Page 109 - Textos de Matemática Vol. 39

ρ ≥ max{p1 + √ 3q1,p2+√
3q1,...,pk +√
NONNEGATIVE INVERSE EIGENVALUE PROBLEM 99
We can realize (pi +qi√
3,pi +iqi,pi −iqi) by a matrix Ai, say, and Ai has entries 3pi+qi√3 along the diagonal. The solution to the NIEP for n = 3 tells
3
us that this is the smallest Perron eigenvalue that makes this list realizable. Theorem 4.2 gives us the following estimate for the Perron eigenvalue.
Theorem 4.3. Let pi and qi, i = 1,...,k, be nonnegative real numbers that satisfy the following inequality:
3q2+2√
3qk + 2√
3(p1 +...+qk−1)}. Then the list (ρ,p1 +iq1,p1 −iq1 ...,pk −iqk,pk +iqk) is realizable.
33
3qi, pi + iqi, pi − iqi) be the spectrum of a nonnegative
Proof. Let (pi + √
matrix Bi. Then Bi has all its diagonal elements equal to pi + √3 qi. We insert
√√3
ρi=pi+ 3qiandci=pi+ 3qiinTheorem4.2tofinishtheproof.
3
More careful analysis of the diagonal elements of the realizing matrices leads to a better estimation for the Perron eigenvalue.
Let us assume that the elements in the list σ are ordered in the following
way:
We define:
and inductively:
ρi = ρi−1 + max{0, pi + √
3q1 ≥ p2 + √
p1 + √ 333
3 q2 ≥ . . . ≥ pk + √ 3qk.
ρ1 = p1 + √
3q i+1 }, ⌊3⌋ 3⌊3⌋
3√
for i = 1,...,k. Let D1 = D1′ . For i = 2,...,k we define the lists Di inductively.
3q1
3qi − p i+1 − √
for i = 2,...,k.
We define the lists: √
3√ 3
Di′ =(pi+ 3 qi,pi+ 3 qi,pi+ 3 qi)
List Di is obtained from the list Di−1 in the following way. Remove one instance
ofp i+1 +√3q i+1 fromthelistDi−1 andaddthelistD′ toit. ⌊3⌋ 3⌊3⌋ i
Theorem 4.4. The list
(ρk,p1 +iq1,p1 −iq1 ...,pk −iqk,pk +iqk)
is the spectrum of a nonnegative matrix with the diagonal elements Dk.
Proof. We will prove the proposition by induction on k. For k = 1 the proposi-
tion follows from the solution to the NIEP for lists with three elements. It is easy
to check that p i+1 + √3q i+1 is an element of the list Di−1 for i = 2,...,k. ⌊3⌋ 3⌊3⌋