POSITIVE MATRICES WITH PRESCRIBED SINGULAR VALUES

We consider the problem of constructing positive matrices with prescribed singular values. In particular, we show how to construct an m × n positive matrix, m ≥ n, with prescribed singular values σ1 ≤ σ2 ≤ · · · ≤ σn. AMS classification : 15A18; 15A29.


Introduction
A singular value decomposition of a matrix A ∈ C m×n is a factorization A = UΣV * , where Σ = diag(σ 1 , σ 2 , . . ., σ r ) ∈ R m×n , r = min{m, n}, σ 1 ≥ σ 2 ≥ . . .≥ σ r ≥ 0 and both U ∈ C m×m and V ∈ C n×n are unitary.The diagonal entries of Σ are called the singular values of A. The columns u j of U are called left singular vectors of A and the columns v j of V are called right singular vectors of A. Every A ∈ C m×n has a singular value decomposition A = U ΣV * and the following relations hold: Av j = σ j u j , A * u j = σ j v j and u * j Av j = σ j .If A ∈ R m×n , then U and V may be taken to be real (see [3]).
In the case m = n, the problem has always a solution if σ 1 > σ 2 .We construct such a solution from the following two results, due to Fiedler [2], and related with the nonnegative inverse eigenvalue problem: Lemma 1. [2] Let α > β ≥ 0, > 0. Then there exists a number c > 0 such that the matrix Ã α c c β ! has eigenvalues α + , β − .Lemma 2. [2] Let A be a symmetric m × m matrix with eigenvalues α 1, . . ., α m , and let u, kuk = 1, be a unit eigenvector corresponding to α 1 .Let B be a symmetric n × n matrix with eigenvalues β 1, . . ., β n , and let v, kvk = 1, be a unit eigenvector corresponding to β 1 .Then for any ρ, the matrix ! has eigenvalues α 2, . . ., α m , β 2, . . ., β n , γ 1 , γ 2 , where γ 1 , γ 2 are eigenvalues of the matrix b To construct a positive matrix, no necessaryly symmetric, with prescribed singular values, we first use the below result, Lemma 4, which shows how to construct an n × n nonnegative semibordered matrix with prescribed singular values and then, to obtain the required positive matrix, we apply a Brauer [1] type singular value perturbation result given in [4,Corollary 14], which shows how to modify a single singular value of an n × n matrix without changing any of the remaining singular values.
Lemma 3. The characteristic polynomial of the arrow matrix Proof.By expanding is an n × n arrow symmetric matrix with characteristic polynomial and eigenvalues σ 2 1 , σ 2 2 , ..., σ 2 n .We define Then, by taking λ = Hence, Lemma 5. [4] Let A be an m × n matrix with singular values , respectively, the left and right singular vectors corresponding to The paper is organized as follows: In section 2 we construct an n × n positive matrix with prescribed singular values by using results related with the symmetric nonnegative inverse eigenvalue problem.In section 3 we construct a positive matrix, not necessarily symmetric, from Lemmas 4 and 5.In section 4 we consider the rectangular case m × n, m > n.Finally, in section 5 we give some examples to illustrate the results.

Positive matrices with prescribed singular values I
In this section we construct a positive symmetric matrix with prescribed nonnegative eigenvalues, which in this case, are also its singular values.
Theorem 1.Given the real numbers in the last diagonal position if n is odd, is an n × n nonnegative symmetric matrix with singular values σ 1 , σ 2 , . . ., σ n .In order to obtain a positive matrix with prescribed singular values let us consider Now we apply the Lemma 2, n−2 2 times if n is even ( n−1 2 times if n is odd).In the first step of this process we apply Lemma 2 to the matrices with singular values µ 1 , σ n and µ 2 , σ n−1 , respectively.From Lemma 1, the matrix , has eigenvalues µ 1 + 2 and µ 2 − 2 = σ 2 .The unitary vectors are the Perron eigenvectors of B 1 and B 2 , respectively.Then the 4 × 4 matrix In the second step of the process we again apply Lemma 2 to the matrices B 1 and which has singular values µ 3 and σ n−2 .The matrix has eigenvalues µ 1 + 2 + 3 and 2 ) T .Then the 6 × 6 matrix The process is continued until in the last step we apply Lemma 2 to the matrices B n−4 for odd n).The matrix , for odd n), respectively, is positive and since

Positive matrices with prescribed singular values II
In this section we construct an n × n positive matrix A, no necessarily symmetric, with prescribed singular values σ 1 , σ 2 , . . ., σ n .First, by applying Lemma 4 we construct a nonnegative semibordered diagonal matrix C of the form and then, by applying Lemma 5, we add to the matrix C the positive matrix αuv T , where α > 0 and u and v are the left and right singular vectors of C corresponding to its maximal singular value.

Proof.
Let α > 0 such that e σ 1 = σ 1 − α > σ 2 and consider e Then, from Lemma 4 there exists an n × n nonnegative matrix C of the form (3.1), with singular values e σ 1 , σ 2 , ..., σ n .The arrow matrix CC T is nonnegative, irreducible, with Perron eigenvector equal to u 1 corresponding to the Perron root e σ 2 1 (left singular vector of C corresponding to the singular value e σ 1 ).Then u 1 is an entrywise positive vector and so is the right singular vector v 1 .Therefore the matrix αu 1 v T 1 is positive and from Lemma 5, A = C+αu 1 v T 1 is positive with the prescribed singular values σ 1 , σ 2 , ..., σ n . 2 Lemma 4 and the form of the matrix in (3.1) allow us to construct a nonnegative matrix , say C, whose exponential matrix e C has prescribed singular values: To show this, we consider the matrix C in (3.1).Then a straight forward calculation shows that is also a semibordered diagonal matrix.Now, for i = 2, 3, . . ., n, we have Finally, from we have

Positive matrices with prescribed singular values III
In this section we consider the construction of a rectangular positive matrix with prescribed singular values.First we construct an m × n nonnegative matrix B, m > n, of the form and then we apply the perturbation result given by Lemma 5 to fill out B with positive entries.
and we choose a n+1 , a n+2 . . ., a m in such a way that Since we want to construct the matrix B in (4.1), then we observe that where and from the well known formula det(I + xy T ) = 1 + y T x, we have Thus, for σ = a 2 i , i = 1, . . ., n, we have

Proof.
Let α > 0 such that e σ 1 = σ 1 − α > σ 2 .Then from Theorem ?? there exists an m × n nonnegative matrix B of the form (4.1) with singular values e σ 1 , σ 2 , .., σ n .The matrix BB T is irreducible nonnegative and therefore its Perron eigenvector u 1 (left singular vector of B corresponding to e σ 1 ) is positive and so is the right singular vector v 1 .Hence, from Lemma 5, A = B + αu 1 v T 1 is an m × n positive matrix with singular values σ 1 , σ 2 , ..., σ n . 2 We apply Lemma 2 to the matrices B 1 and

2 ! 2 ! 2 (B n− 1 2
for odd n, where u and v are the unitary eigenvectors of B n−2 for odd n) and B n 2 (B n+1 2

== e b 1 = e b 1 and
e a i , i = 2, ..., n.Hence, given the real positive numbersσ 1 > σ 2 ≥ • • • ≥ σ n > 0,from Lemma 4, we may construct a matrix e C of the form (3.1), with singular values σ 1 , . . ., σ n , where the e b i are given by (1.3) and (1.4) and can be chosen as positive.Clearly, the e a i can also be chosen positive.Then, if e C = e C, we have ³ e C ´1i = b i ³ e b 1 − e a i b1 − a i = e b i , i = 2, .., n and b i = (b 1 − a i ) (e b 1 − e a i ) e b i i = 2, ..., n.