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Multiplicative convolution of real asymmetric and real anti-symmetric matrices

  • Mario Kieburg EMAIL logo , Peter J. Forrester and Jesper R. Ipsen

Abstract

The singular values of products of standard complex Gaussian random matrices, or sub-blocks of Haar distributed unitary matrices, have the property that their probability distribution has an explicit, structured form referred to as a polynomial ensemble. It is furthermore the case that the corresponding bi-orthogonal system can be determined in terms of Meijer G-functions, and the correlation kernel given as an explicit double contour integral. It has recently been shown that the Hermitised product XMX2X1AX1TX2TXMT, where each Xi is a standard real Gaussian matrix and A is real anti-symmetric, exhibits analogous properties. Here we use the theory of spherical functions and transforms to present a theory which, for even dimensions, includes these properties of the latter product as a special case. As an example we show that the theory also allows for a treatment of this class of Hermitised product when the Xi are chosen as sub-blocks of Haar distributed real orthogonal matrices.

MSC 2010: 15A52; 42C05

Award Identifier / Grant number: DP170102028

Award Identifier / Grant number: CRC 1283

Funding statement: We acknowledge support by the Australian Research Council through grant DP170102028 (PJF), the ARC Centre of Excellence for Mathematical and Statistical Frontiers (PJF,JRI,MK), and the German research council (DFG) via the CRC 1283: “Taming uncertainty and profiting from randomness and low regularity in analysis, stochastics and their applications”.

A Appendix

In this appendix we want to prove the explicit expression of the spherical function Φ as stated in Theorem 2.4. We do this in three steps. First we consider the action of a corank 2 projection on even-dimensional anti-symmetric matrices, see Section A.1. We use this projection to construct a recurrence relation in the dimension for the group integral in the denominator of equation (2.8), see Section A.2, which is solved in Section A.3.

A.1 Eigenvalue PDF for a corank 2 projection

For C,XH both 2n×2n real anti-symmetric matrices, we define the matrix-valued Fourier transform of X by

(A.1)PH(C)=exp[ı2TrXC]PH(X)𝑑X.

Let the singular values of C and X be denoted by cA and xA, respectively. Consider the circumstance that PH(C)=PH(kCkT) for all kK=O(2n) and write f^X(c)=PH(ıcτ2). With pA(x) denoting the jPDF of the singular values of X, by making use of the Harish-Chandra group integral (1.3) for K=O(2n), it was shown in [18, Proposition 3.4] (with an extra factor of 1/n! for the convention that the eigenvalues are not ordered) that

(A.2)pA(x)=j=0n-1(-1)j2π(2j)!Δn(x2)1n!0𝑑c10𝑑cnf^X(c)Δn(c2)j=1ncos(xjcj).

We will use equation (A.2), with nn-1, to deduce the jPDF for the corank 2 projection given by the (2n-2)×(2n-2) random matrix

(A.3)X=Π2n-2,2nk(ıaτ2)kTΠ2n-2,2nT,kK=O(2n),

where a=(a1,,an)A is fixed.

Remark A.1.

We want to underline that to be fully rigorous we have to introduce a regularizing function in equation (A.2) like a Gaussian or a function similar to equation (2.15) with a vanishing parameter ϵ0 to guarantee the absolute integrability, since f^X(c)Δn(c2) is not necessarily an L1-function. For general PH, it can grow like polynomial on n. The particular problem to be overcome is the interchange of the inverse Laplace transform and the expectation value (A.1). In the present case, where we want to compute the distribution of the projection (A.3) with a fixed a, the function f^X(c)Δn(c2) is an L1-function on , see below. Thus, we can neglect these regularizing terms.

Proposition A.2.

Let x={xj}j=1n-1A denote the singular values of the random matrix (A.3), and let pA(x|a) denote the corresponding jPDF. We have

pA(x|a)=(2n-2)!(n-1)!Δn-1(x2)Δn(a2)det[11(ak-xj)Θ(ak-xj)]j=1,,n-1k=1,,n,

where Θ denotes the Heaviside step function.

Proof.

According to definition (A.1), with X given by equation (A.3), upon use of the cyclic property of the trace and the definition of Π2n-2,2n, we have

(A.4)PH(C)=Kexp[-12Trk(aτ2)kT[Cdiag(0,0)]]d*k.

The variables c=(c1,,cn-1)A are the singular values of C. The matrix integral corresponding to the average (A.4) is an example of the Harish-Chandra group integral (1.3) for O(2n), where Cdiag(0,0) is to be thought of as C[0ϵ-ϵ0] with ϵ0. As such, it has the evaluation

(A.5)f^X(c)=PH(ıcτ2)=j=0n-1(2j)!(-1)n(n-1)/2Δn(a2)Δn-1(c2)l=1n-1cl2det[11cos(cjak)]j=1,,n-1k=1,,n.

Substituting equation (A.5) in equation (A.2), the latter with nn-1, we obtain

pA(x|a)=(2n-2)!(n-1)!Δn-1(x2)Δn(a2)(j=1n-1-2π)Adcdet[11cos(cjak)]j=1,,n-1k=1,,nj=1n-1cosxjcjcj2.

In preparation for further simplification, we subtract the first row of the determinant from each of the subsequent rows. The integrations can then be carried out row-by-row to give

pA(x|a)=(2n-2)!(n-1)!Δn-1(x2)Δn(a2)det[112π0cos(xjc)(1-cos(akc))c2𝑑c]j=1,,n-1k=1,,n.

Evaluating the integral with the help of the residue theorem gives

2π0cosxjc(1-cosakc)c2𝑑c=(ak-xj)Θ(ak-xj),

which concludes the proof. ∎

Remark A.3.

(a) Let us order a and x as a1<a2<<an and x1<x2<<xn-1. Examining the determinant, we see that the intervals [0,a1] and [an-1,an] can be maximally occupied by a single singular value, namely, x1 and xn-1, respectively. More singular values would yield that either the first three rows or last two rows become linearly dependent. Furthermore, in each interval [aj,aj+1] we cannot have more than two singular values xj because of the same reason (three or more rows become linearly dependent). The same also applies to the intervals [xj,xj+1], which can maximally comprise two singular values of a (three or more columns become linearly dependent when violated). A detailed analysis tells us that those orders, where the determinant does not vanish, create a matrix which is upper triangular with maximally one additional lower diagonal. Once a combined order of a and x is given as 0<α1<α2<<α2n-2<α2n-1, with α a permutation of (a,x), the determinant evaluates to j=1n-1(α2j+1-α2j), which directly follows from the particular form of the matrix. The reason why α1 does not appear in the product follows from the following case discussion. We have either x1<a1, which means that we can subtract the first row times x1 with the second row canceling x1, or a1<x1, where a1 immediately drops out because of the Heaviside step-function, which vanishes in this case.

(b) Let b=diag(b1,,bn)n, uK=U(n), and consider the corank 1 projection given by the (n-1)×(n-1) random matrix Πn-1,nubuΠn-1,nT. The method used for proving Proposition A.2, with the role of the Harish-Chandra group integral for O(2n) now played by its unitary (U(n)) counterpart, can be adapted to show that the jPDF of the eigenvalues of this random matrix is equal to

Δn-1(x)Δn(b)det[11Θ(bk-xj)]j=1,,n-1k=1,,n=Δn-1(x)Δn(b)χb1<x1<b2<<xn-1<bn,

where χJ is the indicator function for the condition J. This result is well known [7].

A.2 A recurrence relation

We denote the numerator in equation (2.8) by

fn(s,x)=K=O(2n)d*kj=1n[detΠ2j,2nkxkTΠ2j,2nT](sj-sj+1)/2-1.

We first restrict to sn, with Re(sj-sj+1)2 for all j=1,,n-1, to have a bounded integrand, and then analytically continue at the end. Our present interest is in establishing a recurrence in the matrix dimension n.

Lemma A.4 (Recursion of fn).

Let n>2, let aA=R+n be the singular values of xH=o(2n), assumed non-degenerate, and let sCn with Re(sj-sj+1)2 for all j=1,,n-1. We have

fn(s,ıaτ2)=(2n-2)!(n-1)!(deta)sn+n-1+n-1𝑑a~Δn-1(a~2)Δn(a2)det[11(ac-a~b)Θ(ac-a~b)]b=1,,n-1c=1,,n
(A.6)×fn-1(diag(s1-sn-n,,sn-1-sn-n),ıdiag(a~1,,a~n-1)τ2).

Proof.

Since the function fn is K-invariant, i.e., fn(s,x)=fn(s,kxkT) for all kK and xH, we can replace x by ıaτ2, with aA its singular values. Then the j=n term in the product of the integrand evaluates to

(detΠ2n,2nkxkTΠ2n,2nT)(sn+n+1)/2-1=(deta)sn+n-1.

Being independent of the orthogonal matrix k, this term can be taken outside of the integral. Since Π2j,2n=Π2j,2(n-1)Π2(n-1),2n for each j=1,,n-1, for the remaining terms, we can write

j=1n-1(detΠ2j,2nkxkTΠ2j,2nT)(sj-sj+1)/2-1
=j=1n-1(detΠ2j,2(n-1)Π2(n-1),2nkxkTΠ2(n-1),2nTΠ2j,2(n-1)T)(sj-sj+1)/2-1.

We then introduce an orthogonal matrix k~O(2n-2) by the multiplicative shift kdiag(k~,𝟙2)k and integrate k~ via the Haar measure on O(2n-2) to obtain

fn(s,ıaτ2)=(deta)sn+n-1K=O(2n)d*kK=O(2n-2)d*k~
×j=1n-1[detΠ2j,2(n-1)k~Π2(n-1),2nk(ıaτ2)kTΠ2(n-1),2nTk~TΠ2j,2(n-1)T](sj-sj+1)/2-1
=(deta)sn+n-1K=O(2n)d*k
×fn-1(diag(s1-sn-n,,sn-1-sn-n),Π2(n-1),2nk(ıaτ2)kTΠ2(n-1),2nT).

Since also fn-1 is O(2n-2)-invariant we can replace x~=Π2(n-1),2nk(ıaτ2)kTΠ2(n-1),2nT by ıa~τ2, with a~+n-1 the singular values of the [2(n-1)×2(n-1)]-dimensional real anti-symmetric random matrix Π2(n-1),2nk(ıaτ2)kTΠ2(n-1),2nT. The jPDF of a~ is given in Proposition A.2, which yields the recursion (A.6), completing the proof. ∎

A.3 Proof of Theorem 2.4

Let aA=+n be the singular values of xH=o(2n), assumed non-degenerate, and let sn with Re(sj-sj+1)2 for all j=1,,n-1. The spherical function (2.8) is given in terms of fn according to

Φ(s;x)=fn(s,x)fn(s,ı𝟙nτ2),

and so the explicit knowledge of fn implies the explicit form of Φ.

Regarding the latter, we will use complete induction to show

(A.7)fn(s;ıaτ2)=cn(s)det[acsb+n-1]b,c=1,,nΔn(a2),

where

(A.8)cn(s)=(-1)n(n-1)/2j=0n-1(2j)!Δn(s)1k<ln(sk-sl-1).

For n=1, we have f1(s1;ıa1τ2)=a1s1, agreeing with ansatz (A.7) with c1(s1)=1 as is consistent with equation (A.8).

For the induction step we substitute ansatz (A.7) for fn-1 in the recursion (A.6) to obtain

fn(s,ıaτ2)=cn-1(diag(s1-sn-n,,sn-1-sn-n))(2n-2)!(n-1)!(deta)sn+n-1
×+n-1𝑑a~Δn-1(a~2)Δn(a2)det[11(ac-a~b)Θ(ac-a~b)]b=1,,n-1c=1,,ndet[a~csb-sn-2]b,c=1,,n-1Δn-1(a~2)
=cn-1(diag(s1-sn-n,,sn-1-sn-n))(2n-2)!(deta)sn+n-1Δn(a2)
×det[110ac𝑑t(ac-t)tsb-sn-2]b=1,,n-1c=1,,n
=(-1)n-1(2n-2)!l=1n-1(sl-sn-1)(sl-sn)cn-1(diag(s1-sn-n,,sn-1-sn-n))
(A.9)×det[acsb+n-1]b,c=1,,nΔn(a2).

In the second equality we used a variant of Andréief’s integration identity [29, Appendix C.1], and in the third we integrated by parts using the fact that the boundary terms vanish for b=1,,n-1 because Re(sb-sn)2(n-b). Comparison of this result with ansatz (A.7) yields the recursion in the constants

cn(s)=(-1)n-1(2n-2)!l=1n-1(sl-sn-1)(sl-sn)cn-1(diag(s1-sn-n,,sn-1-sn-n)),

which, with the initial condition c1(s)=1, implies equation (A.8).

For a=𝟙n, with the help of l’Hôpital’s rule, we obtain

(A.10)fn(s;ı𝟙nτ2)=j=0n-1(2j)!/j!1k<ln2(sk-sl-1).

Dividing equation (A.9) by equation (A.10) yields equation (2.10).

The result (A.9) can be analytically continued in the complex variables zj=sj-sj+1-2 with j=1,,n-1 when dividing the spherical function Φ(s,x) by maxj=1,,n{ajl=1nsl}. This division is equivalent with restricting all aj to the interval [0,1]. Then the singular values of each matrix Π2j,2nkxkTΠ2j,2nT are also restricted to the interval [0,1] and fn(s,ıaτ2)1k<ln(sk-sl-1)(sl-sk) is a bounded analytic function in each zj on the complex half-hyperplanes Rezj0. Carlson’s theorem can be applied to equation (A.9) for each zj to uniquely analytically continue to the whole complex plane, which concludes the proof of Theorem 2.4.

Acknowledgements

The first author wants to thank the University of Melbourne for its hospitality where this project was carried out.

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Received: 2018-03-01
Revised: 2018-11-19
Accepted: 2018-12-08
Published Online: 2019-01-20
Published in Print: 2019-10-01

© 2019 Walter de Gruyter GmbH, Berlin/Boston

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