Page 164 - Textos de Matemática Vol. 39

154 ION ZABALLA
Theorem 2.6. Let A = A1 ⊕ A2 be a simple decomposition of A ∈ Cn×n with A1 ∈Cp×p andA2 ∈Cq×q.Letρ>0besuitableforA.Thenforallε>0there e x i s t s δ > 0 s u c h t h a t i f ∥ A − A˜ ∥ < δ t h e r e i s a n i n v e r t i b l e m a t r i x T ∈ C n × n such that the following properties hold:
( i ) T − 1 A˜ T = A˜ 1 ⊕ A˜ 2 ; ˜ ˜
(ii) max{∥In −T ∥,∥A1 −A1 ∥,∥A2 −A2 ∥}<ε.
(iii) Λ(A˜1) ⊂ Vρ(A1) and Λ(A˜2) ⊂ Vρ(A2).
Proof. Assume that ε > 0 is a fixed positive real number.
T2 = 0 Iq and∥P ∥<δ2 then∥In−T1 ∥<δ1 and∥In−T2 ∥<δ1.
(a) By the continuity of the matrix product, there is δ1 > 0 such that if ∥In − T1∥ < δ1 and ∥ In − T2 ∥< δ1 then ∥ In − T1T2 ∥< ε.
(b)Forthisδ1 thereisδ2 > 0suchthatifT1 = Ip 0and Ip P P Iq
In particular, we can choose δ2 = δ1 for the l1 norm.
(c) For this δ2 there is δ3 > 0 such that if max{∥ A1 − B1 ∥, ∥ A2 − B2 ∥, ∥C∥}<δ3 thenΛ(B1)⊂Vρ(A1),Λ(B2)⊂Vρ(A2)andifZisthe solution of the Sylvester equation B1Z − ZB2 = C then ∥ Z ∥< δ2. In fact, Λ(B1) ⊂ Vρ(A1) and Λ(B2) ⊂ Vρ(A2) because of the continuity of the eigenvalues of A1 and A2, respectively. And ∥ Z ∥< δ2 because Y = 0 is the unique solution of A1Y −Y A2 = 0 and the solution of this equation continuously depends on their coefficients and independent
term (Cramer’s rule).
(d) Let E = E11 E12 and ε′ = min{δ3, ε}. For this positive real number
E21 E22
thereisδ4 >0suchthatifmax{∥E∥,∥B∥}<δ4,A˜1 =A1+E11+ E12BandA˜2 =A2+E22−BE12 then∥A1−A˜1 ∥<ε′,∥A2−A˜2 ∥<ε′ and ∥ E12 ∥< ε′.
(e) Finally, if δ4′ = min{δ2, δ4} there is δ5 > 0 such that if ∥ E ∥< δ5 then, byTheorem2.4thereisamatrixP ∈Cp×q suchthat∥P ∥<δ4′.
Now, let δ = min{δ4,δ5} and let A˜ be any matrix satisfying ∥ A−A˜ ∥< δ. PutE=A−A˜.Byitem(e),∥P ∥<δ4′ andsomax{∥E∥,∥P ∥}<δ4.Byitem (d), if A˜1 = A1 +E11 +E12P and A˜2 = A2 +E22 −PE12 then ∥ A1 −A˜1 ∥< ε′, ∥A2−A˜2 ∥<ε′ and∥E12 ∥<δ3.Thus∥A1−A˜1 ∥<ε,∥A2−A˜2 ∥<εandby item (c), Λ(A˜1) ⊂ Vρ(A1), Λ(A˜2) ⊂ Vρ(A2) and ∥ Z ∥< δ2, Z being the unique solution of A˜1Z − ZA˜2 = E12. Furthermore, since ∥ P ∥< δ2 and ∥ Z ∥< δ2, byitem(b),∥In−T1 ∥<δ1 and∥In−T2 ∥<δ1.Finally,ifT =T1T2 then,by i t e m ( a ) ∥ I − T ∥ < ε a n d − 1 ˜ A˜ 1 0
This concludes the proof.

T AT= 0 A˜ . 2