Page 86 - Textos de Matemática Vol. 39
P. 86
76 JUAN-MIGUEL GRACIA
which implies
max{σn(z1I−D),σn(z2I−D)}=|z2 −λj|.
Consequently,
(2.11) minz1,z2∈Λ(Y )∥Y − D∥ ≤ ∥Y0 − D∥ = max{σn(z1I − D), σn(z2I − D)}. From (2.10) and (2.11),
minz1,z2∈Λ(Y )∥Y − D∥ = max{σn(z1I − D), σn(z2I − D)}.
Case 2. Let us suppose that there is a simple eigenvalue λk of A that is
simultaneously the eigenvalue of A nearest to z1 and z2. Therefore, |z1 −λk|<min|z1 −λl|:l∈{1,...,n}\{k},
|z2 −λk|<min|z2 −λl|:l∈{1,...,n}\{k}.
Now either |z1 − λk| < |z2 − λk| or |z1 − λk| = |z2 − λk|.
Case 2.1
Let us suppose that |z1 − λk| < |z2 − λk|. In this case there are two
possibilities: either ∃p ∈ {1,...,n}\{k} such that |z1−λk| < |z1−λp| < |z2−λk|, case 2.1.1 (see Figure 2(a)), or ∀l ∈ {1,...,n} \ {k}, |z2 − λk| ≤ |z1 − λl|, case 2.1.2, (see Figure 2(b)).
22 11 00
−1 −1 −2 −2 −3 −3 −4 −4 −5 −5 −6 −6
−7−20246 −7−20246
λ1
z1
λp λk
|z2 −λk|
λ3
λ4
z2
|z2 −λk|
λ1
z1
λk
λ3
λ4
z2
(a) Case 2.1.1.
(b) Case 2.1.2.
Figure 2. Case 2.1.
