Page 87 - Textos de Matemática Vol. 39

NEAREST TO A NORMAL MATRIX 77 In Case 2.1.1 define the matrix Y1 := D except the entries λk and λp that
are replaced by z2 and z1, respectively. Then,
∥ Y 1 − D ∥ = ⎜ ⎜ ⎜
⎜ ⎜ ⎜ ⎜⎝
0
⎟=|z2−λk|. ⎟
⎟ ⎟ ⎟ ⎟⎠
⎛0
⎞ ⎟ ⎟
⎜ ⎜ . . .
⎜
0
z2 − λk
0
...
0
z1 −λp
0
...
⎟
⎜ ⎜ ⎜
⎟ ⎟ ⎟
This implies
minz1,z2∈Λ(Y )∥Y − D∥ = |z2 − λk| = σn(z2I − D)
and consequently
minz1,z2∈Λ(Y )∥Y − D∥ = max{σn(z1I − D), σn(z2I − D)}.
In Case 2.1.2,
{σ2n−1R(0), σ2nR(0)} = {σ1Γk(0), σ2Γk(0)} = {|z2 − λk|, |z1 − λk|}.
Hence, f(t) = σ1Γk(t) in an small interval 0 ≤ t ≤ ε, ε > 0; as σ1Γk(t) is an increasing function on [0,∞) and f(t) → 0 when t → ∞ there must be a t0 > 0 from which the function f(t) coincides with the second singular value of a block Γi(t) with i ̸= k. So, f(t) attains its maximum value at t0 > 0; and, consequently,
minz1,z2∈Λ(X)∥X − A∥ = f(t0). and the hypotheses of Theorem 2.1 fail. See Figure 3(a).
4 3.5 3 2.5
2 σ1(Γk(t)) 1.5
|z2 − λk| 1 |z1 −λk| 0.5
f (t)
4.5 4 3.5 3
2.5 f (t) 2
σ1 (Γk (t)) 1.5
1
0.5 |z1 −λk|=|z2 −λk|
00 01234t501234t5
t0 t0 (a) Case 2.1.2. (b) Case 2.2.
Figure 3. Cases 2.1.2 and 2.2.