Page 113 - Textos de Matemática Vol. 39

NONNEGATIVE INVERSE EIGENVALUE PROBLEM 103
and
60+18t+6t2 ≤−(10+6t−a3)(t−a)+ 4 .
(15+3t+a(2t−a))2 We analyze the function on the right side of the inequality.
(15+3t+a(2t−a))2 F (a) = −(10 − 6t − a3)(t − a) + 4
Writinga=t(1−p),where 1 ≤p≤1,wefind: 2
.
G(p)= 1(15+3t−(−1+p2)t2)2 +pt(−10−6t−(−1+p)3t3). 4
We compute the derivative of G(p) :
(5.2) G′(p) = −t(10+6t+15pt+3pt2 −t3 +7pt3 −9p2t3 +3p3t3).
We see that G′(p) is negative for all sufficiently small t > 0. Since we are interested in small values of t, it follows that the maximum value of G(p) occurs when p = 0 and a = t. Hence the best value for t for realizability by a doubly companion matrix is
t1 = 1(−3+−3+8√
is realizable is

√
t2 = 16 6 − 39 = 0.43799...
6) = 0.536904...
2
We have already remarked that the smallest known t that makes σ1 real-
izable is t0 = 0.51931098... That shows us that not all realizable lists with five elements can be realized by a doubly companion matrix.
6. The example (3+t,3−t,−2,−2,−2)
The solution to the NIEP for lists with five elements and trace zero [13]
tells us that the smallest t for which
σ2 =(3+t,3−t,−2,−2,−2)
If we add a zero to σ2, we can show that the list is realizable for smaller valuesoft.Letσ3 =(3+t,3−t,−2,−2,−2,0).
Let
let cA(x) be the characteristic polynomial of the matrix A and let g(x) = (x − 3 − t)(x − 3 + t)(x + 2)3x.
⎡⎢0 1 0 0 0 0⎤⎥ ⎢a1 0 1 0 0 0⎥
A=⎢b1 0 0 1 0 0⎥, ⎢⎣c1 0 a3 0 1 0⎥⎦
000001 0 0 c1 b1 a1 0