Solvability of Inverse Eigenvalue Problem for
Dense Singular Symmetric Matrices
Anthony Y. Aidoo1*, Kwasi Baah Gyamfi2, Joseph Ackora-Prah2, Francis T. Oduro2 1
Department of Mathematics and Computer Science, Eastern Connecticut State University, Willimantic, USA 2
Department of Mathematics, Kwame Nkrumah University of Science and Technology, Kumasi, Ghana Email: *[email protected]
Received September 6,2012; revised October 23, 2012; accepted November 6, 2012
ABSTRACT
1, , n
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 ma-trix is derived for a given list and dependency parameters. Explicit computations are performed for n5 and
to illustrate the result.
4
r
, ,
Keywords: Inverse Eigenvalue Problem; Dense; Nonnegative; Singular; Symmetric
1. Introduction
The inverse eigenvalue problem (IEP) involves the re-construction of a matrix from prescribed spectral data. In order to limit the number of usually infinitely many solu-tions that are usually possible, if a solution does exists, it is usually required that the matrix constructed preserves a specific structure in addition to the spectral requirement. The nonnegative inverse eigenvalue problem (NIEP) is a special case of the IEP first posed by Kolmogorov in 1937. It involves the determination of the existence of entrywise nonnegative matrices when given a set of complex numbers that make up the spectrum of the yet to be determined matrix. The extensions and related condi-tions have been solved in several different ways. We give a brief review of results for the general inverse eigen-value problem (IEP) and NIEP, as related to our work. A detailed treatment can be found in [1]. We begin with a realization criterion provided by Soto [2] for the NIEP. This result is based on conditions derived by Fiedler [3] and Borobia [4]. A list of real numbers 1 n
1, , n
is said to be realizable if it forms the spectrum of an en-try wise nonnegative matrix.
Theorem 1. Let
1
0 .
be a set of real nu- mbers such that
1 2 p p n
1 2
If there exits a partition s
, ,
k , ,s
with
1 k2 kpk
k k
1, 2,
1
1 1,
1 0,
k 1 2 ,
pk
k k k
1, 2, 3, , ,
2
j j pk
pk
k k k j
k
S j
k
p
k
and for odd
1 1
2 2
min , 0 , 1, 2, , ,
pk pk
k k
S k s
1
0
, 1, 2, , ,
pk j
j
k k k k
Sk
T S k s
1 1
1 1
2 0 1
max p j; max
j
k k s S
L S
and
1
0, 2
, k
s k
T k
L T
satisfying
then is realizable (by a nonnegative matrix with constant row sums).
It has been proved that the above theorem is sufficient for the existence of an symmetric nonnegative matrix with real spectrum
n n
1, 2, ,n
, see [2]. Recently Wu studied the IEP and gave the solvability criterion for its solution given a spectrum of complex numbers in the following two theorems [5].
Theorem 2. For a given list of complex numbers,
1, 2, ,n
, if it has closed property under
plex conjugation, that is contains an element and its complex conjugation, then there must be at least one real matrix A with spectrum .
1
n i i
x
Proof. By the closed property of under complex conjugation, construct the polynomial:
f x .
1 a x an1 n
,a
,
(1)
By multiplying out, combining similar terms, and sim- plifying, the above equation can be written as:
xna xn1f x
1, 2,
a a
n
a
, ,
where n are real numbers. It follows that
1 2 are real numbers and the polynomial (1) has zeros 1 2 n
, ,
a a ,
. Thus, using a a1, 2,,an, the matrix
A can be constructed as:
1 2
0 1 0
0 0 1
0 0 0
n n n
a a a
A
2 1
0 0
0 0
.
0 1
a a
It is straightforward to show that A has spectrum
1, 2,, n
, ,
,
.
Theorem 3. Given a list of real numbers 1 2 n
1
, ,
1
n i i j
i j n i
odd number even number.
n n
, the sufficient condition that there exists at least one nonnegative matrix A with spectrum
is that has the following properties
, 1
1
0,
0,
0,
0, is an
0, is an
i j
i j k
i n
i j n
i j k k
i
n n
Several other results have been obtained for the in-verse eigenvalue problem for Hermitian matrices (see for example [6]). It is obvious from the above theorems and the references that the solution of the IEP or even the NIEP is not unique. In addition, most of the solutions provided in the literature yield only sparse matrices. In this paper, we provide conditions for the solvability of the IEP for singular symmetric matrices. We show that unique solutions exist for matrices of rank one and give the solutions for dense matrices of rank greater than one.
2. Preliminaries
In what follows, we denote the singular
symmet-ric matrix
s a i j, , 1,,n by WiA
R kR
ij
a uch that aij ji An.
thout loss of generality, singularity is achieved by multiplying the first row by prescribed scalars. We de-note a singular symmetric matrix of rank r by n r, . In this case we write i1 i to denote that the
row is k times the first row, where . Finally, we denote the spectrum of n by 1
th i k
A n
,,n
0i
. If the rank of A is r, then we assume for i1,,r
0
, but i , i r 1,,n.
Lemma 1. Let A be a singular symmetric matrix of
rank r. Then there exits an isomorphism between the ele-ments of A and its distinct non-zero eigenvalues if and
only if rank
A 1 2,1
A A 2,1
2,1 11 211 11 2
11 11
1 .
a ka k
a
ka k a k k
A
.
Corollary: The inverse eigenvalue problem has a unique solution for singular symmetric matrices of rank 1 and any prescribed row multiples.
We begin by considering . By definition, is of the form:
, 2
.2 1
Let A
2 0
Since 2,1 is singular of rank 1, it follows that . We have:
2,1
11
1 2
Tr A a k . Therefore 11 2 1 a
k
.
Hence:
2,1 2 2
1 . 1
k k k k
A
A
(2)
Thus 2,1 has been reconstructed for given and prescribed scalar k.
We see from this formula that for any given and parameter k, we can generate any 2 × 2 singular symmet-ric matrix of rank one. For example if k2, 5, we have
2,1
1 2 . 2 4
A
n n
3. IEP for Dense Singular Symmetric
Matrices
We generalize the method above in the following two theorems, first for an singular symmetric matrix of rank 1 and then of rank r, where 1 r n.
R k R
Theorem 4.Given the spectrum and the row
depend-ence relations i1 i i, , where the i’s are nonzero real numbers, the inverse eigenvalue
problem for a
1, , 1 i n k
n n singular symmetric matrix of rank 1 is solvable.
Proof. Given the spectrum n
1, 2,,n
1
, since rank An
0
, it follows from our notation above that
1 and i 0, i 2,,n. Let ki, i 1, ,kn1
be he row multiples. Letting t
1 2 1 2
1 2 1 2 2
2 3 1
2 2 2 2 2 3 1
.
n n n
n
a k k k
k k
k k k k
k k k
2 2 2
1 2 1
1 k k kn .
11 11 1 11 1 2 11
2 2
11 1 11 1 11 1 2 11
2 2 2
11 1 2 11 1 2 11 1 2 3 11 1 ,1
2 2 2
11 1 2 1 11 1 2 1 11 1 2 3 1 11 1
n
i i i
a a k a k k
a k a k a k k a k
a k k a k k a k k k a
a k k k a k k k a k k k k a k
A
Then
,1
2 2 2 2 2 2 11 1 1 2 1 2 3
n
Tr
a k k k k k k
A
Hence
11 2 2 2
1 1 2
1 a
k k k
2 2 2
1 2 1
.
n
k k k
2 n
We note that for , Equation (2) shows that
2 1 k
An1,1
A r
n r, aij
A n2
2 , , , s s s
k i j
k i j
k i j
. Thus by writing the matrix , the
result follows by induction on n. We state the following theorem for the general case where n has rank .
Theorem 5.The inverse eigenvalue problem for an n
× n singular symmetric matrix of rank r is solvable pro-vided that n − r arbitrary parameters are prescribed.
Proof. Let , where . It is obvious that, 1 j 0 1 11 0 1 0 s i ij s i s a a
where 111 2 2 2
1 1 2
1 a
k k k
2 2 2
1 2 n r
k k k
, where 2
a i j n r
and 2 ij nn r a
1, 2, , r
(3)
such that Rik Ri1 i1 where kiR,
and is the row of 2, ,
i n r 1 Ri ith A.
12 1 i n k
1 1 s s s 1
r
Tr A . We see that for :
1
11 2 2 2 2 2 2
1 1 2 1 2 1
.
1 k k k k k kn
2 r a : 1
11 2 2 2
1 1 2
1 a
k k k
2 2 2
1 2 2
.
n
k k k
3
r :
1
11 2 2 2 2 2 2
1 1 2 1 2 3
.
1 n
a
k k k k k k
4
r :
1
11 2 2 2 2 2 2
1 1 2 1 2 4
.
1 n
a
k k k k k k
1 r n
3 n 5
The result follows for any rank .
4. Numerical Examples
In this section, we illustrate the results above with small matrices of size r 4
3,1
A
of rank 1 . We begin with the singular 3 × 3 symmetric matrix of rank 1,
, which is of the form:
1 1 2 2 2 11 1 1 1 2 11
3,1 2 2 2
2 2 2 1 1 2
1 2 1 2 1 2
1
, 1
k k k
a k k k k a
k k k
k k k k k k
A 1 1 k .
, k22,
For 6
3,1
1 1 2
1 1 2 .
2 2 4
A k k
1 1 2 1 2 3
2 2 2
1 1 1 2 1 2 3
11 2 2 2 2 2
4,1
1 2 1 2 1 2 1 2 3 2 2 2 2 2 2 1 2 3 1 2 3 1 2 3 1 2 3
1 k k k k k k
k k k k k k k
a
k k k k k k k k k
k k k k k k k k k k k k
A
, we have:
Similarly, given 1, 2 and k3, we obtain the fol- lowing singular symmetric matrix:
In this case, 11 2 2 2 2 2 2 1 1 2 1 2 3 1
a
k k k k k k
. When 2,
1 3
k , k22 and k34, we obtain the following 4 × 4 singular symmetric matrix of rank one:
4,1
1 3 6 24
311 311 311 311
3 9 18 72
311 311 311 311 .
6 18 36 144
311 311 311 311 24 72 144 576 311 311 311 311
Obviously, the two examples given above can easily be verified for density, symmetry, rank, and that they satisfy the prescribed spectrum. Only the nonzero eigen-value and the k’s have to be known for the solution of this type of IEP. As such, for given nonzero eigenvalue and k’s, the solution is unique.
The examples below clearly illustrate Theorem 5. For an n × n singular symmetric matrix of rank r, n − r arbi-trary values must be prescribed in addition to the nonzero spectrum for the IEP to be solvable. This clearly leads to infinitely many solutions. We illustrate this with the fol-lowing examples beginning with the IEP for n × n singu-lar symmetric matrices of rank 2. A 3,2 is of the form:
11 13 2
11 13 13 33
a ka a
k a ka
a ka a
11 11 3,2 13 ka A
Here,
2
33 1k a
11 1
a k
1 2 11Tr A a
2
and
a331 2
. Hence 1 2
2 11 1 a k
2
1 20. 33a . Thus
2
2 2 2
11 1 11 1 1
a k a k
which yields 1 2 1 k 11 a
and 2 a33. Therefore a13 becomes a free variable. When 1 2, 2 3, k4 and 13 , for example, we obtain the following singular symmetric matrix:
5 a
3,2
17 17 A 2 8 5 17 8 32 20 . 17
5 20 3
A 11 a 1 2 1 3 1 1 1 . r r r i i
In general, the solution of the IEP for n r, leads to the solution of an rth degree polynomial equation in of the form:
2 2 11 12 2 1
1 1 1 1 2 2 1 1 1 2 2 2 1 1 1 0 1 1 1 r n r r r
i n r
i k r
r
i n
k i k k r
r
i n
k i k
a k k
k k a
k k a
k k a
4 r2
2 2 2 2 2 11 1 1 2
2 2 2
1 2 1 1 2 11 1 2
1
1 0.
a k k k
k k k a
(4)To solve the case for and , we deduce from the general polynomial equation above that the following quadratic in holds:
n
11 a
This yields: 1
11 2 2 2
1 1 2
1 a
k k k
, a442
6
and a14 be- comes a free variable. For 1 , 25, k14,
2 3
k and a14 2, we obtain a singular symmetric matrix below:
4,2
6 24 72
2 161 161 161
24 96 288 8
. 161 161 161
72 288 864 24 161 161 161
2 8 24 5
A
5,2
A
2 2 2 2 2 2 2 2 11 1 1 2 1 2 3
2 2 2 2 2 2 1 2 1 1 2 1 2 3 11 1 2
1
1 0.
a k k k k k k
k k k k k k a
Similarly, for , we obtain the following quadratic in a11:
The solution gives: 1
11 2 2 2 2 2 2
1 1 2 1 2 3
1 k k k k k k
2 a
a and
55
where a15 is a free variable. When 1 2
,
2 5
, k13, k25, 3 and a , we obtain a singular symmetric matrix below:
7
k 154
5,2
2 6 30 210
4
11260 11260 11260 11260
6 18 90 630
12
11260 11260 11260 11260
.
30 90 450 3150
60
11260 11260 11260 11260
210 630 3150 22050
420
11260 11260 11260 11260
4 12 60 420 5
A
4,3
A
3 2
3 2 2 2
11 1 2 3 11
2
11 1 2 1 3 2 3 1 2 3
1 1
1 0.
a k a k
a k
By the same method, leads to the following cubic equation:
Solving the above cubic equation we obtain the follow-
ing roots. 2 1
1 11 1 11 2
1
a k a
k
, 2a33 and 3 a44, where a , and a are free variables.
13 14 34
Finally, we want to consider 5 × 5 singular symmetric matrix of rank 4. Using Equation (2), we obtain the
fol-a
lowing quartic equation in a11 where 1, 2, 3 and 4
are the nonzero members of the spectrum.
4
4 2 3
11 1 2 3 4 11
1 2 1 3 1 4 2 3 2 4 3 4 1 2 3 1 2 4 1 3 4 2 3 4 1 2 3 4
1
0.
a k
3 22
2 2
11 2 11 1
1
1
a k
a k
a k
Factoring the above quartic equation, we obtain the
following results: 1
11 2
1 a
k
, 2 a33, 3 a44 and 4 a55
14 a15 a35
34 a
. The free variables are , , , , and a .
13 a a
4, 45
As an example, if we let 2 3 7, 4 11,
, , , , ,
2 a13
k 8 a14 4 a15 5 a35 14 a346 and a45 17
5,4
1.3 2.7 2.6 5.4 8 16
4 8
5 10
A
8 4 5
16 8 10 . 4 6 14 6 7 17 14 17 12
1 r n
1 r n
We provide in the appendix, a program that generates any dense n × n singular symmetric matrix of rank 1 for given row multipliers. The program could be easily modified for rank .
5. Summary and Discussion
The IEP has a variety of applications which include con-trol design, system identification, principal component analysis, geophysics and mechanical systems. Several interpretations and applications of the IEP have been considered. However, the direct IEP involving full ma-trices has received relatively less attention owing to the intractability of the problem for even small sized matri-ces. In this paper, we focused on the IEP for dense sin-gular symmetric matrices. We showed that while unique
solutions exists for an n × n matrices of rank 1, infinitely many solution exist for n × n singular symmetric matri-ces of rank r, where . With our method, an ex-tension to nonsingular matrices for small n, could be-come feasible via a numerical analytic interpretation of the IEP.
6. Acknowledgements
The authors Kwasi Baah Gyamfi and Joseph Ackora- Prah are grateful to Eastern Connecticut State University for the resources provided at the Department of Mathe-matics and Computer Science to enable them to complete this research. Anthony Y. Aidoo acknowledges with grati-tude the support received from 2001 CSU Research Grant for this work.
REFERENCES
[1] M. T. Chu and G. H. Golub, “Inverse Eigenvalue Prob-lem: Theory, Algorithms and Applications,” Oxford Uni-versity Press, Oxford, 2005.
doi:10.1093/acprof:oso/9780198566649.001.0001 [2] R. L. Soto, “Realizability Criterion for Symmetric
Non-negative Inverse Eigenvalue Problem,” Linear Algebra
and Its Applications, Vol. 416, No. 2-3, 2006, pp. 783- 794. doi:10.1016/j.laa.2005.12.021
[3] M. Fiedler, “Eigenvalues of Nonnegative Symmetric Ma-trices,” Linear Algebra and Its Applications, Vol. 9, No. 1, 1974, pp. 119-142. doi:10.1016/0024-3795(74)90031-7 [4] A. Borobia, “On Nonnegative Inverse Eigenvalue
Prob-lem,” Linear Algebra and Its Applications, Vol. 223/224, 1995, pp. 131-140. doi:10.1016/0024-3795(94)00343-C [5] J. Wu, “On Open Problems of Nonnegative Inverse
Ei-genvalues Problem,” Advances in Pure Mathematics, Vol. 1, No. 4, 2011, pp. 128-132.
doi:10.4236/apm.2011.14025
[6] R. Fernandes and C. M. da Focenca, “The Inverse Eigen-value Problem for Hermitian Matrices,” Linear Multilin-ear Algebra, Vol. 57, No. 7, 2009, pp. 673-682.
Appendix
The following program generates an n × n singular symmetric matrix of rank 1
function ()
%UNTITLED2 Summary of function goes here % Detailed explanation goes here
lambda=input('Enter a positive number as trace of the generalized matrix')
k = input('1.Enter m positive numbers that characterize the generalided matrix')
m = length(k);h(1)=1;ss=[]; for i=1:m
h(i+1)=h(i)*k(i); end
h;kk=1;
for i=m:-1:1
kk=kk*(k(i)^2)+1; end
kk;
format short eng a=lambda/kk ss=a*h; mm(1,:)=ss; for i=2:(m+1);
mm(i,:)=k(i-1)*mm((i-1),:); end
mm
%fprintf('The required singular matrix of rank 1 is:\n %d \n',mm)
tr=trace(mm)
