Page 110 - Textos de Matemática Vol. 39

100 THOMAS LAFFEY AND HELENA SˇMIGOC
Now we assume that the proposition holds for k − 1. In this case the list (ρk−1,p1 +iq1,p1 −iq1 ...,pk−1 −iqk−1,pk−1 +iqk−1)
is the spectrum of a nonnegative matrix with a diagonal element p[k+1] + √√3
3q k+1 . We join this list with the list (pk + 3qk,pk + iqk,pk − iqk) using 3⌊3⌋
Theorem 4.1, to finish the proof. The example below illustrates that Theorem 4.4 gives us a better estimate
for the Perron eigenvalue than Theorem 4.3.
Example 4.5. Let
σ = (ρ, 3 + i√ 3,3−i√
We want to find a good estimate for the Perron eigenvalue ρ that will make the list σ realizable.
Theorem 4.3 gives us:
ρ = max{6, 9, 10} = 10.
The numbers in the list are already ordered as we have assumed in Theorem 4.4. Now we compute ρi as described above: ρ1 = 6, ρ2 = 9 and ρ3 = 9. We conclude that the list σ is realizable for ρ = 9.
5. The example (3 + t, 3, −2, −2, −2)
As mentioned in the introduction, the NIEP is not yet solved for general
lists of size n = 5. A particular test case has been the list σ1 =(3+t,3,−2,−2,−2).
The Perron-Forbenius theory implies that σ1 cannot be realized if t = 0. For t>0, it is a consequence of the Boyle-Handelman theory that σ1 with suffi- ciently many zeros appended can be realized. Loewy proved that σ1 is realiz- able by a nonnegative symmetric matrix if and only if t ≥ 1 and Laffey and Meehan have shown that σ is realizable for all t ≥ t0 where
t0 = 0.519310982048...
3,1+i2√
3,1−i2√
3,1+i√3,1−i√
3).
is a root of the polynomial where
h0(t) = discriminant(w(q, t), q)
w(q,t) = q4 +(t2 −6t−30)q3 −(t4 +6t3 +15t2 −136t−345)q2
+(15t4 + 90t3 + 57t2 + 1050t − 1800)q −60t4 − 360t3 − 36t2 + 2760t + 3600.