In this paper, we explore the inverse spectral problem for a kind of pseudo-symmetric matrix A(ε,c,s) that corresponds to a graph of a banana tree when all ε=1. The problem of reconstructing pseudo-symmetric matrices from their spectra was considered. The sufficient and necessary conditions for the existence of these matrices are provided, and an algorithm and numerical example were provided to illustrate their characteristics. In addition, the effectiveness of the solution was validated using MatlabR2021b.
The inverse eigenvalue problem for pseudo-symmetric matrices (PIEP) is the issue of reconstructing a matrix with a prespecified structure from given eigendata. For example, Xu [
Given an n × n symmetric matrices A = ( a i j ) n × n , the graph of A, denoted as G ( A ) , is a graph that comprises vertex set V ( G ) = { 1 , 2 , ⋯ , n − 1 , n } and edge set E ( G ) = { i j : a i j ≠ 0 , i ≠ j } . For a graph G with n vertices, S ( G ) denotes the set of all n × n symmetric matrices which have G as their graph. An acyclic matrix is a matrix whose graph forms a double starlike tree [
The focal point of this paper is to investigate the inverse eigenvalue problem for pseudo-symmetric matrices with the graph as an ( c , s ) -banana tree when all ϵ = 1 and referred to as PIEPB ( ϵ , c , s ) . An ( c , s ) -banana tree, is a graph created by attaching one leaf of each of c copies of a s-star graph with a single root vertex that is distinct from all the stars, where c ≥ 1 , s ≥ 3 and n = c s + 1 , marked as follows: the root vertex is labeled by 1, the vertices of distance 1 from the root vertex as the intermediate vertices is labeled by ( i − 1 ) s + 2 , the center of every ( S s ) is labeled by ( i − 1 ) s + 3 and leaves of the center is labeled by j = ( i − 1 ) s + 4 , ⋯ , i s + 1 , i = 1 , 2 , ⋯ , c (
A 9 = [ a 1 b 1 , 2 b 1 , 6 b 1 , 2 a 2 b 2 , 3 b 2 , 3 a 3 b 3 , 4 b 3 , 5 b 3 , 4 a 4 b 3 , 5 a 5 b 1 , 6 a 6 b 6 , 7 b 6 , 7 a 7 b 7 , 8 b 7 , 9 b 7 , 8 a 8 b 7 , 9 a 9 ] ,
Based on the inverse spectral problem for matrices whose graph is a banana tree, we improve their matrices and continue to study the inverse spectral problem for the pseudo-symmetric matrices. For example pseudo-symmetric matrix A 9 ∈ A ( ϵ , 2 , 4 ) is of the following form
A 9 = [ a 1 b 1 , 2 b 1 , 6 ϵ 1 , 2 b 1 , 2 a 2 b 2 , 3 ϵ 2 , 3 b 2 , 3 a 3 b 3 , 4 b 3 , 5 ϵ 3 , 4 b 3 , 4 a 4 ϵ 3.5 b 3 , 5 a 5 ϵ 1 , 6 b 1 , 6 a 6 b 6 , 7 ϵ 6 , 7 b 6 , 7 a 7 b 7 , 8 b 7 , 9 ϵ 7 , 8 b 7 , 8 a 8 ϵ 7 , 9 b 7 , 9 a 9 ] ,
where main diagonal a = ( a 1 , a 2 , ⋯ , a 9 ) ∈ ℝ and the two first lower and upper off-diagonals b − 1 = ( b 1 , 2 , ⋯ , b 7 , 9 ) and b 1 = ( ϵ 1 , 2 b 1 , 2 , ⋯ , ϵ 7 , 9 b 7 , 9 ) , with all b > 0 and all ϵ ∈ { 1 , − 1 } . If all ϵ = 1 , the pseudo-symmetric matrix A ( ϵ , c , s ) changed to the symmetric matrix A ( c , s ) .
PIEPB ( ϵ , c , s ) . Given different numbers
{ λ n ( 1 ) , ⋯ , λ j ( 1 ) , ⋯ , λ 1 ( 1 ) , ⋯ , λ j ( j ) , ⋯ , λ n ( n ) } ∈ ℝ ,
construct an n × n matrix A n ∈ A ( ϵ , c , s ) such that λ j ( 1 ) and λ j ( j ) are the eigenvalues of A j , j = 1 , 2 , ⋯ , n , the leading principal submatrix of A, respectively.
The structure of this work is organized as follows. In Sect.2, some properties of pseudo-symmetric matrix A ( ϵ , c , s ) were given. In Sect.3, the necessary and sufficient conditions for the solvability of an inverse eigenvalue problems of pseudo-symmetric matrix A ( ϵ , c , s ) . In Sect.4, the evolutionary plots of the distribution of eigenvalues of low-order leading principal submatrix were drawn through algorithm and numerical simulation, indicating the accuracy of the result. Finally, in Sect.5, the conclusions are presented.
Lemma 2.1. Let A n ∈ A ( ϵ , c , s ) and let A j be the j × j leading principal submatrix of A, with characteristic polynomial φ j ( λ ) = det ( λ I j − A j ) , j = 1 , ⋯ , n . Then the sequence { φ j ( λ ) } j = 1 n satisfies the recurrence formulae
φ 1 ( λ ) = λ − a 1 ,
φ j ( λ ) = ( λ − a j ) φ j − 1 ( λ ) − ϵ 1 , j b 1 , j 2 det ( B j λ ) , j = ( i − 1 ) s + 2 , i = 1 , 2 , ⋯ , c ,
φ j ( λ ) = ( λ − a j ) φ j − 1 ( λ ) − ϵ ( j − 1 ) , j b ( j − 1 ) , j 2 φ j − 2 ( λ ) , j = ( i − 1 ) s + 3 , i = 1 , 2 , ⋯ , c ,
φ j ( λ ) = ( λ − a j ) φ j − 1 ( λ ) − ϵ ( i − 1 ) s + 3 , j b ( i − 1 ) s + 3 , j 2 φ ( i − 1 ) s + 2 ( λ ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ − a k ) ,
j = ( i − 1 ) s + 4 , ⋯ , i s + 1 , i = 1 , 2 , ⋯ , c .
where B j λ is the submatrix row and columns 2 , 3 , ⋯ , n − 1 of the matrix λ I n − A n , φ 0 ( λ ) = 1 and ∏ k = ( i − 1 ) s + 4 j − 1 ( λ − a k ) = 1 when j = ( i − 1 ) s + 4 .
Lemma 2.2. Let A n ∈ A ( ϵ , c , s ) , then φ j ( λ ) and φ j − 1 ( λ ) have no common root.
Proof. Supposing that λ 0 is a common root of φ j ( λ ) and φ j − 1 ( λ ) , by Lemma 2.1, we get that λ 0 is also a root of φ j − 2 ( λ ) , and so we have φ 0 ( λ ) = 0 ≠ 1 , Which is a contradiction. Hence φ j ( λ ) and φ j − 1 ( λ ) have no common root.
Lemma 2.3. ( [
(1) If μ < λ 1 , we have that ( − 1 ) n φ ( μ ) > 0 ,
(2) If μ > λ n , we have that φ ( μ ) > 0 .
Lemma 2.4. ( [
λ n ( 1 ) < λ n − 1 ( 1 ) < ⋯ < λ 2 ( 1 ) < λ 1 ( 1 ) < λ 2 ( 2 ) < λ 3 ( 3 ) < ⋯ < λ n − 1 ( n − 1 ) < λ n ( n ) ,
and
λ j ( 1 ) < a i < λ j ( j ) , i = 2 , ⋯ , j ; j = 2 , ⋯ , n .
Lemma 2.5. The sufficient and necessary conditions for the existence of symmetric matrix A n ∈ A ( c , s ) whose graph is an ( c , s ) -banana tree, such that λ j ( 1 ) and λ j ( j ) are the smallest and largest eigenvalue of A j , the leading principal submatrix of A, respectively, is
λ n ( 1 ) < λ n − 1 ( 1 ) < ⋯ < λ 3 ( 1 ) < λ 2 ( 1 ) < λ 1 ( 1 ) < λ 2 ( 2 ) < ⋯ < λ n − 1 ( n − 1 ) < λ n ( n ) . (1)
Proof. Sufficiency: suppose that { λ j ( 1 ) , λ j ( j ) } j = 1 n satisfies (1) for j = 1 , ⋯ , n . To prove the existence of a symmetric matrix A n ∈ A ( c , s ) with the required properties is equivalent to going to prove that the system
{ P j ( λ j ( 1 ) ) = 0 P j ( λ j ( j ) ) = 0 , j = 1 , ⋯ , n , (2)
For j = 1 , it is immediate that a 1 = λ 1 ( 1 ) .
According to Lemma 2.1, ϵ 1 , j = 1 for j = ( i − 1 ) s + 2 , i = 1 , 2 , ⋯ , c , system (2) can be written as
{ φ j ( λ j ( 1 ) ) = ( λ j ( 1 ) − a j ) φ j − 1 ( λ j ( 1 ) ) − b 1 , j 2 det ( B j λ j ( 1 ) ) = 0 φ j ( λ j ( j ) ) = ( λ j ( j ) − a j ) φ j − 1 ( λ j ( j ) ) − b 1 , j 2 det ( B j λ j ( j ) ) = 0 , j = ( i − 1 ) s + 2 , i = 1 , 2 , ⋯ , c ,
by condition (1) and Lemma 2.3, we get ( − 1 ) j − 1 φ j − 1 ( λ j ( 1 ) ) > 0 , det ( B j λ j ( 1 ) ) > 0 , φ j − 1 ( λ j ( j ) ) > 0 , and ( − 1 ) j − 2 det ( B j λ j ( 1 ) ) > 0 , then
a j = λ j ( 1 ) φ j − 1 ( λ j ( 1 ) ) det ( B j λ j ( j ) ) − λ j ( j ) φ j − 1 ( λ j ( j ) ) det ( B j λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) det ( B j λ j ( j ) ) − φ j − 1 ( λ j ( j ) ) det ( B j λ j ( 1 ) ) ,
b 1 , j 2 = ( λ j ( j ) − λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 1 ( λ j ( j ) ) φ j − 1 ( λ j ( 1 ) ) det ( B j λ j ( j ) ) − φ j − 1 ( λ j ( j ) ) det ( B j λ j ( 1 ) ) > 0 .
According to Lemma 2.1, ϵ ( j − 1 ) , j = 1 for j = ( i − 1 ) s + 3 , i = 1 , 2 , ⋯ , c , system (2) can be written as
{ φ j ( λ j ( 1 ) ) = ( λ j ( 1 ) − a j ) φ j − 1 ( λ j ( 1 ) ) − b ( j − 1 ) , j 2 φ j − 2 ( λ j ( 1 ) ) = 0 φ j ( λ j ( j ) ) = ( λ j ( j ) − a j ) φ j − 1 ( λ j ( j ) ) − b ( j − 1 ) , j 2 φ j − 2 ( λ j ( j ) ) = 0 , j = ( i − 1 ) s + 3 , i = 1 , 2 , ⋯ , c
by condition (1) and Lemma 2.3, we have ( − 1 ) j − 1 φ j − 1 ( λ j ( 1 ) ) > 0 and φ j − 1 ( λ j ( j ) ) > 0 then
a j = λ j ( 1 ) φ j − 1 ( λ j ( 1 ) ) φ j − 2 ( λ j ( j ) ) − λ j ( j ) φ j − 1 ( λ j ( j ) ) φ j − 2 ( λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 2 ( λ j ( j ) ) − φ j − 1 ( λ j ( j ) ) φ j − 2 ( λ j ( 1 ) ) ,
b ( j − 1 ) , j 2 = λ j ( j ) − λ j ( 1 ) φ j − 1 ( λ j ( 1 ) ) φ j − 1 ( λ j ( j ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 2 ( λ j ( j ) ) − φ j − 1 ( λ j ( j ) ) φ j − 2 ( λ j ( 1 ) ) > 0 .
According to Lemma 2.1, ϵ ( i − 1 ) s + 3 , j = 1 for j = ( i − 1 ) s + 4 , ⋯ , i s + 1 , i = 1 , 2 , ⋯ , c , system (2) can be written as
{ φ j ( λ j ( 1 ) ) = ( λ j ( 1 ) − a j ) φ j − 1 ( λ j ( 1 ) ) − b ( i − 1 ) s + 3 , j 2 φ ( i − 1 ) s + 2 ( λ j ( 1 ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( 1 ) − a k ) = 0 φ j ( λ j ( j ) ) = ( λ j ( j ) − a j ) φ j − 1 ( λ j ( j ) ) − b ( i − 1 ) s + 3 , j 2 φ ( i − 1 ) s + 2 ( λ j ( j ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( j ) − a k ) = 0 ,
by condition (1), Lemma 2.3 and Lemma 2.4, we get ( − 1 ) j − 1 φ j − 1 ( λ j ( 1 ) ) > 0 , φ j − 1 ( λ j ( j ) ) > 0 , ( − 1 ) ( i − 1 ) s + 2 φ ( i − 1 ) s + 2 ( λ j ( 1 ) ) > 0 , φ ( i − 1 ) s + 2 ( λ j ( j ) ) > 0 , ( − 1 ) j − ( i − 1 ) s − 4 ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( 1 ) − a k ) > 0 , ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( j ) − a k ) > 0 , then
a j = λ j ( 1 ) φ j − 1 ( λ j ( 1 ) ) φ ( i − 1 ) s + 2 ( λ j ( j ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( j ) − a k ) − λ j ( j ) φ j − 1 ( λ j ( j ) ) φ ( i − 1 ) s + 2 ( λ j ( 1 ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( 1 ) − a k ) φ j − 1 ( λ j ( 1 ) ) φ ( i − 1 ) s + 2 ( λ j ( j ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( j ) − a k ) − φ j − 1 ( λ j ( j ) ) φ ( i − 1 ) s + 2 ( λ j ( 1 ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( 1 ) − a k ) ,
b ( i − 1 ) s + 3 , j 2 = ( λ j ( h ) − λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 1 ( λ j ( j ) ) φ j − 1 ( λ j ( 1 ) ) φ ( i − 1 ) s + 2 ( λ j ( j ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( j ) − a k ) − φ j − 1 ( λ j ( j ) ) φ ( i − 1 ) s + 2 ( λ j ( 1 ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( 1 ) − a k ) > 0 .
Necessity: Assuming that there exists a unique matrix A n ∈ A ( c , s ) such that λ j ( 1 ) and λ j ( j ) are the smallest and maximal eigenvalues of their corresponding principal submatrices, respectively, and by Cauchy's alternating theorem and Lemma 2.4, (1) holds.
Theorem 3.1. Let different numbers { λ n ( 1 ) , λ n − 1 ( 1 ) , ⋯ , λ 1 ( 1 ) , λ 2 ( 2 ) , ⋯ , λ n ( n ) } ∈ ℝ , if
φ j − 1 ( λ j ( 1 ) ) det ( B j λ j ( j ) ) − φ j − 1 ( λ j ( j ) ) det ( B j λ j ( 1 ) ) ≠ 0 , (3)
φ j − 1 ( λ j ( 1 ) ) φ j − 2 ( λ j ( j ) ) − φ j − 1 ( λ j ( j ) ) φ j − 2 ( λ j ( 1 ) ) ≠ 0 , (4)
φ j − 1 ( λ j ( 1 ) ) φ ( i − 1 ) s + 2 ( λ j ( j ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( j ) − a k ) − φ j − 1 ( λ j ( j ) ) φ ( i − 1 ) s + 2 ( λ j ( 1 ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( 1 ) − a k ) ≠ 0 , (5)
where j = 2 , 3 , ⋯ , n , then there exists a matrix A n ∈ A ( ϵ , c , s ) , such that λ j ( 1 ) and λ j ( j ) are, respectively, the eigenvalues of A j , j = 1 , ⋯ , n .
Proof. Sufficiency: suppose the conditions (3) (4) and (5) hold. To prove the existence of a pseudo-symmetric matrix A n ∈ A ( ϵ , c , s ) is equivalent to proving that
{ φ j ( λ j ( 1 ) ) = 0 φ j ( λ j ( j ) ) = 0 , j = 1 , ⋯ , n , (6)
When j = 1 , it is immediate that a 1 = λ 1 ( 1 ) .
According to Lemma 2.1 for j = ( i − 1 ) s + 2 , i = 1 , 2 , ⋯ , c , system (6) can be written as
{ φ j ( λ j ( 1 ) ) = ( λ j ( 1 ) − a j ) φ j − 1 ( λ j ( 1 ) ) − ϵ 1 , j b 1 , j 2 det ( B j λ j ( 1 ) ) = 0 φ j ( λ j ( j ) ) = ( λ j ( j ) − a j ) φ j − 1 ( λ j ( j ) ) − ϵ 1 , j b 1 , j 2 det ( B j λ j ( j ) ) = 0 , j = ( i − 1 ) s + 2 , i = 1 , 2 , ⋯ , c ,
by Lemma 2.3, we have
a j = λ j ( 1 ) φ j − 1 ( λ j ( 1 ) ) det ( B j λ j ( j ) ) − λ j ( j ) φ j − 1 ( λ j ( j ) ) det ( B j λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) det ( B j λ j ( j ) ) − φ j − 1 ( λ j ( j ) ) det ( B j λ j ( 1 ) ) , (7)
according to Lemma 2.2, ( λ j ( j ) − λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 1 ( λ j ( j ) ) ≠ 0 , added to condition (3), we get α ≠ 0 , and
b 1 , j = | α | > 0 , ϵ 1 , j = α | α | . (8)
According to Lemma 2.1 for j = ( i − 1 ) s + 3 , i = 1 , 2 , ⋯ , c , system (6) can be written as
{ φ j ( λ j ( 1 ) ) = ( λ j ( 1 ) − a j ) φ j − 1 ( λ j ( 1 ) ) − ϵ ( j − 1 ) , j b ( j − 1 ) , j 2 φ j − 2 ( λ j ( 1 ) ) = 0 φ j ( λ j ( j ) ) = ( λ j ( j ) − a j ) φ j − 1 ( λ j ( j ) ) − ϵ ( j − 1 ) , j b ( j − 1 ) , j 2 φ j − 2 ( λ j ( j ) ) = 0 , j = ( i − 1 ) s + 3 , i = 1 , 2 , ⋯ , c ,
by Lemma 2.3, we get
a j = λ j ( 1 ) φ j − 1 ( λ j ( 1 ) ) φ j − 2 ( λ j ( j ) ) − λ j ( j ) φ j − 1 ( λ j ( j ) ) φ j − 2 ( λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 2 ( λ j ( j ) ) − φ j − 1 ( λ j ( j ) ) φ j − 2 ( λ j ( 1 ) ) , (9)
ϵ ( j − 1 ) , j b ( j − 1 ) , j 2 = ( λ j ( j ) − λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 1 ( λ j ( j ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 2 ( λ j ( j ) ) − φ j − 1 ( λ j ( j ) ) φ j − 2 ( λ j ( 1 ) ) = β .
according to Lemma 2.2, ( λ j ( j ) − λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 1 ( λ j ( j ) ) ≠ 0 , added to condition (4), we get β ≠ 0 , and
b ( j − 1 ) , j = | β | > 0 , ϵ ( j − 1 ) , j = β | β | . (10)
According to Lemma 2.1 for j = ( i − 1 ) s + 4 , ⋯ , i s + 1 , i = 1 , 2 , ⋯ , c , system (6) can be written as
{ φ j ( λ j ( 1 ) ) = ( λ j ( 1 ) − a j ) φ j − 1 ( λ j ( 1 ) ) − ϵ ( i − 1 ) s + 3 , j b ( i − 1 ) s + 3 , j 2 φ ( i − 1 ) s + 2 ( λ j ( 1 ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( 1 ) − a k ) = 0 φ j ( λ j ( j ) ) = ( λ j ( j ) − a j ) φ j − 1 ( λ j ( j ) ) − ϵ ( i − 1 ) s + 3 , j b ( i − 1 ) s + 3 , j 2 φ ( i − 1 ) s + 2 ( λ j ( j ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( j ) − a k ) = 0 ,
by Lemma 2.3, we have
a j = λ j ( 1 ) φ j − 1 ( λ j ( 1 ) ) φ ( i − 1 ) s + 2 ( λ j ( j ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( j ) − a k ) − λ j ( j ) φ j − 1 ( λ j ( j ) ) φ ( i − 1 ) s + 2 ( λ j ( 1 ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( 1 ) − a k ) φ j − 1 ( λ j ( 1 ) ) φ ( i − 1 ) s + 2 ( λ j ( j ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( j ) − a k ) − φ j − 1 ( λ j ( j ) ) φ ( i − 1 ) s + 2 ( λ j ( 1 ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( 1 ) − a k ) , (11)
ϵ ( i − 1 ) s + 3 , j b ( i − 1 ) s + 3 , j 2 = ( λ j ( j ) − λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 1 ( λ j ( j ) ) φ j − 1 ( λ j ( 1 ) ) φ ( i − 1 ) s + 2 ( λ j ( j ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( j ) − a k ) − φ j − 1 ( λ j ( j ) ) φ ( i − 1 ) s + 2 ( λ j ( 1 ) ) ∏ k = ( i − 1 ) s + 4 j − 1 ( λ j ( 1 ) − a k ) = θ .
according to Lemma 2.2, ( λ j ( j ) − λ j ( 1 ) ) φ j − 1 ( λ j ( 1 ) ) φ j − 1 ( λ j ( j ) ) ≠ 0 , added to condition (5), we get θ ≠ 0 , and
b ( i − 1 ) s + 3 , j = | θ | > 0 , ϵ ( i − 1 ) s + 3 , j = θ | θ | . (12)
Conversely. Suppose that PIEPB ( ϵ , c , s ) has a unique solution, i.e., the system (6) has solutions a j , b and satisfies all b > 0 . In the same way that we proved the sufficiency, we get (7)-(12). Thus (3), (4) and (5) hold.
| j | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|
| λ j ( 1 ) | 1.0000 | −2.0000 | −3.0000 | −4.0000 | −5.0000 | −6.0000 | −7.0000 | −8.0000 | −9.0000 |
| λ j ( j ) | 1.0000 | 2.5000 | 4.0000 | 4.5000 | 5.0000 | 6.0000 | 7.0000 | 8.0000 | 9.0000 |
A 9 = [ 1.0000 1.9364 0 0 0 5.7852 0 0 0 1.9364 0 2.8722 0 0 0 0 0 0 0 2.8722 1.0000 3.1378 3.5966 0 0 0 0 0 0 3.1378 0 0 0 0 0 0 0 0 3.5966 0 − 1.2364 0 0 0 0 5.7852 0 0 0 0 − 1.4279 4.1556 0 0 0 0 0 0 0 4.1556 2.0115 5.7833 4.9181 0 0 0 0 0 0 5.7833 − 2.2485 0 0 0 0 0 0 0 4.9181 0 − 2.0778 ] .
Recompute that: the eigenvalues of its the leading principal submatrices A j , j = 1 , ⋯ , 9 , are:
Λ ( A 1 ) = { 1.0000 _ } ,
Λ ( A 2 ) = { − 1.5000 _ , 2.5000 _ } ,
Λ ( A 3 ) = { − 3.0000 _ , 1.0000 , 4.0000 _ } ,
Λ ( A 4 ) = { − 4.0000 _ , − 0.9478 , 1.9478 , 5.0000 _ } ,
Λ ( A 5 ) = { − 5.5000 _ , − 1.4288 , − 0.4516 , 2.1439 , 6.0000 _ } ,
Λ ( A 6 ) = { − 6.5000 _ , − 5.2765 , − 0.7170 , − 0.0754 , 5.4045 , 6.5000 _ } ,
Λ ( A 7 ) = { − 7.5000 _ , − 5.3941 , − 0.8648 , − 0.2904 , 2.0710 , 5.8255 , 7.5000 _ } ,
Λ ( A 8 ) = { − 8.5000 _ , − 5.5735 , − 4.7300 , − 0.7334 , − 0.1191 , 4.2519 , 6.0028 , 8.5000 _ } ,
Λ ( A 9 ) = { − 9.5000 _ , − 5.8133 , − 5.0707 , − 2.1488 , − 0.7265 , − 0.1021 , 4.7635 , 6.1190 , 9.5000 _ } .
Given different real number λ 6 ( 1 ) = 4 , λ 5 ( 1 ) = 5 , λ 4 ( 1 ) = − 0.1 , λ 3 ( 1 ) = 2 , λ 2 ( 1 ) = 0.1 , λ 1 ( 1 ) = 1 , λ 2 ( 2 ) = − 8 , λ 3 ( 3 ) = 3 , λ 4 ( 4 ) = 0.6 , λ 5 ( 5 ) = − 7 , λ 6 ( 6 ) = − 9 . construct a matrix A 6 ∈ A ( ϵ , 1 , 5 ) , such that λ j ( 1 ) and λ j ( j ) are the eigenvalues of A j , j = 1 , 2 , ⋯ , 6 , respectively. In particular, this matrix corresponds to the graph is an (1, 5)-banana tree when all ϵ = 1 (
By applying algorithm 1 and matlabR2021b, we compute the following matrix
A 6 = [ 1.0000 2.8460 0 0 0 0 − 2.8460 − 8.9000 9.9680 0 0 0 0 9.9680 − 3.2295 2.7638 2.3389 4.4658 0 0 2.7638 0.0157 0 0 0 0 2.3389 0 − 6.9523 0 0 0 − 4.4658 0 0 − 8.8135 ] .
Recompute that: the eigenvalues of its the leading principal submatrices A j , j = 1 , ⋯ , 6 , are:
Λ ( A 1 ) = { 1.0000 _ } .
Λ ( A 2 ) = { 0.1000 _ , − 8.0000 _ } .
Λ ( A 3 ) = { 3.0000 _ , 2.0000 _ , − 16.1295 } .
Λ ( A 4 ) = { 4.6961 , 0.6000 _ , − 0.1000 _ , − 16.3100 } .
Λ ( A 5 ) = { 5.0000 _ , 0.5656 , − 0.0974 , − 7.0000 _ , − 16.5343 } .
Λ ( A 6 ) = { 4.0000 _ , 0.6886 , − 0.1053 , − 7.0048 , − 9.0000 _ , , − 15.4581 } .
Based on matrices whose graph is a banana tree, we study the inverse spectral problem for pseudo-symmetric matrices. It is obviously difficult to reconstruct the pseudo-symmetric matrix given the eigenvalues, and in order to achieve this deficiency, other constraints are added on top of the eigendata, which in turn enables the reconstruction of the pseudo-symmetric matrix. We obtained sufficient necessary conditions to solve the problem.
This research was supported by the National Nature Science Foundation of China and Shanxi Province Basic Research Program Grant (202203021211088).
The authors declare no conflicts of interest.
Su, R., Lei, Y.J. and Li, F.H. (2023) An Inverse Eigenproblems for Pseudo-Symmetric Matrices. Open Access Library Journal, 10: e11067. https://doi.org/10.4236/oalib.1111067