Solvability of Inverse Eigenvalue Problem for Dense Singular Symmetric Matrices

Abstract

Given a list of real numbers      , we determine the conditions under which  will form the spectrum of a dense n × n singular symmetric matrix. Based on a solvability lemma, an algorithm to compute the elements of the matrix is derived for a given list and dependency parameters. Explicit computations are performed for  5 n  and to illustrate the result.

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“…The free variables are a 13 , a 14 , a 15 , a 34 , a 35 and a 45 trA (5,4) = λ 1 + λ 2 + λ 3 + λ 4 = a 11 (1 + |k| 2 ) + a 33 + a 44 + a 55 . Thus a 11 =λ 1 /(1 + |k| 2 ), λ 2 = a 33 , λ 3 = a 44 and λ 4 = a 55 .…”

“…For instance, given λ 1 = 5, λ 2 = 2, λ 3 = -1, λ 4 = 3, k = -2i, a 13 = 1 + i, a 14 = i, a 15 = 2i, a 34 = 1-2i, a 35 = 1 + 2i, and a 45 = 1 + 3i, a singular Hermitian matrix A (5,4) is generated below:…”

“…A (4,3) = Case 4: Lastly we present 5 × 5 singular Hermitian matrices of rank 3. Then A (5,3) will be of the form: tr(A (5,3) ) = λ 1 + λ 2 + λ 3 = a 11 (1 + |k 1 | 2 + |k 1 | 2 |k 2 | 2 ) + a 44 + a 55 . This implies that a 11 = λ 1 /1 + |k 1 | 2 + |k 1 | 2 |k 2 | 2 , λ 2 = a 44 and λ 3 = a 55 .…”

Solution of the Inverse Eigenvalue Problem for Certain (Anti-) Hermitian Matrices Using Newton's Method

“…Most research effort have been directed at solving the inverse eigenvalue problem for nonsingular symmetric matrices (Chu & Golub, 2005;Gladwell, 2004;Deakin & Luke, 1992;Chu, 1995). Recently, however, the case of singular symmetric matrices of arbitrary order and rank has been virtually solved provided linear dependency relations are specified Aidoo, Gyamfi, Ackora-Prah, & Oduro, 2013). It is plausible to expect solutions of the inverse eigenvalue problem in a sufficiently small neighborhood of any such Hermitian matrix.…”

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“…A singular Hermitianmatrix A (4,3)is obtained if we let λ 1 = 2, λ 2 = -1, λ 3 = 5, k = -i, a 13 = 2 + i, a 14 = 2i and a 34 = 1-3i:A (4,3) = Lastly we present 5 × 5 singular Hermitian matrices of rank 3. Then A(5,3) will be of the form: tr(A (5,3) ) = λ 1 + λ 2 + λ 3 = a 11 (1 + |k 1 | 2 + |k 1 | 2 |k 2 | 2 ) + a 44 + a 55 . This implies that a 11 = λ 1 /1 + |k 1 | 2 + |k 1 | 2 |k 2 | 2 , λ 2 = a 44 and λ 3 = a 55 .…”