Page 111 - Textos de Matemática Vol. 39

NONNEGATIVE INVERSE EIGENVALUE PROBLEM 101 h0(t) is an integer polynomial of degree 20 and it is of the form t4h(t),
where
h(t) = 25t16+660t15+6510t14+27816t13+33401t12−96276t11
−252616t10 + 35136t9 + 823536t8 + 2717600t7 + 5803932t6 + 5375328t5 +1632228t4+1688520t3+1769700t2−830000t−750000.
We note that h(t) is irreducible mod 31 and hence it is irreducible over the field of rational numbers. We conjecture that t = t0 is the least real number for which σ1 is realizable. One can realize (3 + t0, 3, −2, −2, −2) by a nonnegative matrix of the form
(5.1)
⎡⎢t0 1 0 0 0⎤⎥
⎢p 0 1 0 0⎥ A 0 = ⎢⎣ 0 k 0 1 0 ⎥⎦ .
00001
00sr0
One can find a nonnegative nonzero matrix Y satisfying
A0Y = Y A0, trace(A0Y ) = 0.
This property always occurs for Perron extreme spectra [11], but the converse is not true in general. Let σ be a Perron extreme spectrum of size n, A a corresponding realizing matrix and Y a matrix having the property (5.1). In the case when Y is nonderogatory, A is a polynomial in Y :
n−1
i=0
for some real yi. It is worth remarking that it has been observed that at least one of the yi is 0. It would be of interest to prove this fact in general or to find a counterexample. Originally, we had expected that Y corresponding to the matrix A0 above would provide a counterexample, but for the specific t = t0,
one of the yi is 0.
Now we consider the question what is the smallest t for which the list σ1
is realized by a doubly companion matrix. In particular, we want to know if we can realize σ1 with t0 defined above by a doubly companion matrix.
Let g(x) = (x−3−t)(x−3)(x+2)3,
⎡⎢ a 1 0 0 0⎤⎥
⎢ b 0 1 0 0⎥ A1=⎢⎣c 0010⎥⎦
d0001 e+zywvu
and let cA1 (x) be the characteristic polynomial of A1. We want to find nonneg- ative entries for the matrix A, such that it will have the spectrum σ1 with the smallest possible t.
A =
yiY i