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On constructing Hermitian unitary matrices with prescribed moduli 1 | OMICS International
ISSN: 1736-4337
Journal of Generalized Lie Theory and Applications
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On constructing Hermitian unitary matrices with prescribed moduli 1

Alicja SMOKTUNOWICZ1* and Wojciech TADEJ2

1Faculty of Mathematics and Information Science, Warsaw University of Technology, Plac Politechniki 1, 00-661 Warsaw, Poland,E-mail: [email protected]

2Faculty of Mathematics and Natural Sciences, Cardinal Stefan Wyszynski University, ul. Dewajtis 5, 01-815 Warsaw, Poland, E-mail: [email protected]

*Corresponding Author:
Alicja SMOKTUNOWICZ
Faculty of Mathematics and Information Science, Warsaw University of Technology,
Plac Politechniki 1, 00-661 Warsaw, Poland
E-mail: [email protected]

Received date: December 16, 2007; Revised date: March 07, 2008

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Abstract

The problem of reconstructing the phases of a unitary matrix with prescribed moduli is of a broad interest to people working in many applications, e.g in the circuit theory, phase shift analysis, multichannel scattering, computer science (e.g in the theory of error correcting codes, design theory). We propose efficient algorithms for computing Hermitian unitary matrices for given symmetric bistochastic matrices A(n × n) for n = 3 and n = 4. We mention also some results for matrices of arbitrary size n.

Introduction

We will study the set of unistochastic matrices which is a subset of the set of bistochastic matrices. We say that a matrix image is bistochastic (doubly stochastic) if all its entries are nonnegative real numbers and all its row sums and column sums are equal to 1. A unistochastic matrix is a bistochastic matrix whose entries are the squares of the absolute values of the entries of some unitary matrix U. We recall that a matrix image is Hermitian if B = B*, i.e.image for i, j = 1, . . . , n. A matriximage is unitary ifimage A matriximageis orthogonal if image We say that a symmetric bistochastic matriximage is H-unistochastic if there exists a Hermitian unitary matrix image such thatimageIf U is real (so U is orthogonal) then A is called H-orthostochastic.

Perhaps the van der Waerden matrix (Wn) is the most famous unistochastic matrix. Its elements are all equal to image. For example, for n = 4 there exists an orthogonal preimage U (2U is called an Hadamard matrix):

image

so W4 is even orthostochastic (and also H- orthostochastic). However, it is easy to verify that W3 is not H-unistochastic!

Hadamard’s Conjecture (still open!) says that for n > 2 the Hadamard matrices exist when n = 4k and only for such n, see W.Tadej et al. [5] for explicit examples of the Hadamard matrices.

We consider the following research problems.

• I. Given a bistochastic matrix image check if there exists a unitary matriximage such thatimage (so A is unistochastic).

• II. Given a symmetric bistochastic matrix image check if there exists a Hermitian unitary matrix image such thatimage (so A is H-unistochastic).

For n = 2 every bistochastic matrix A is symmetric and is orthostochastic (U can be chosen to be orthogonal). We have

image

Given a 3×3 bistochastic matrix A it is easy to check whether it is unistochastic or not (see e.g [2]). We get

image

Therefore the problem is to form a triangle from 3 line segments of given lengths imageimage Then A is unistochastic if and only if the chain-link conditions are fulfilled:image In this case

image

Some methods for constructing unitary preimages to 3×3 bistochastic matrices are discussed in [3].

However, for a given 4×4 bistochastic matrix A it is not easy to check whether it is unistochastic or not! There are only partial results, so it is reasonable to try to develop efficient algorithms to check whether a given bistochastic matrix A(n × n) is unistochastic or not. We focus our attention only on symmetric bistochastic matrices and their Hermitian unitary preimages.

New results

For simplicity define a matrix

image

Notice that without loss of generality we can seek a Hermitian unitary matrix U(n × n) in the dephased form, i.e. such that the first row and the first column of U are the same as the first row and the first column of M. It is obvious because if U is not dephased, we can find a unitary diagonal matrix D such that the matrix image satisfies these conditions and it is still Hermitian.

Given a 3×3 symmetric bistochastic matrix A it is easy to check whether it is H-unistochastic or not.

Notice that we can assume that the diagonal elements of A are ordered in such a way that image for some permutation {p1, p2, p3} of {1, 2, 3}, then we can permute rows and columns of A. Define a permutation matrix image where I = [e1, e2, e3]. Then image has the desired property. Note also that A is H-unistochastic iff image is H-unistochastic. That is, U is a unitary preimage for A iff PTUP is a unitary preimage for image.

As it was said above, we can assume that a Hermitian unitary U has the dephased form

image

Theorem 2.1. Let A(3 × 3) be a symmetric bistochastic matrix, image for all i, j andimage Then A is H-unistochastic if and only if the following matrix

image

where s = −1 or s = 1, is orthogonal.

Now we consider the case n = 4. Assume that all the elements of a symmetric bistochastic matrix A are nonzero. We show that our problem can be reduced to the linear system of equations.

Write A(4 × 4) and U as follows

image

where image andimage Hereimage is assumed in order to avoid trivial cases.

By the orthogonality of the first column of U and the columns 2, 3, 4 we obtain

image

We can assume that the signs sk are prescribed, in an algorithm we have to check all the combinations of signs (±1).

Let image Thenimage can be computed as a unique solution of the following linear system of equation Bx = f, Bx = f, where x = [x1, x2, x3]T and f = [f2, f3, f4]T , where fk = −ak(m1 + mk(sk)) for k = 2, 3, 4. Here

image

Then det(B) = −2 (a2a3a4)(b3b4c4) 6= 0, so there exists a unique solution x of the linear system Bx = f. We can compute it as follows x = B−1f, where

image

Now it is easy to compute yk. We should check the conditions: image Then we can compute yk from the formulae

image

Finally, we have to verify the orthogonality of the computed matrix U.

We have only some partial results for arbitrary size n. Notice that all the eigenvalues of a Hermitian unitary matrix imageare real and equal to −1 or 1. There exists a unitary matrix image such that U = QDQ*, Q*Q = I, where D = diag(1, 1, . . . , 1,−1,−1, . . . ,−1).

If we impose an additional condition on U, a Hermitian unitary preimage of A to be found, namely that U = QDQ* with Q*Q = I and D = diag(1, 1, . . . , 1,−1). Now it is not difficult to solve our problem! Notice that writing D = I − 2 diag(0, . . . , 0, 1) = I − 2 image we obtainimageso U is a reflection (Householder transformation).

Let image Thenimage whereimageWe assume that a1,1 ≠ 1 (for otherwise the problem reduces to the case (n − 1) × (n − 1)). Then we can choose z1 being real and positive because U does not depend on scaling of z (if z = αu with u*u = 1 and |α| = 1, then U = I − 2 uu*). Moreover, if D is a unitary diagonal matrix then DUD* = I −2 (Du)(Du)* is also a Householder matrix, so we can search for U in the dephased form. Then the desired Householder matrix U is real and must have the following form to have the correct moduli in the first row of U

image

Notice that

image

so U is orthogonal image The only thing to do is to compute U and check the condition image the orthogonality of the columns of U we propose reorthogonalization. We apply QR decomposition to U. To compute QR decomposition we can use the Householder or Givens methods or special versions of Gram- Schmidt orthogonalization methods (see eg. [6]). Here is a code for MATLAB using the function qr (the Householder method):

[Q,R]=qr(U); M=A.^(1/2); Z=Q./abs(Q); U=M.*Z; I=eye(n); error_U=norm(I-U’*U)); error_A=norm(A-abs(U).*abs(U));

The numerical tests in MATLAB confirm the advantage of the proposed algorithms.

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