Constructive methods for spectra with three nonzero elements in the nonnegative inverse eigenvalue problem

0.15 MB·

by  Anthony Cronin

Constructive methods for spectra with three nonzero
elements in the nonnegative inverse eigenvalue problem
7
Anthony G. Cronin
1 ∗
0
2

n
a
J Abstract

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1 We present and compare three constructive methods for realizing non-real spectra

with three nonzero elements in the nonnegative inverse eigenvalue problem.
]
A
We also providesome necessary conditions for realizability andnumerical examples.
R
In particular, we utilise the companion matrix.
.
h
t
a AMS Subject Classification: 15A18; 15A29; 47A75; 58C40
m
Keywords: Nonnegative matrices; Inverse eigenvalue problem; Spectral theory
[

1
v 1 Introduction
8
0
7
We say that a list of n complex numbers σ = (λ ,λ ,...,λ ) is realizable if σ is the list
4 1 2 n
0 of eigenvalues (including multiplicities) of an n n entry-wise nonnegative matrix A and
. ×
1 that σ realizes A or that A is a realizing matrix for σ.
0
Let
7
1 s := λk +λk + +λk, k = 1,2,3,...
: k 1 2 ··· n
v
i Then
X
r
a s = tr(Ak) 0, k = 1,2,3,... (1)
k
≥
is a necessary condition for σ to be realizable.
The JLL conditions discovered independently by Johnson [3] and Loewy and London [10]
state that
nm−1s sm, k,m = 1,2,3,... (2)
km ≥ k
A further necessary condition for realizability of σ comes from the Perron-Frobenius the-
orem [1], that is that there exists j with λ real and λ λ , for all i. Such a λ is called
j j i j
≥ | |
the Perron root of σ, usually denoted ρ. The nonnegative inverse eigenvalue problem (or
NIEP) asks for necessary and sufficient conditions on σ in order that σ be the spectrum
of an entry-wise nonnegative n n matrix.
×
∗School of Mathematics and Statistics, University College Dublin, Ireland, [email protected]
The same problem in which we may augment the list σ by adding an arbitrary number
N of zeros was solved by Boyle and Handelman [2]. Using dynamical systems theory and
ergodic theory, they proved the remarkable result that if
(a) σ has a Perron root λ > λ (all j > 1) and
1 j
| |
(b) s 0 for all positive integers k (and s = 0 for some m implies s = 0 for all positive
k m d
≥
divisors d of m)
then
σ := (λ ,...,λ ,0,...,0) (N zeros added)
N 1 n
is realizable for all sufficiently large N.
In this paper we compare three constructive techniques for realizing non-real spectra
with exactly three non-zero elements. Section 2 examines the companion matrix, gives
a surprising result for a realizing companion matrix and provides some examples and
remarks. Section 3 examines a construction due to Laffey [5]. Finally in section 4 we
compare three constructive methods including a multi-block companion matrix in an
attempt to minimize the dimension of realizing matrices with three non-zero element
spectra.
2 Known constructions
While Boyle and Handelman’s result was an existence one, in recent years more attention
has been paid to trying to find a constructive realizing matrix for given spectra [5].
This paper outlines three such constructions and compares each in turn. Our first con-
struction involves the companion matrix.
Definition 1. If f(x) = xn +p xn−1 +p xn−2 + +p , then the companion matrix of
1 2 n
···
f(x) is
0 1 0 ... 0
 0 0 1 ... ... 
A(f) =  ... ... ... ... 0 .
 
 
 0 0 ... 0 1 
 
 p p ... p p 
n n−1 2 1
− − − − 
Note that the companion matrix has the property that det(xI A(f)) = f(x).
−
Clearly, A(f) is nonnegative if and only if p 0 for i = 1,...,n.
i
≤
It is known that spectra of the type σ = (ρ,a ib) where b = 0, ρ a+ib , are realizable
± 6 ≥ | |
if and only if they are realizable by a matrix of the form αI +C where α = ρ+2a 0, I
3 3 ≥ 3
is the 3 3 identity matrix and C is a nonnegative trace zero companion matrix.
×
This was first proved by Loewy and London in [10].
However thisconstructiondoesnotholdforn = 4withspectraoftheformσ = (ρ,a ib,µ)
±
where µ = 0, since for example the list (4,2+i,2 i,3) is realizable by
6 −
8 1 0 0
3
 0 8 1 0 
3
52 1 8 0
 
27 3 3
 
 0 0 0 3 
 
but the list σ 11 = (5, −3 +i, −3 i, 1) is not realizable by C where C is a nonnegative
− 4 4 4 4 − 4
trace zero companion matrix since
5 1 3 2
f(x) = x x x+ +1
− 4 − 4 (cid:18) 4 (cid:19)
(cid:16) (cid:17)(cid:16) (cid:17) (cid:16) (cid:17)
3x2 15x 125
= x4 +
− 8 − 8 256
has its constant term positive and so the companion matrix of f(x) is not nonnegative.
However for µ = 0 we do have the following:
Proposition 2. Suppose σ = (ρ,a ib,0) is realizable, then σ is realizable by αI + C
4
±
where α 0 and C is a nonnegative trace zero companion matrix if and only if ρ 2a.
≥ ≥
Proof. Let
S = ρk +(a+ib)k +(a ib)k, and
k
−
s = (ρ α)k +(a α+ib)k +(a α ib)k +( α)k, for k = 1,2,...
k
− − − − −
where α = ρ+2a.
4
Then s = 0, and by the Newton identities [11], the characteristic polynomial of C is
1
s s
x4 2x2 3x+d
− 2 − 3
where d = det(C) = (ρ α)((a α)2 +b2)α 0.
− − − ≤
Note s = 4S2−S12 0 (using α = ρ+2a), by (2) for n = 4,m = 2,k = 1.
2 4 ≥ 4
So C 0 if and only if s 0.
3
≥ ≥
But s = 3(ρ 2a)(ρ2 +4b2).
3 8 −
Hence σ is realizable as αI +C if and only if ρ 2a.
4
≥
2.1 Examples
One can easily find examples of the form (ρ,a ib) with ρ < 2a, which are realizable.
±
For example (12,9 i) is realizable by
±
10 1 0
A =  0 10 1 
4 2 10
 
and so (12,9 i,0) is realizable by the 4 4 nonnegative matrix A 0 .
1
± × ⊕
Note that we can also find lists (ρ,a ib) such that (ρ,a ib,0) is not realizable but
± ±
(ρ,a ib,0,0) is realizable. The example σ = (21,8 12i,0) is not realizable since it fails
± ±
the JLL condition 4s s2 0. In fact 4s s2 = 245.
2 − 1 ≥ 2 − 1 −
However, σ = (21,8 12i,0,0) is realizable, for example, by αI +C where α = 37 and
1 ± 5 5
C is the trace zero companion matrix of the polynomial
18x3 9892x2 2532243x 335969028
x5 .
− 5 − 25 − 125 − 3125
Similarly σ = (5,2 3i) has 4s s2 = 4(15) 92 = 21 < 0 and 5s s2 = 6
± 2 − 1 − − 2 − 1 −
(while 6s s2 > 0) but σ with three zeros added is realizable by 3I +C where C is the
2− 1 2 6
nonnegative 6 6 trace zero companion matrix of
×
3x4 1197x2 351x 6993
x6 2x3 .
− 4 − − 16 − 2 − 64
2.2 Remarks
Additional remarks on non-real spectra with exactly three non-zero entries
1. Suppose that σ = (ρ,a ib), where ρ,a,b are real, i = √ 1, b > 0 and ρ √a2 +b2.
± − ≥
Write s = ρk +(a+ib)k +(a ib)k.
k
−
If a 0, then σ with N zeros appended is realizable if and only if (N+3)s s2 where N
≤ 2 ≥ 1
is the smallest positive integer with this property. In this case σ is realizable by αI +C
N+3
where α = ρ+2a and C is a nonnegative (N +3) (N +3) companion matrix with trace
N+3 ×
zero [8].
1
2. Suppose a > 0. By replacing σ by σ′ = (kρ,ka ikb) where k = 1 2, we may
± |a2+b2|
assume that a2+b2 = 1, since X is a realizing matrix for σ if and only if kX is a realizing
matrix for σ′. So assume a > 0, ρ 1 and σ = (ρ,a ib) where b > 0 and a2 +b2 = 1.
≥ ±
We may write a = cosθ, b = sinθ where 0 θ < π. Then σ with sufficiently many zeros
≤ 2
added is the spectrum of a nonnegative matrix if and only if ρk + 2coskθ > 0 for all
positive integers k by Boyle and Handelman [2]. Because of the periodicity of the cosine
function, this condition can be checked in finitely many steps. In particular, when θ = π
l
1
for some positive integer l, let ρ > ρ0 = max0≤k<⌊2l⌋+1(2cos klπ)l−k.
If σ is realized by an M M nonnegative matrix then the following JLL condition must
×
hold
kπ π
Ml−k−1(ρl−k 2cos ) (ρ+2cos )l−k.
− l ≥ l
Choosing ρ ρ to be a sufficiently small positive number, we can realize such σ by an
0
−
M M nonnegativematrixwiththeminimumdimension M satisfyingthisJLLinequality.
×
3. Suppose that σ = (ρ,a ib), where a > 0,b > 0,a2 + b2 = 1, is realizable with
±
sufficiently many zeros added. One can ask whether σ (with sufficiently many zeros
added) is realizable by αI +C where α > 0 and C is a nonnegative N N trace zero
N N N
×
companion matrix for some positive integer N. When N = 3, it is well-known ([10]) that
this realization is possible if and only if ρ a+b√3 (since s = 0 here).
1
≥
When N = 4, Proposition 2 says ρ 2a is necessary and sufficient.
≥
We now present:
Proposition 3. Suppose that σ = (ρ,a ib), a > 0,b > 0,a2 + b2 = 1, is realizable
±
with sufficiently many zeros added. If σ (with sufficiently many zeros added) is realizable
by αI + C where C is a nonnegative N N trace zero companion matrix for all
N N N
×
sufficiently large N, then ρ 2a (so σ with only one zero added is realizable).
≥
Proof. Let σ = (ρ,a+ib,a ib,0,...,0 .
N
− }
N−3 zeros
Let α = ρ+2a, and S = tr(C )k, s|o {z }
N k N
S = (ρ α)k +(a α+ib)k +(a α ib)k +( α)k +( α)k + +( α)k.
k
− − − − − − ··· −
In order that σ with N 3 zeros added be realizable in the desired form for all sufficiently
−
large N, we require C to be nonnegative i.e. the coefficients P must be non-positive for
N j
j = 1,2,...N, in det(xI C )
N
−
= xN +P xN−1 +P xN−2 + +P
1 2 N
···
2((N 2)a ρ) ρ+2a
= (x τ)(x2 − − x+1)(x+ )N−3
− − N N
where τ = (N−1)ρ−2a.
N
The coefficient of xN−1 is zero since C has trace 0.
N
Consider the coefficient of xN−4.
The second Newton identity [11] states that S +S P +2P = 0, but S = 0 and so we
2 1 1 2 1
have that P = S2.
2 − 2
The fourth Newton identity states that S +S P +S P +S P +4P = 0 which simplifies
4 3 1 2 2 1 3 4
to S S22 +4P = 0.
4 − 2 4
Hence the coefficient P of xN−4 in det(xI C ) is 1 S S22 and for realizability
4 − N −4 4 − 2
(cid:16) (cid:17)
by a companion matrix, for sufficiently large N, this coefficient must not be positive i.e.
S S22 must be nonnegative.
4 − 2
(cid:16) (cid:17)
Expanding S S22 we get (2a+ρ)(N−3)(f(N)) where f(N) is a quadratic polynomial in
4 − 2 2N3
(cid:16) (cid:17)
N in terms of ρ,a and b which equals
(ρ 2a)(ρ2+4b2)N2+((4ρa2+8ρb2+16ab2 (3ρ3+2ρ2a+8a3))N+12ρ2a+24ρa2+2ρ3+16a3.
− −
The leading coefficient (ρ 2a)(ρ2 +4b2) is nonnegative when ρ 2a.
− ≥
But from Proposition 2 this is exactly the condition for σ 0 to be realizable in the
∪ { }
desired form and this establishes the result.
Remark 4. The negativity of the coefficient of xN−4 in Proposition 3 turns out to be the
most restrictive condition in establishing the result. This proposition demonstrates that
when adding lots of zeros to σ to aid realizability, the method αI +C for nonnegative
N N
companion matrices C is not the best strategy.
N
The realizability by matrices of the form (see the next section)
x 1 0 ... 0
1
 x x 2 0 ... 0 
2 1
x x x 3 0 0
 3 2 1 
XN =  ... ... ... ... ... 
 
 
x x ... x x N 1
 N−1 N−2 2 1 
 − 
 xN xN−1 ... x3 x2 x1 
 
is possible for all sufficiently large N, but leads to very high dimensional realizations.
3 Laffey’s construction
In [5] Laffey gives a constructive method for the Boyle-Handelman theorem. He shows
that if τ = (µ ,...,µ ),
1 n
x : = µk + +µk, k = 1,2,3,...,
k 1 ··· n
q(x) : = (x µ ) (x µ )
1 n
− ··· −
= xn +q xn−1 +q xn−2 + +q x+q
1 2 n−1 n
···
then the matrix
x 1 0 ... 0
1
 x x 2 0 ... 0 
2 1
x x x 3 0 0
 3 2 1 
Xn =  ... ... ... ... ... ... 
 
 
x x ... x x n 1
 n−1 n−2 2 1 
 − 
 xn xn−1 ... x3 x2 x1 
 
has characteristic polynomial
Q(x) = xn +nq xn−1 +n(n 1)q xn−2 + +n!q .
1 2 n
− ···
Thus it follows that the spectrum of Q(x) is realizable if the x , (i = 1,2,...,n), are
i
nonnegative. Hence if we wish to realize a given spectrum σ = (λ ,λ ,...,λ ), let
1 2 n
q(x) = xn +q xn−1 +q xn−2 + +q x+q
1 2 n−1 n
···
with
p
i
q = for i = 1,2,...,n
i
n(n 1) (n i+1)
− ··· −
then q(x) has its corresponding
Q(x) = f(x) = xn +p xn−1 + +p x+p
1 n−1 n
···
= (x λ ) (x λ ).
1 n
− ··· −
LaffeygivesalowerboundforN, thenumber ofzerosrequiredtobeaddedforrealizability,
in the proof of his main result [5] depending on a number of parameters.
This bound is not optimal in general however.
For example the list σ = (1.1,e±iθ) for θ = 0.0188π cannot be realized by a 3 3 matrix
×
by Remark 3 of Section 2.2 since ρ (cid:3) a+b√3.
This list with one zero added can however be realized by the nonnegative 4 4 matrix
×
1.013005334 1 0 0
 0 1.041605274 1 0 
.
0 0 1.041605274 1
 
 
 0.000326227 0 0 0.000296825 
 
This construction arises from the 4 4 solution of the NIEP via examination of the
×
coeffiecients of the characteristic polynomial of realizing matrices given in [12].
Thusσ 0 isrealizablewhereastheminimumnumberofzerosN requiredforrealizability
∪{ }
in Laffey’s construction is N = 198. The condition that all the main diagonal entries
of X are equal is quite restrictive and allowing these entries to be any nonnegative
n
entries that add up to the sum of the elements in σ is less restrictive in this sense.
Here cos(θ) = 99825635046 and note that realization becomes more difficult when the
···
spectral gap i.e. the distance between the Perron root and the second biggest eigenvalue
- decreases.
4 Three constructive methods
Suppose that σ = (ρ,λ ,λ ), where ρ > λ and λ ,λ are non-real complex conjugates,
2 2 2 2 2
| |
and that all the power sums s > 0. By the Boyle-Handelman theorem [2], there exists a
k
positive integer N such that σ N zeros is realizable. In this section, we consider finding
∪
such an N and a corresponding realizing matrix for σ N zeros.
∪
Method 1: αI +C
As we saw in the previous sections one method is to try to realize σ N zeros by αI +C,
m
∪
where m = N +3,α 0, and C is a nonnegative trace zero companion matrix.
≥
For instance, in the Examples section we saw that for σ = (5,2+3i,2 3i), σ 3 zeros
− ∪
is realizable in this way and that (5,2+3i,2 3i,0,0) is not realizable. So this method
−
yields the best possible result in this case.
However, this method need not always work.
For example, let σ = (1.4,e±iθ) = (ρ,a+ib,a ib) where cosθ = 0.95 (note that ρ (cid:3) 2a).
−
Then
f(x) = (x 1.4)(x2 1.9x+1)
− −
= x3 3.3x2 +3.66x 1.4
− −
= x3 (ρ+2a)x2 +(1+2aρ)x ρ
− −
If xNf(x) is to be realizable by αI +C, where m = N +3, then for x = y + 2a+ρ, the
m N+3
polynomial
2a+ρ N+3 2a+ρ N+2
g(y) = y + (ρ+2a) y +
(cid:18) N +3(cid:19) − (cid:18) N +3(cid:19)
2a+ρ N+1 2a+ρ N
+(1+2aρ) y + ρ y +
(cid:18) N +3(cid:19) − (cid:18) N +3(cid:19)
must have all its coefficients non-positive.
Using the substitution y 1 and multiplying g(y) by tN+3 we get
→ t
2a+ρ N+3 2a+ρ N+2
1+ t (ρ+2a)t 1+ t
(cid:18) N +3 (cid:19) − (cid:18) N +3 (cid:19)
2a+ρ N+1 2a+ρ N
+(1+2aρ)t2 1+ t ρt3 1+ t .
(cid:18) N +3 (cid:19) − (cid:18) N +3 (cid:19)
We note that the companion matrix of F(x) = xNf(x) is nonnegative if and only if
f(t) = tN+3F(1) has all its coefficients non-positive. Expanding this expression we see
t
that at least one of the coefficients of t is positive.
Note that if cosθ < 0 then σ N zeros is realizable for the smallest nonnegative integer
∪
N satisfying (N +3)s s2, see [8].
2 ≥ 1
Method 2: X
m
Laffey [5] showed that when λ = ρ > λ ,j > 1 for σ = (ρ,λ ,...,λ ) and all s > 0,
1 j 2 n k
| |
there exists N 0 such that σ N zeros is realizable by an m m matrix of the form
≥ ∪ ×
x 1 0 ... 0
1
 x x 2 0 ... 0 
2 1
x x x 3 0 0
 3 2 1 
Xm =  ... ... ... ... ... 
 
 
x x ... x x m 1
 m−1 m−2 2 1 
 − 
 xm xm−1 ... x3 x2 x1 
 
with m = N +3.
The x are recovered inductively from the equations trXk = s ,k = 1,2,3,...,m.
j m k
For example, for σ = (1.4,e±iθ) where cosθ = 0.95, this method succeeds with N = 9.
The minimum N for which σ N zeros is realizable is not known in this case; N = 9 is
∪
the minimum for which realizability by a matrix of the form X is possible. While this
m
method is systematic and easy to carry out, it is not guaranteed to yield the minimum N
for which σ N zeros is realizable however.
∪
Method 3: Multi-block companion matrices
Another method which may beattempted, isto realize σ N zeros by a matrix of theform
∪
C N 0 0 0
1 1
 
0 C N 0 0
 2 2 
 
M =  ... ... ... ... ... 
 
 
 0 0 0 ... N 
 r 1 
 − 
 
R R R C
 
 1 2 r 1 r 
· · ·
 − 
where
the matrix blocks C are nonnegative companion matrices of size k k for i =
i i i
• ×
1,2,...,r 1
≥
the matrices N are size k k (for i = 1,2,...,r 1), which have their (k ,1)
i i i+1 i
• × −
entry equal to one and all other entries equal to zero and
the matrix R is the k k matrix whose rows are all equal to zero, except the last
i r i
• ×
row which is equal to (r r r ).
i1 i2 ··· iki
Such matrices were first studied by Laffey [6], and by Laffey and Smigoc [9], in work on
spectra with at most two positive elements and all other elements having negative real
parts. They can also be employed in obtaining realizations over the real numbers of the
spectra realized by Kim, Ormes and Roush [4]. Unfortunately there is no known “good”
algorithm for choosing r or the companion matrices C .
i
A result (Theorem 5) of Laffey, Loewy and Smigoc in [7], implies that if σ = (λ ,λ ,λ )
1 2 2
satisfies the hypotheses given earlier in this section, i.e.
1. s (σ) > 0
k
2. λ > λ
1 2
| |
then if
f(t) = (1 λ t)(1 λ t)(1 λ t)
1 2 2
− − −
e
1
we have that 1 f(t)N has positive coefficients for sufficiently large integers N.
−
The N arising in the proof is the minimum N for which σ (N 3) zeros is realizable by
e ∪ −
a matrix of the form X above.
m
Writing f(t)N1 = 1 l1t l2t2 lNtN where li 0, one choice is to take r = N,
− − −···− −··· ≥
C the companion matrix of
i e
e(x) := xN l xN−1 l xN−2 l for i = 1,2,...,N
1 2 N
− − −···−
andthendeterminetheentriesinthelastrowofM inorderthatdet(xI M) = xN2−3f(x)
−
where f(x) = (x λ )(x λ )(x λ ).
1 2 2
− − −
This method succeeds precisely when the entries in position
(N2,1),(N2,2),...,(N2,N2 N)
−
turn out to be nonnegative. This can be done algorithmically as follows:
The Algorithm
Calculate the remainder r of xN2−3f(x) upon division by e(x), then the quotient q of
1 1
xN2−3f(x) upon division by e(x), then the remainder r of q upon division by e(x), the
2 1
quotient q of q upon division by e(x), and in general, for i > 1, r the remainder r of
2 1 i i
q upon division by e(x) and the quotient q of q upon division by e(x) concluding
i−1 i i−1
with r the remainder of q upon division by e(x).
N−1 N−2
The entries in position (N2,j),j = 1,2,...,N2 N are the minus of the coefficients of
−
the powers of xi (i = 0,1,...,N 1) occurring in r ,u = 1,2,...,N 1.
u
− −
For the example given earlier of f(x) = x3 3.3x2 + 3.66x 1.4 corresponding to
σ = (1.4,e±iθ) where cosθ = 0.95, we find tha−t 1 f(t)14 has t−he first four coefficients
−
positive (in fact it can be shown that all the coefficients of the Taylor series expansion of
e
1
1 f(t)4 are positive) and the corresponding 16 16 matrix M is found to have nonneg-
− ×
ative entries. So one obtains a realization of σ with 13 zeros added in this case.
e
While in this example, the realization of σ by a matrix of type X requires fewer (nine)
N
zeros, it is unknown which method provides the best bound for other given σ.