Page 84 - Textos de Matemática Vol. 39

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JUAN-MIGUEL GRACIA
6 5 4 3 2 1
00 1 2 3 4t5
σ1 (Γ(t))
σ2 (Γ(t))
6 5 4 3 2 1
Γ(t)x = a1 t 0 = t ; 0 a2 1 a2
hence,
This proves (2.6), and it implies
σ1 (Γ(t))
σ2 (Γ(t))
(2.7)
(2.8) As
Analogously, if a1 ̸= 0, ∀t > 0,
lim σ1Γ(t) = ∞. t→∞
(a) Case |a1| < |a2|.
Figure 1. Singular values of Γ(t).
Taking x = (0, 1)T we have
Γ(t) 01 = at = t2 + |a2|2. 222
00 1 2 3 4 t5 (b) Case |a1| = |a2|.
σ2Γ(t) ≤ σ2Γ(t)= 1
|a1a2| . |a1|2 +t2
and
inequality (2.6) implies
whence
σ2Γ(t) ≤
σ1 Γ−1 (t) Γ−1(t)= 1 a2 −t,
a1a2 0 a1
|a1|2 +t2
σ1 Γ−1(t) ≥ |a1a2| ,
|a1a2| . |a1|2 + t2