} } F(l) d^2 z_1\ldots d^2 z_n, \end{aligned}$$

  (5.3)

  where $l=\sum _{j=1}^n {{\mathrm{Im}}}z_j$.

Proof

By Theorem 1(i), each of the eigenvalues $z_1,\ldots ,z_n$ lies in ${\mathbb {C}}_+$. Moreover, by the result of Arlinskiĭ–Tsekanovskiĭ [1, Thm. 5.1], the mapping

$$\begin{aligned}&\{a_j\}_{j=1}^{n-1}, \{b_j\}_{j=1}^n \mapsto z_1,\ldots , z_n \nonumber \\&(0,\infty )^{n-1}\times {\mathbb {R}}^n \rightarrow ({\mathbb {C}}_+)^n \end{aligned}$$

(5.4)

is one-to-one and onto the set $\{(z_j)\in ({\mathbb {C}}_+)^n: \sum _{j=1}^n {{\mathrm{Im}}}z_j =l \}$ (see 5.17 below). Then so is the mapping $\{\lambda _j\}_{j=1}^{n}, \{w_j\}_{j=1}^{n-1}\mapsto z_1,\ldots , z_n$, where $\mu$ (2.5) is the spectral measure of ${\mathcal J}$. Let us compute the Jacobian of this transformation.

Lemma 6

$$\begin{aligned} \left| \det \frac{\partial \left( {{\mathrm{Re}}}z_1,\ldots , {{\mathrm{Re}}}z_n,{{\mathrm{Im}}}z_1,\ldots , {{\mathrm{Im}}}z_{n-1} \right) }{\partial \left( \lambda _1,\ldots ,\lambda _n,w_1,\ldots ,w_{n-1} \right) } \right| = l^{n-1} \prod _{j<k} \frac{|\lambda _j-\lambda _k|^2}{|z_j-z_k|^2}. \end{aligned}$$

(5.5)

Proof

Let $m(z)=\langle e_1, ({\mathcal J}-z)^{-1} e_1\rangle = \sum _{j=1}^n \frac{w_j}{\lambda _j-z}$. Denote the characteristic polynomial as $\sum _{j=0}^n \kappa _j z^j = \det (z-{\mathcal J}_{l})=\prod _{j=1}^n (z-z_j)$, where $\kappa _n=1$. Let us first compute the Jacobian of the transformation of ${{\mathrm{Re}}}\kappa _0,\ldots ,{{\mathrm{Re}}}\kappa _{n-1},{{\mathrm{Im}}}\kappa _0,\ldots ,{{\mathrm{Im}}}\kappa _{n-2}$ with respect to $\lambda _1,\ldots ,\lambda _n,w_1,\ldots ,w_{n-1}$. Note that ${{\mathrm{Im}}}\kappa _{n-1} = -\sum _{j=1}^n {{\mathrm{Im}}}z_j = -l$ is fixed.

Observe that

$$\begin{aligned} \sum _{j=0}^n \kappa _j z^j = \det (z-{\mathcal J}) \det ( I -(z-{\mathcal J})^{-1} il I_{1\times 1}) = \left( 1+il m(z)\right) \prod _{j=1}^n (z-\lambda _j). \end{aligned}$$

(5.6)

By taking the real parts for $z\in {\mathbb {R}}$, and then using analytic continuation, we obtain

$$\begin{aligned} \tfrac{1}{2} \prod _{j=1}^n (z-z_j)+\tfrac{1}{2} \prod _{j=1}^n (z-\bar{z}_j) = \sum _{j=0}^n ({{\mathrm{Re}}}\kappa _j) z^j = \prod _{j=1}^n (z-\lambda _j). \end{aligned}$$

(5.7)

This implies that the Jacobian submatrix $\frac{\partial \left( {{\mathrm{Re}}}\kappa _0,\ldots , {{\mathrm{Re}}}\kappa _{n-1} \right) }{\partial \left( w_1,\ldots ,w_{n-1} \right) }$ is equal to the $n\times (n-1)$ zero matrix, while

$$\begin{aligned} \left| \det \frac{\partial \left( {{\mathrm{Re}}}\kappa _0,\ldots , {{\mathrm{Re}}}\kappa _{n-1} \right) }{\partial \left( \lambda _1,\ldots ,\lambda _n \right) } \right| = \prod _{j<k} |\lambda _j-\lambda _k|. \end{aligned}$$

(5.8)

Thus we just need to evaluate $|\det \frac{\partial \left( {{\mathrm{Im}}}\kappa _0,\ldots , {{\mathrm{Im}}}\kappa _{n-2} \right) }{\partial \left( w_1,\ldots ,w_{n-1} \right) }|$, regarding $\lambda _j$’s as constants.

The imaginary parts of (5.6) for $z\in {\mathbb {R}}$ give

$$\begin{aligned} \sum _{j=0}^{n-1} ({{\mathrm{Im}}}\kappa _j) z^j= & {} l m(z) \prod _{j=1}^n (z-\lambda _j) = -l \sum _{j=1}^n w_j \prod _{\begin{array}{c} 1 \le k \le n \\ k\ne j \end{array}} (z-\lambda _k) \nonumber \\= & {} -l \left[ \sum _{j=1}^{n-1} w_j (\lambda _j-\lambda _n)\prod _{\begin{array}{c} 1 \le k \le n-1 \\ k\ne j \end{array}} (z-\lambda _k) \right] - l \prod _{k=1}^{n-1} (z-\lambda _k) \end{aligned}$$

(5.9)

Denote the polynomial in the square brackets as $s(z)=\sum _{j=0}^{n-2} s_j z^j$. Then by (5.9),

$$\begin{aligned} \det \frac{\partial \left( {{\mathrm{Im}}}\kappa _0,\ldots , {{\mathrm{Im}}}\kappa _{n-2} \right) }{\partial \left( s_0,\ldots ,s_{n-2}\right) } = (-l)^{n-1}. \end{aligned}$$

(5.10)

Now note that s(z) can be rewritten as

$$\begin{aligned} s(z)=\sum _{j=1}^{n-1} \widetilde{w}_j \prod _{\begin{array}{c} 1 \le k \le n-1 \\ k\ne j \end{array}} \frac{z-\lambda _k}{\lambda _j-\lambda _k}, \end{aligned}$$

where

$$\begin{aligned} \widetilde{w}_j = w_j (\lambda _j-\lambda _n) \prod _{\begin{array}{c} 1 \le k \le n-1 \\ k\ne j \end{array}} (\lambda _j-\lambda _k). \end{aligned}$$

(5.11)

One can now recognize that s(z) is the interpolating polynomial $s(\lambda _k) = \widetilde{w}_k$ for $k=1,\ldots ,n-1$. This implies

$$\begin{aligned} \left| \det \frac{\partial \left( \widetilde{w}_1,\ldots , \widetilde{w}_{n-1} \right) }{\partial \left( s_0,\ldots ,s_{n-2} \right) } \right| = \prod _{1 \le j<k \le n-1} |\lambda _j - \lambda _k|. \end{aligned}$$

(5.12)

Finally, from (5.11),

$$\begin{aligned} \det \frac{\partial \left( \widetilde{w}_1,\ldots , \widetilde{w}_{n-1} \right) }{\partial \left( w_1,\ldots ,w_{n-1} \right) } = \prod _{j=1}^{n-1} (\lambda _j - \lambda _n) \prod _{1 \le j<k \le n-1} |\lambda _j - \lambda _k|^2. \end{aligned}$$

(5.13)

Combining (5.10), (5.12), (5.13), we get

$$\begin{aligned} \left| \det \frac{\partial \left( {{\mathrm{Im}}}\kappa _0,\ldots , {{\mathrm{Im}}}\kappa _{n-2} \right) }{\partial \left( w_1,\ldots ,w_{n-1} \right) } \right| = l^{n-1} \prod _{1\le j<k \le n} |\lambda _j-\lambda _k|. \end{aligned}$$

(5.14)

Using (5.8), we get

$$\begin{aligned} \left| \det \frac{\partial \left( {{\mathrm{Re}}}\kappa _0,\ldots ,{{\mathrm{Re}}}\kappa _{n-1},{{\mathrm{Im}}}\kappa _0,\ldots , {{\mathrm{Im}}}\kappa _{n-2} \right) }{\partial \left( \lambda _1,\ldots ,\lambda _n,w_1,\ldots ,w_{n-1} \right) } \right| = l^{n-1} \prod _{1\le j<k \le n} |\lambda _j-\lambda _k|^2. \end{aligned}$$

(5.15)

Finally, observe that if we have to restriction on $\kappa _j$’s and $z_j$’s, then

$$\begin{aligned} \prod _{j<k} |z_j-z_k|^2&= \left| \det \frac{\partial \left( {{\mathrm{Re}}}\kappa _0,\ldots , {{\mathrm{Re}}}\kappa _{n-1},{{\mathrm{Im}}}\kappa _0,\ldots , {{\mathrm{Im}}}\kappa _{n-1} \right) }{\partial \left( {{\mathrm{Re}}}z_1,\ldots ,{{\mathrm{Re}}}z_n,{{\mathrm{Im}}}z_1,\ldots ,{{\mathrm{Im}}}z_{n} \right) }\right| \\&= \left| \det \frac{\partial \left( {{\mathrm{Re}}}\kappa _0,\ldots , {{\mathrm{Re}}}\kappa _{n-1},{{\mathrm{Im}}}\kappa _0,\ldots , {{\mathrm{Im}}}\kappa _{n-1} \right) }{\partial \left( {{\mathrm{Re}}}z_1,\ldots ,{{\mathrm{Re}}}z_n,{{\mathrm{Im}}}z_1,\ldots ,{{\mathrm{Im}}}z_{n-1},{{\mathrm{Im}}}\kappa _{n-1} \right) }\right| \\&= \left| \det \frac{\partial \left( {{\mathrm{Re}}}\kappa _0,\ldots , {{\mathrm{Re}}}\kappa _{n-1},{{\mathrm{Im}}}\kappa _0,\ldots , {{\mathrm{Im}}}\kappa _{n-2} \right) }{\partial \left( {{\mathrm{Re}}}z_1,\ldots ,{{\mathrm{Re}}}z_n,{{\mathrm{Im}}}z_1,\ldots ,{{\mathrm{Im}}}z_{n-1} \right) }\right| . \end{aligned}$$

The first equality is the standard fact; second equality comes from the change of variables ${{\mathrm{Im}}}\kappa _{n-1} = -\sum _{j=1}^n {{\mathrm{Im}}}z_j$; the last equality comes from Laplace expansion for the determinants (under the condition ${{\mathrm{Im}}}\kappa _{n-1} = {\text {const}}$).

Combining the last Jacobian with (5.15), we obtain the statement of the lemma. $\square$

The joint distribution of $\{\lambda _j\}_{j=1}^{n}, \{w_j\}_{j=1}^{n-1}$ is

$$\begin{aligned} \tfrac{1}{g_{\beta ,n} c_{\beta ,n}} \prod _{j<k} |\lambda _j-\lambda _k|^\beta \prod _{j=1}^n e^{- \lambda _j^2/2} \prod _{j=1}^n w_j^{\beta /2-1} d\lambda _1\ldots d\lambda _n dw_1\ldots dw_{n-1}. \end{aligned}$$

Using this and Lemma 6, we obtain that the distribution of $z_j$’s is

$$\begin{aligned} \tfrac{1}{g_{\beta ,n} c_{\beta ,n}} l^{-(n-1)} \prod _{j<k} |\lambda _j-\lambda _k|^{\beta -2} \prod _{j=1}^n e^{- \lambda _j^2/2} \prod _{j=1}^n w_j^{\beta /2-1} \prod _{j<k} |z_j-z_k|^2 d^2 z_1\ldots d({{\mathrm{Re}}}z_n). \end{aligned}$$

(5.16)

Note that

$$\begin{aligned} l=-{{\mathrm{Im}}}\kappa _{n-1}&= \sum _{j=1}^n {{\mathrm{Im}}}z_j, \end{aligned}$$

(5.17)

$$\begin{aligned} \sum _{j=1}^n \lambda _j&= \sum _{j=1}^n {{\mathrm{Re}}}z_j, \end{aligned}$$

(5.18)

$$\begin{aligned} \sum _{j \ne k} \lambda _j \lambda _k&= \sum _{j\ne k} {{\mathrm{Re}}}(z_j z_k). \end{aligned}$$

(5.19)

The first equation comes from (5.9), while the latter two follow from (5.7). Then

$$\begin{aligned} \sum _{j=1}^n \lambda _j^2= & {} \left( \sum _{j=1}^n {{\mathrm{Re}}}z_j\right) ^2 - \sum _{j\ne k} {{\mathrm{Re}}}(z_j z_k) = \sum _{j=1}^n ({{\mathrm{Re}}}z_j)^2 + \sum _{j\ne k} ({{\mathrm{Im}}}z_j)({{\mathrm{Im}}}z_k) \nonumber \\= & {} \sum _{j=1}^n {{\mathrm{Re}}}\left( z_j\right) ^2 + l^2. \end{aligned}$$

(5.20)

Finally, from (5.6),

$$\begin{aligned} -i lw_j = il \mathop {{{\mathrm{Res}}}}\limits _{z=\lambda _j} m(z) = \mathop {{{\mathrm{Res}}}}\limits _{z=\lambda _j} \prod _{k=1}^n \frac{z-z_k}{z-\lambda _k} = \frac{\prod _{k=1}^n (\lambda _j-z_k)}{\prod _{k\ne j} (\lambda _j-\lambda _k)}, \end{aligned}$$

(5.21)

so

$$\begin{aligned} \prod _{j=1}^n w_j = (\tfrac{i}{l})^n \frac{\prod _{j,k} (\lambda _j-z_k)}{\prod _{j<k} |\lambda _j-\lambda _k|^2} = (\tfrac{i}{l})^n \tfrac{1}{2^n} \frac{\prod _{j,k} (\bar{z}_j-z_k)}{\prod _{j<k} |\lambda _j-\lambda _k|^2} = \tfrac{1}{(2l)^n} \frac{\prod _{j,k} |\bar{z}_j-z_k|}{\prod _{j<k} |\lambda _j-\lambda _k|^2}, \end{aligned}$$

(5.22)

where we used (5.7) with $z=z_k$, $k=1,\ldots ,n$. Combining (5.17), (5.20), (5.22) with (5.16), we obtain (5.1).$\square$

5.1.1 Example

Since $\Gamma$ in Theorem 2 has rank 1, we can decompose it as $\Gamma = L^* L$, where $L = (l_{1j})_{j=1}^n$ is an $1\times n$ matrix. Assuming the entries $l_{1j}$ of L are independent and normal $N(0,\sigma \mathbf {I}_\beta )$, then $l=\sum _{j=1}^n |l_{1j}|^2 \sim \sigma ^2 \chi ^2_{\beta n}$, that is l is distributed on $(0,\infty )$ according to F(l)dl with $F(l) = \tfrac{1}{(\sqrt{2} \sigma )^{\beta n} \Gamma (\beta n/2)} l^{\beta n/2-1}e^{-l/(2\sigma ^2)}$. In this special case, eigenvalues $\{z_1,\ldots ,z_n\}$ are distributed on $({\mathbb {C}}_+)^n$ according to

$$\begin{aligned}&\tfrac{1}{(\sqrt{2} \sigma )^{\beta n} \Gamma (\beta n/2) c_{\beta ,n}g_{\beta ,n}} \, e^{-\frac{1}{2} \sum _{j=1}^n {{\mathrm{Re}}}(z_j^2) } \prod _{j,k=1}^n |z_j-\bar{z}_k |^{\tfrac{\beta }{2} -1} \prod _{j<k} |z_j-z_k|^2 \nonumber \\&\quad \times \, e^{-\frac{l^2}{2} -\frac{l}{2\sigma ^2}} d^2 z_1\ldots d^2 z_n. \end{aligned}$$

(5.23)

5.2 Perturbations of Laguerre $\beta$-Ensembles

Proof of Proposition 2

We use the same notation as in the previous section: let $z_j$’s be the eigenvalues of ${\mathcal J}_l$; let $\lambda _j$’s and $w_j$’s be the eigenvalues and eigenweights of the spectral measure of ${\mathcal J}$ (which is of the form (2.5) with (2.10) for the case (i) and (2.12) with (2.14) for the case (ii)). By [1], $z_j\in {\mathbb {C}}_+$ for every j.

Consider now case (i). Equations (5.7) and (5.9) imply

$$\begin{aligned} {{\mathrm{Re}}}s_k(z_1,\ldots ,z_n)&= s_k(\lambda _1,\ldots ,\lambda _n), \quad k=1,2,\ldots ,n; \end{aligned}$$

(5.24)

$$\begin{aligned} {{\mathrm{Im}}}s_k(z_1,\ldots ,z_n)&= l \sum _{j=1}^n w_j s_{k-1}(\{\lambda _t\}_{t\ne j}), \quad k=1,2,\ldots ,n, \end{aligned}$$

(5.25)

where $s_0:=1$, and $s_k$ ($k\ge 1$) is the k-th elementary symmetric polynomial

$$\begin{aligned} s_k(z_1,\ldots ,z_n) := \sum _{1\le j_1<j_2<\ldots <j_k\le n} z_{j_1}\ldots z_{j_k}. \end{aligned}$$

(5.26)

Since for each j, $\lambda _j> 0, w_j>0, l>0$, we obtain that $z_1,\ldots ,z_n$ must belong to

$$\begin{aligned} \left\{ (z_j)_{j=1}^n \in ({\mathbb {C}}_+)^n : s_k(z_1,\ldots ,z_n) \in Q_1, \quad k=1,2,\ldots ,n \right\} , \end{aligned}$$

(5.27)

where $Q_1:=\{z: 0<\mathrm{Arg}z<\pi /2 \}$. Conversely, take a collection of points from (5.27). Since it belongs to $({\mathbb {C}}_+)^n$, we know from [1, Thm. 5.1] that there exists a unique matrix of the form ${\mathcal J}+ il I_{1\times 1}$ with $l>0$ and $a_j>0$, $j=1,\ldots ,n-1$. Equation (5.7) along with the positivity of (5.24) implies that $\lambda _1,\ldots ,\lambda _n$ are the real roots of the polynomial $\prod _{j=1}^n (z-\lambda _j)$ with alternating signs of the coefficients. By Descartes’ rule of signs, such a polynomial cannot have negative zeros. This means that all $\lambda _j$’s are positive. Therefore (5.27) is precisely the space of all possible eigenvalue configurations of $H_{eff}$. Let us now show that it coincides with (3.2).

It is elementary that (3.2) is a subset of (5.27). To see the converse, take any sequence from (5.27). Since $s_n(z_1,\ldots ,z_n)= z_1 z_2 \ldots z_n \in Q_1$, we must have that

$$\begin{aligned} 0+2k\pi< \mathrm{Arg}z_1 + \mathrm{Arg}z_2 +\cdots + \mathrm{Arg}z_n < \pi /2 + 2k\pi \end{aligned}$$

(5.28)

for some integer $k\ge 0$. We already know that these $z_1,\ldots ,z_n$ are the eigenvalues of ${\mathcal J}+ il I_1$, where ${\mathcal J}$ is positive definite. Let us now fix ${\mathcal J}$ and view $z_1,\ldots ,z_n$ as functions of $l\ge 0$ only. Each of these functions is continuous and never passes through 0. For any $0< l<\infty$, we have (5.28) for some k. But when $l=0$ the sum of the arguments is zero. By continuity $k=0$ for any l, i.e.,  (5.27) = (3.2).

To deal with the case (ii), we use similar arguments with $m+1$ instead of n and $\lambda _1,\ldots ,\lambda _m,0$ as the eigenvalues (with $\lambda _j>0, j=1,\ldots , m$). Then Eqs. (5.24) and (5.25) imply that the eigenvalues $z_1,\ldots ,z_{m+1}$ of ${\mathcal J}+ilI_{1\times 1}$ belong to

$$\begin{aligned} \big \{ (z_j)_{j=1}^{m+1} \in ({\mathbb {C}}_+)^{m+1} : s_{m+1}(z_1,\ldots ,z_{m+1}) \in i{\mathbb {R}}_+; \big . \nonumber \\ \big . s_k(z_1,\ldots ,z_{m+1}) \in Q_1, \quad k=1,2,\ldots ,m \big \}, \end{aligned}$$

(5.29)

where ${\mathbb {R}}_+ = \{z\in {\mathbb {R}}: z>0\}$. Conversely, by [1, Thm. 5.1], any configuration of point from (5.29) coincides with eigenvalues of some ${\mathcal J}+ilI_{1\times 1}$, $l>0$. The eigenvalues $\lambda _1,\ldots ,\lambda _{m+1}$ of ${\mathcal J}$ satisfy $s_{k}(\lambda _1,\ldots ,\lambda _{m+1})>0$ for $k=1,\ldots ,m$ and $s_{m+1}(\lambda _1,\ldots ,\lambda _{m+1})=0$. This implies $\lambda _j>0$ for all j except for one zero eigenvalue.

Finally, let us show that (5.29) coincides with (3.3). The inclusion (3.3)$\subseteq$(5.29) is easy. Conversely, take any configuration $\{z_j\}_{j=1}^{m+1}$ from (5.29). By the above, these points are the eigenvalues of some ${\mathcal J}+ilI_{1\times 1}$ with $l>0$, where ${\mathcal J}$ has eigenvalues $\{0,\lambda _1,\ldots ,\lambda _m\}$ with $\lambda _j>0$ for $1\le j\le m$. Since $s_{m+1} \in i{\mathbb {R}}_+$ in (5.29), we have

$$\begin{aligned} \mathrm{Arg}\, z_1 + \mathrm{Arg}\, z_2 +\ldots + \mathrm{Arg}\, z_{m+1} = \pi /2 + 2k\pi \end{aligned}$$

(5.30)

for some integer $k\ge 0$. After reordering, we can assume that $z_j \rightarrow \lambda _j$, $1\le j\le m$, and $z_{m+1}\rightarrow 0$ when $l\rightarrow 0$ (while ${\mathcal J}$ is fixed). Therefore $\mathrm{Arg}\, z_j \rightarrow 0$ as $l\rightarrow 0$ for $1\le j\le m$, while $0\le \mathrm{Arg}\, z_{m+1}\le \pi /2$ for any l. This proves that $k=0$, and so (5.29) $\subseteq$ (3.3), finishing the proof. $\square$

In the next theorem we compute the joint distribution of eigenvalues of rank one perturbations of the Laguerre $\beta$-ensembles.

Theorem 4

Fix a deterministic $l>0$, and for any $\beta >0$ and any integer $m,n>0$, let ${\mathcal J}$ be the $n\times n$ matrix from $L\beta E_{(m,n)}$ ensemble.

• (i) If $m\ge n$, then the eigenvalues $\{z_1,\ldots ,z_n\}$ of ${\mathcal J}_l = {\mathcal J}+ ilI_{1\times 1}$ are distributed on

  $$\begin{aligned} \Big \{ (z_j)_{j=1}^n \in ({\mathbb {C}}_+)^n : \sum _{j=1}^n \mathrm{Arg}\,z_j < \tfrac{\pi }{2}, \, \sum _{j=1}^n {{\mathrm{Im}}}z_j = l \Big \} \end{aligned}$$

  (5.31)

  according to

  $$\begin{aligned}&\tfrac{1}{q_{\beta ,n,a,l}} \prod _{j,k=1}^n |z_j-\bar{z}_k |^{\tfrac{\beta }{2} -1} \prod _{j<k} |z_j-z_k|^2 \nonumber \\&\quad \times \, e^{-\frac{1}{2} \sum _{j=1}^n {{\mathrm{Re}}}z_j } \left( {{\mathrm{Re}}}\prod _{j=1}^n z_j \right) ^{\tfrac{\beta a}{2}} d^2 z_1\ldots d^2 z_{n-1} d({{\mathrm{Re}}}z_n), \end{aligned}$$

  (5.32)

  where $a=m-n+1-2/\beta$ and

  $$\begin{aligned} q_{\beta ,n,a,l} = 2^{n(\beta /2-1)} h_{\beta ,n,a} c_{\beta ,n} l^{\tfrac{\beta n}{2}-1}, \end{aligned}$$

  where $h_{\beta ,n,a}$ and $c_{\beta ,n}$ are as in (2.11) and (2.8).

• (ii) If $m\le n-1$, then the $m+1$ nonzero eigenvalues of ${\mathcal J}_l = {\mathcal J}+ ilI_{1\times 1}$ are distributed on

  $$\begin{aligned} \left\{ (z_j)_{j=1}^{m+1} \in ({\mathbb {C}}_+)^{m+1} : \sum _{j=1}^{m+1} \mathrm{Arg}\,z_j = \tfrac{\pi }{2}, \, \sum _{j=1}^{m+1} {{\mathrm{Im}}}z_j= l \right\} \end{aligned}$$

  (5.33)

  according to

  $$\begin{aligned}&\tfrac{1}{t_{\beta ,m,n,l}} \prod _{j,k=1}^{m+1} |z_j-\bar{z}_k |^{\tfrac{\beta }{2} -1} \prod _{1\le j<k\le m+1} |z_j-z_k|^2 \nonumber \\&\quad \times \, e^{-\frac{1}{2} \sum _{j=1}^{m+1} {{\mathrm{Re}}}z_j } \prod _{j=1}^{m+1} |z_j|^{\frac{\beta (n-m-1)}{2}} \left( {{\mathrm{Re}}}\prod _{j=1}^m z_j \right) ^{-1} d^2 z_1\ldots d^2 z_m, \end{aligned}$$

  (5.34)

  where

  $$\begin{aligned} t_{\beta ,m,n,l} = (m+1) 2^{(m+1)(\beta /2-1)} h_{\beta ,m,a} d_{\beta ,m,n} l^{\tfrac{\beta n}{2}-1}, \end{aligned}$$

  (5.35)

  where $a=n-m+1-2/\beta$, and $h_{\beta ,m,a}$ and $d_{\beta ,m,n}$ are as in (2.11) and (2.15).

Remarks

• 1. Distributions (5.32) and (5.34) with $\beta =1,2,4$ are the eigenvalue distribution of rank one perturbations of $LOE_{(m,n)}$, $LUE_{(m,n)}$, $LSE_{(m,n)}$, respectively.

• 2. In (ii), $z_{m+1}$ is determined from $z_1,\ldots ,z_m$ because of (5.33).

• 3. Similarly to the remark 2 after Theorem 3, we can also assume that $l>0$ is random (independent of ${\mathcal J}_l$) with a distribution $\gamma$. Then (5.32) and (5.34) are the conditional distributions of $z_j$’s given l. The joint distribution of $z_j$’s and l is then equal to the product with $d\gamma (l)$ and can be calculated as in the case of Gaussian ensembles above.

Proof

• (i) We can take the known joint distribution of the eigenvalues $\lambda _j$’s and the eigenweights $w_j$’s (see Lemma 4) and change the variables to $z_j$’s (by Proposition 2(i) it is one-to-one and onto (5.31), so the Jacobian (5.5) applies). Using (5.22), (5.17), (5.18), (5.24) (with $k=n$), we obtain the resulting distribution (5.32).

• (ii) By Proposition 2(ii), the map from the spectral measures of the form (2.12), (2.14) to the eigenvalues of ${\mathcal J}+il I_{1\times 1}$: $\lambda _1,\ldots ,\lambda _m,w_1,\ldots ,w_m\mapsto z_1,\ldots ,z_{m+1}$ is one-to-one and onto (5.33) (if we impose some natural ordering on $\lambda _j$’s and $z_j$’s; we will remove it in the end of the proof). Its Jacobian is different from (5.5) computed earlier. Similar to the notation in the proof of Lemma 6, let $m(z)=\langle e_1, ({\mathcal J}-z)^{-1} e_1\rangle = -\frac{w_0}{z} + \sum _{j=1}^m \frac{w_j}{\lambda _j-z}$ and $\sum _{j=0}^{m+1} \kappa _j z^j = \det (z-{\mathcal J}_{l})=\prod _{j=1}^{m+1} (z-z_j)$, where $\kappa _{m+1}=1$. Because of $\det {\mathcal J}= 0$, we obtain ${{\mathrm{Re}}}\kappa _0 = 0$. Following similar reasoning as in the proof of Lemma 6, we first obtain the value of the Jacobian

  $$\begin{aligned} \left| \det \frac{\partial \left( {{\mathrm{Re}}}\kappa _1,\ldots , {{\mathrm{Re}}}\kappa _{m},{{\mathrm{Im}}}\kappa _0,\ldots ,{{\mathrm{Im}}}\kappa _{m-1} \right) }{\partial \left( \lambda _1,\ldots ,\lambda _m,w_1,\ldots ,w_{m} \right) } \right| = l^m\prod _{j=1}^m \lambda _j \prod _{1\le j<k\le m} |\lambda _j-\lambda _k|^2. \end{aligned}$$

  (5.36)

  Since ${{\mathrm{Re}}}(z_1\ldots z_{m+1}) = (-1)^{m+1} {{\mathrm{Re}}}\kappa _0 = 0$ and ${{\mathrm{Im}}}\kappa _m = -\sum _{j=1}^{m+1} {{\mathrm{Im}}}z_j = -l$, we have that $z_{m+1}$ is determined by $z_1,\ldots ,z_m$. Therefore we have a one-to-one map ${\mathbb {R}}^{2m}\rightarrow {\mathbb {R}}^{2m}$ taking $z_1,\ldots ,z_{m}$ to ${{\mathrm{Re}}}\kappa _1,\ldots , {{\mathrm{Re}}}\kappa _{m},{{\mathrm{Im}}}\kappa _0,\ldots ,{{\mathrm{Im}}}\kappa _{m-1}$. We need its Jacobian on the manifold ${{\mathrm{Re}}}(z_1\ldots z_{m+1}) =0, \sum _{j=1}^{m+1} {{\mathrm{Im}}}z_j = l$. If we have no restrictions on $z_j$’s or $\kappa _j$’s, then

  $$\begin{aligned} \prod _{1\le j<k\le m+1} |z_j-z_k|^2&= \left| \det \frac{\partial \left( {{\mathrm{Re}}}\kappa _0,{{\mathrm{Im}}}\kappa _0, \ldots ,{{\mathrm{Re}}}\kappa _{m},{{\mathrm{Im}}}\kappa _{m}\right) }{\partial \left( {{\mathrm{Re}}}z_1,{{\mathrm{Im}}}z_1, \ldots ,{{\mathrm{Re}}}z_{m+1},{{\mathrm{Im}}}z_{m+1} \right) } \right| \\&= \left| \det \frac{\partial \left( {{\mathrm{Re}}}\kappa _0,{{\mathrm{Im}}}\kappa _0, \ldots ,{{\mathrm{Re}}}\kappa _{m},{{\mathrm{Im}}}\kappa _{m}\right) }{\partial \left( {{\mathrm{Re}}}z_1,{{\mathrm{Im}}}z_1, \ldots ,{{\mathrm{Re}}}z_{m},{{\mathrm{Im}}}z_{m},{{\mathrm{Re}}}\kappa _0,{{\mathrm{Im}}}\kappa _m \right) } \right| \\&\quad \times \left| \det \frac{\partial \left( {{\mathrm{Re}}}z_1,{{\mathrm{Im}}}z_1, \ldots ,{{\mathrm{Re}}}z_{m},{{\mathrm{Im}}}z_{m},{{\mathrm{Re}}}\kappa _0,{{\mathrm{Im}}}\kappa _m \right) }{\partial \left( {{\mathrm{Re}}}z_1,{{\mathrm{Im}}}z_1, \ldots ,{{\mathrm{Re}}}z_{m+1},{{\mathrm{Im}}}z_{m+1} \right) } \right| \end{aligned}$$

  The last determinant is equal to $|{{\mathrm{Re}}}(z_1\ldots z_m)|$, so

  $$\begin{aligned} \frac{ \prod _{1\le j<k\le m+1} |z_j-z_k|^2 }{ |{{\mathrm{Re}}}(z_1\ldots z_m)|}&{=} \left| \det \frac{\partial \left( {{\mathrm{Re}}}\kappa _0,{{\mathrm{Im}}}\kappa _0, \ldots ,{{\mathrm{Re}}}\kappa _{m},{{\mathrm{Im}}}\kappa _{m}\right) }{\partial \left( {{\mathrm{Re}}}z_1,{{\mathrm{Im}}}z_1, \ldots ,{{\mathrm{Re}}}z_{m},{{\mathrm{Im}}}z_{m},{{\mathrm{Re}}}\kappa _0,{{\mathrm{Im}}}\kappa _m \right) } \right| \\&{=} \left| \det \frac{\partial \left( {{\mathrm{Re}}}\kappa _1,\ldots , {{\mathrm{Re}}}\kappa _{m},{{\mathrm{Im}}}\kappa _0,\ldots ,{{\mathrm{Im}}}\kappa _{m-1}\right) }{\partial \left( {{\mathrm{Re}}}z_1,{{\mathrm{Im}}}z_1, \ldots ,{{\mathrm{Re}}}z_{m},{{\mathrm{Im}}}z_{m} \right) } \right| , \end{aligned}$$

  where in the last determinant we are assuming that ${{\mathrm{Re}}}\kappa _0 ={\text {const}}$ and ${{\mathrm{Im}}}\kappa _m = {\text {const}}$. Combining this with (5.36), we get that on ${{\mathrm{Re}}}\kappa _0 = 0,{{\mathrm{Im}}}\kappa _m=-l$,

  $$\begin{aligned} \left| \det \frac{\partial \left( {{\mathrm{Re}}}z_1,{{\mathrm{Im}}}z_1, \ldots ,{{\mathrm{Re}}}z_{m},{{\mathrm{Im}}}z_{m} \right) }{\partial \left( \lambda _1,\ldots ,\lambda _m,w_1,\ldots ,w_{m} \right) } \right| {=} l^{m} \big |{{\mathrm{Re}}}\prod _{j=1}^m z_j\big | \prod _{j=1}^m \lambda _j \frac{ \prod _{1\le j<k\le m} |\lambda _j-\lambda _k|^2}{\prod _{1\le j<k\le m+1} |z_j-z_k|^2}. \end{aligned}$$

  (5.37)

  Repeating the arguments from (5.21) and (5.22), we obtain

  $$\begin{aligned} w_0 = \frac{\prod _{j=1}^{m+1}|z_j|}{l \prod _{j=1}^m|\lambda _j|} ,\quad \text{ and } \prod _{j=1}^m w_j = \frac{1}{l^{m} 2^{m+1}} \frac{\prod _{j,k=1}^{m+1}|z_j-\bar{z}_k|}{\prod _{j=1}^{m+1} |z_j| \prod _{j=1}^m |\lambda _j| \prod _{j<k} |\lambda _j-\lambda _k|^2}. \end{aligned}$$

  Finally, just as in (i), we still have $\sum _{j=1}^m \lambda _j = \sum _{j=1}^{m+1} {{\mathrm{Re}}}z_j$. Now, starting from the joint distribution of $\lambda _1,\ldots ,\lambda _m,w_1,\ldots ,w_m$ (see Proposition 1), applying the Jacobian (5.37), and using these substitutions (note that terms with $\prod |\lambda _j|$ cancel out in the process), we arrive at the distribution (5.34). Note that the factor $(m+1)$ in (5.35) comes from removing the ordering of $z_j$’s and $\lambda _j$’s (there are $(m+1)!$ of permutations for $\{z_j\}_{j=1}^{m+1}$, and only m! for $\{\lambda _j\}_{j=1}^m$).

$\square$