Page 112 - Textos de Matemática Vol. 39

102 THOMAS LAFFEY AND HELENA SˇMIGOC
First we consider the case when u = 0. Comparing coefficients of g(x) and cA1 (x) gives us the following equations:
b+v = 15+3t c+w = 10−6t+tv
60+28t = bv+tw−d−y. 60+28t ≤ bv+tw
≤ 10t−6t2 +t2v+bv
≤ 10t−6t2 +(15+3t+t2)2,
and this inequality is equivalent to
15−18t−15t2 −6t3 −t4 ≥ 0.
Those equations imply:
2
This inequality holds for
t≤t1 = 1(−3+−3+8√
6) = 0.536904...
Since we have and
(b + a2) + (v + a(t − a)) = 15 + 3t + a(2t − a) (15+3t+a(2t−a))2
2
Now we will show that we cannot obtain a realization with t smaller than
t1 using doubly companion matrices even when we allow u > 0. We can assume that a ≥ u. Comparing polynomials cA1 (x) and g(x) gives us the following equations:
a+u=t
b+v = 15+3t+au
c+w = 10−6t+av+bu 60+28t = cu+bv+aw−d−y.
First we estimate
cu+bv+aw ≤ a(c+w)+bv=(10−6t+av+bu)a+bv=
= (10−6t)a+(b+a2)(v+a(t−a))−a3(t−a). b + v = 15 + 3t + a(t − a)
Now
60 + 28t ≤ (10 − 6t)a + 4 − a3(t − a)
(b + a2)(v + a(t − a)) ≤ 4 .
(15+3t+a(2t−a))2