Localization of pseudospectra of matrix pencils

For simplicity, throughout this chapter, we consider the space of pencilsL^p_w(C^n×n,k · k) and assume that each components ofwis nonzero. This makesL^p_w(C^n×n,k · k) a normed space. We further assume that the normk · k onC^n×n a subordinate norm and has the property thatkdiag(A₁, A₂)k= max(kA₁k, kA₂k).

First, we consider pseudospectra of scalar pencils which will be used in the subsequent development.

Proposition 4.2.1. Let L ∈ L^p_w(C,| · |) be a scalar pencil given by L(z) := α −zβ.

Then the pseudospectrum of L is given by Λ²(L) :={(z ∈C: |α−zβ|

k(1, z)k_w⁻¹_,q ≤²}. Now considering the homogeneous form L(c, s) := cα−sβ, the pseudospectrum is given by Λ_²(L) :=

½

(c, s)∈C²\ {0}: |cα−sβ|

k(c, s)k_w⁻¹_,q ≤²

¾ .

We now consider pseudospectra of diagonal and block diagonal pencils which will be crucial in the later development. We have the following result whose proof is immediate.

Theorem 4.2.2. Let L ∈ L^p_w(C^n×n,k · k) be a regular pencil. Suppose that L is block diagonal and is given by L= diag(L₁,L₂). Then we have Λ_²(L) = Λ_²(L₁)∪Λ_²(L₂).

In particular, suppose that L is a diagonal pencil and is given by L := diag(L_i), where each L_i is a scalar pencil. Then we have Λ_²(L) = ∪ⁿ_i=1Λ_²(L_i).

Let L∈L^p_w(C^n×n,k · k) given by L(z) =A−zB and X ∈C^n×n. Then XL and LX denote the pencils given by XL(z) = XA−zXB and LX(z) = AX−zBX. Then for the backward error functionηw,p(z,L) we have the following.

Theorem 4.2.3. Let L∈L^p_w(C^n×n,k · k) be regular. Then for ∆L∈L^p_w(C^n×n,k · k)we have

η_w,p(λ,L)− |||∆L|||_w,p ≤η_w,p(λ,L+∆L)≤η_w,p(λ,L) +|||∆L|||_w,p.

Let X and Y be invertible matrices in C^n×n and cond(X, Y) := kX⁻¹k kYk. Then η_w,p(λ,L)

cond(X, Y) ≤η_w,p(λ, Y⁻¹LX)≤cond(Y, X)η_w,p(λ,L).

Similar result hold for homogeneous matrix pencils.

Proof: We have

η_w,p(λ,L+∆L) := inf{|||G|||_w,p :λ∈Λ(L+∆L+G)}

= inf{|||(∆L+G)−∆L|||_w,p :λ∈Λ(L+∆L+G)}.

This gives η_w,p(λ,L+∆L)≤η_w,p(λ,L) +|||∆L|||_w,p and η_w,p(λ,L+∆L)≥η_w,p(λ,L)−

|||∆L|||_w,p.Hence the result follows.

Next, we have

ηw,p(λ, Y⁻¹LX) = inf{|||∆L|||w,p :λ∈Λ(Y⁻¹LX+∆L)}

= inf{|||∆L|||_w,p :λ∈Λ(L+Y∆LX⁻¹)}

= inf{|||Y⁻¹Y∆LX⁻¹X|||_w,p :λ∈Λ(L+Y∆LX⁻¹)}

≤ cond(Y, X)η_w,p(λ,L) (4.1)

and

η_w,p(λ,L) = η_w,p(λ, Y Y⁻¹LXX⁻¹)

≤ kYkkX⁻¹kη_w,p(λ, Y⁻¹LX) = cond(X, Y)η_w,p(λ, Y⁻¹LX). (4.2) Hence the desired result follows. ¥

Now we have the following Bauer-Fike type inclusion theorems for pseudospectra.

Theorem 4.2.4. Let L∈L^p_w(C^n×n,k · k) be regular. Then we have the following.

(a)Let X, Y ∈C^n×n be non-singular. Set κ₁ :=kX⁻¹kkYk andκ₂ :=kY⁻¹kkXk. Then Λ_²(L)⊂Λ_κ2²(Y⁻¹LX)andΛ_²(Y⁻¹LX)⊂Λ_κ1²(L).

(b) Let κ:= inf{kY⁻¹k kXk: Y⁻¹ LX = diag(L₁, L₂)}. Then we have Λ²(L)⊆Λκ²(L1)∪Λκ²(L2).

(c) Let X and Y be non-singular such that Y⁻¹LX = diag(L1,L2). Let X :=£

X₁ X₂¤ and (Y⁻¹)^∗ :=£

Y₁ Y₂¤

be conformal partitions. Setκ:=kX₁kkY₁^∗k+kX₂kkY₂^∗k. Then we have

Λ_²(L)⊆Λ_κ²(L₁)∪Λ_κ²(L₂).

(d) Let X and Y be non-singular such that Y⁻¹LX = diag(L₁, . . . ,L_k). Let X :=

£X₁ · · · X_k¤

and (Y⁻¹)^∗ := £

Y₁ · · · Y_k¤

be conformal partitions. Set φ_j(²) :=

kkX_jk kY_j^∗k². Then we have

Λ²(L)⊆ ∪^k_j=1Λφj(²)(Lj).

Proof: Without loss of generality we prove the results for non-homogeneous pencils.

The proof of (a) follows from the fact thatη_w,p(λ, Y⁻¹LX)≤ kY⁻¹k kXkη_w,p(λ,L) and η_w,p(λ,L)≤ kYk kX⁻¹kη_w,p(λ, Y⁻¹LX).

By (a),we have that Λ_²(L)⊆Λ_κ²(Y⁻¹LX), whereκ=kY⁻¹k kXk.Now from Theorem 4.2.2, we have Λ_κ²(Y⁻¹LX) = Λ_κ²(L₁)∪Λ_κ²(L₂). Thus Λ_²(L)⊆Λ_κ²(L₁)∪Λ_κ²(L₂).

Next, we haveL(z) = Ydiag(L₁(z), L₂(z))X⁻¹ ⇒L(z)⁻¹ =Xdiag(L₁(z)⁻¹,L₂(z)⁻¹)Y⁻¹

=X₁L₁(z)⁻¹Y₁^∗+X₂L₂(z)⁻¹Y₂^∗ ⇒ kL(z)⁻¹k ≤max(kL₁(z)⁻¹k,kL₂(z)⁻¹k)(kX₁kkY₁^∗k+ kX₂kkY₂^∗k) = max(kL₁(z)⁻¹k,kL₂(z)⁻¹k)κ. Hence the proof follows .

Finally, L(z)⁻¹ = P_k

j=1XjLj(z)⁻¹Y_j^∗ ⇒ kL(z)⁻¹k ≤ k max

1≤j≤kkXjkkY_j^∗kkLj(z)⁻¹k. This implies that Λ_²(L)⊆Λ_φj(²)(L), where φ_j(²) =kkX_jk kY_j^∗k. ¥

For the special case when L is simple, we have the following result.

Corollary 4.2.5. Let L ∈ L^p_w(C^n×n,k · k) be given by L(z) = A −zB or L(c, s) = cA−sB. Suppose that Y^∗AX = diag(α_i) and Y^∗BX = diag(β_i). Set y_i := Y e_i and x_i :=Xe_i. Then (α_i/β_i, y_i, x_i) is an eigentriple of L(z) = A−zB and ((β_i, α_i), y_i, x_i) is an eigentriple of L(c, s) =cA−sB. Further, we have

Λ²(L)⊆ ∪ⁿ_i=1Λκ²(Li) and Λ²(L)⊆ ∪ⁿ_i=1Λnκi²(Li),

where Li(z) = αi−zβi or L(c, s) = cαi−sβi and κ:=kY^∗kkXk, κi :=kxikkyik.

If L(z) = A−zB is diagonalizable and B is non-singular then we can choose X and Y such that Y^∗BX =I and Y^∗AX = diag(λ1, . . . , λn). Consequently, we have the following special case.

Corollary 4.2.6. Let L ∈ L^p_w(C^n×n,k · k) be given by L(z) = A−zB. Suppose that B is non-singular. Let Y and X be such that Y^∗BX = I and Y^∗AX = diag(λ_i). Let y_i :=Y e_i and x_i :=Xe_i. Then (λ_i, x_i, y_i) is an eigentriple of L. Further, we have

Λ_²(L)⊆ ∪ⁿ_i=1Λ_κ²(L_i) and Λ_²(L)⊆ ∪ⁿ_i=1Λ_nκi²(L_i), where L_i(z) = z−λ_i, κ:=kY^∗k kXk and κ_i :=kx_ik ky_ik.

To obtain further inclusions for pseudospectra, we need the following result.

Lemma 4.2.7. [28, 24] Let

·I₁ M 0 I₂

¸

∈C^n×n, where I₁ and I₂ are the identity matrices of size k and n−k, respectively, and M ∈C^k×(n−k). Then

°°

°°

·I₁ M 0 I2

¸ °°°

°2

= r

1 + 1

2(kMk²₂ + q

kMk⁴₂+ 4kMk²₂).

We now generalize this result and obtain the following.

Lemma 4.2.8. LetZ =

·αI1 M 0 βI₂

¸

∈C^n×n,where I1 andI2 are the identity matrices of sizek andn−k, respectively,M ∈C^k×(n−k) and α, β are nonzero positive real numbers.

Then

σ_max(Z) = s

(α² +β²+kMk²₂) +p

(α²+β²+kMk²₂)²−4α²β²

2 ,

σmin(Z) = s

(α²+β²+kMk²₂)−p

(α²+β²+kMk²₂)²−4α²β²

2 ,

cond(Z) = σ_max(Z)

σ_min(Z) = (α²+β²+kMk²₂) +p

(α²+β²+kMk²₂)²−4α²β²

2αβ and

σ_max(Z)σ_min(Z) =αβ, where σ_min(Z) and σ_max(Z) are the smallest and the largest sin- gular values of Z, respectively.

Proof: We have kZk²₂ =ρ(Z^∗Z),where ρ(Z^∗Z) = sup

k(x,y)k2=1

½£

x^∗ y^∗¤ Z^∗Z

·x y

¸

; x∈R^k, y ∈R^n−k

¾

.Now,

£x^∗ y^∗¤ Z^∗Z

·x y

¸

= £

x^∗ y^∗¤·

α² α M

α M β² +M^∗M

¸ ·x y

¸

= £

x^∗ y^∗¤·

α²x+αM y αM^∗x+ (β²+M^∗M)y

¸

= α²kxk²₂+ β²kyk²₂+ αx^∗My+αy^∗M^∗y+αy^∗M^∗x+y^∗M^∗My

= α²kxk²₂+β²kyk²₂+ 2Re(α x^∗M y) +kyk²₂kMk²₂.

Note that there exists vectorsxandysuch thatkxk₂ = 1 =kyk₂ andx^∗My =kMyk₂ = kMk₂. Let (x₁, x₂) :=

· x

√2, y

√2

¸

. Then k(x₁, x₂)k₂ = 1 and £

x^∗₁ x^∗₂¤ Z^∗Z

·x₁ x2

¸

=

{α²kx₁k²₂+β²kx₂k²₂+ 2αkx₁k₂kMk₂kx₂k₂ +kx₂k²₂kMk²₂}. Consequently, we have ρ(Z^∗Z) = sup

k(x,y)k2=1

{α²kxk²₂+β²kyk²₂+ 2αkxk₂kMk₂kyk₂ +kyk²₂kMk²₂}

= ρ

· α² αkMk₂ αkMk₂ kMk²₂ +β²

¸

= (α²+β²+kMk²₂) +p

(α²+β²+kMk²₂)²−4α²β² 2

⇒ρ(Z^∗Z) = (α²+β²+kMk²₂) +p

(α²+β²+kMk²₂)²−4α²β² 2

⇒σ_max(Z) = kZk₂ = s

(α²+β²+kMk²₂) +p

(α²+β²+kMk²₂)²−4α²β²

2 .(4.3)

Now, we have σ_min(Z) = 1

σ_max(Z⁻¹), where Z⁻¹ =

·α⁻¹I₁ −α⁻¹Mβ⁻¹ 0 β⁻¹I2

¸

. Thus

σ_max(Z⁻¹) =

√2 q

(α²+β² +kMk²₂)−p

(α²+β² +kMk²₂)²−4α²β² .

Therefore, we have σ_min(Z) =

s

(α²+β²+kMk²₂)−p

(α²+β²+kMk²₂)²−4α²β²

2 . (4.4)

From (4.3) and (4.4) we haveσ_min(Z)σ_max(Z) = αβ. Hence the proof. ¥

Now, we have the following result which will be useful in deriving pseudospectra inclusions.

Theorem 4.2.9. Set P :=p

1 +kZ₁k²₂, Q:=p

1 +kZ₂k²₂ and let X :=

·I₁ Z₂ 0 I₂

¸ ·Q^1/2 0 0 P^−1/2

¸ , Y :=

·I₁ Z₁ 0 I₂

¸ ·Q^1/2 0 0 P^−1/2

¸

.Then

kXk2 =p Q/P

q

(P+Q)+√

(P+Q)²−4P/Q

2 , kX⁻¹k2 =

q

(P+Q)+√

(P+Q)²−4P/Q

2 ,

kYk2 = q

(P+Q)+√

(P+Q)²−4Q/P

2 , kY⁻¹k2 =p

P/Q q

(P+Q)+√

(P+Q)²−4Q/P

2 and

cond(X) =p

Q/P^(P+Q)+

√(P+Q)²−4P/Q

2 , cond(Y) =p

P/Q^(P+Q)+

√(P+Q)²−4Q/P

2 .

Further, we have cond(Y, X) = cond(X, Y) = p

cond(X)cond(Y)≤P +Q.

Proof: We have X =

·Q^1/2I1 P^−1/2Z2

0 P^−1/2I2

¸

. Then for α = Q^1/2, β = P^−1/2, M = P^−1/2Z2, by Lemma 4.2.8 we obtain cond(X) = p

Q/P^(P+Q)+

√(P+Q)²−4P/Q

2 , kXk2 =

pQ/P q

(P+Q)+√

(P+Q)²−4P/Q

2 and kX⁻¹k₂ =

q

(P+Q)+√

(P+Q)²−4P/Q

2 .

Similarly, for Y =

·Q^1/2I1 P^−1/2Z1

0 P^−1/2I2

¸

,by Lemma 4.2.8 we obtain cond(Y) =p

P/Q^(P+Q)+

√(P+Q)²−4Q/P

2 ,kYk₂ =

q

(P+Q)+√

(P+Q)²−4Q/P

2 and

kY⁻¹k₂ =p P/Q

s

(P +Q) +p

(P +Q)²−4Q/P

2 .

Again by Lemma 4.2.8, we haveσmin(X)σmax(X) = σmin(Y)σmax(Y) =αβ = qQ

P ⇒

kXk2 kY⁻¹k2 =kYk2 kX⁻¹k2 ⇒cond(Y, X) = cond(X, Y). Now cond(X, Y) = p

cond(X, Y)cond(Y, X)

= p

kX⁻¹k₂kYk₂kY⁻¹k₂kXk₂

= p

cond(X)cond(Y).

Now we show that cond(X, Y)≤P +Q. We have cond(X, Y) =

s

(P +Q) +p

(P +Q)²−4Q/P 2

s

(P +Q) +p

(P +Q)²−4P/Q 2

≤ (P +Q) +p

(P +Q)²−4Q/P

4 + (P +Q) +p

(P +Q)²−4P/Q 4

≤ (P +Q)

2 +P +Q

4 +P +Q

4 =P +Q.

Hence the proof. ¥

LetL∈L^p_w(C^n×n,k·k₂) be a regular pencil given byL =

·L₁ L₁₂ 0 L₂

¸

,whereL₁,L₂,L₁₂ are pencils of size m, n−m and m×n−m, respectively, and Λ(L₁)∩Λ(L₂) = ∅.Now, consider the Sylvester operator

S :C^m×n−m×C^m×n−m →L^p_w(C^m×n−m,k · k)

defined by S(Z1, Z2) := L1Z1 − Z2L2. Then as we shall see, S(Z1, Z2) = −L12 has a unique solution. So let Z1 and Z2 be unique solution of S(Z1, Z2) = −L12. Set P :=p

1 +kZ₁k² and Q:=p

1 +kZ₂k². Define X :=

·Q^1/2I₁ P^−1/2Z₂ 0 P^−1/2I₂

¸

and Y :=

·Q^1/2I₁ P^−1/2Z₁ 0 P^−1/2I₂

¸

. (4.5)

Then we have Y⁻¹LX = diag(L₁,L₂).

Theorem 4.2.10. LetL∈L^p_w(C^n×n,k · k₂)be a regular pencil given by L=

·L₁ L₁₂ 0 L₂

¸ , whereΛ(L₁)∩Λ(L₂) = ∅.LetZ₁ andZ₂ be solution of the generalized Sylvester equation L₁Z₁−Z₂L₂ =−L₁₂. Set P :=p

1 +kZ₁k² and Q:=p

1 +kZ₂k². Define κ₁ :=

s

(P +Q) +p

(P +Q)²−4Q/P 2

s

(P +Q) +p

(P +Q)²−4P/Q 2

and κ₂ :=P +Q. Then we have

Λ_²(L)⊆Λ_κ1²(L₁)∪Λ_κ1²(L₂)⊆Λ_κ2²(L₁)∪Λ_κ2²(L₂).

Proof: DefiningXandY as in (4.5), we haveY⁻¹LX = diag(L1,L2).By Theorem 4.2.9, we have cond(X, Y) =κ1 ≤κ2.Hence the result follows from Theorem 4.2.4. ¥

Now, consider the matrix A :=

·A1 A12

0 A₂

¸

, where A1 ∈ C^m×m, A2 ∈ C(n−m)×(n−m). SetR(z) := k(A−zI)⁻¹k2, R1(z) :=k(A1−zI)⁻¹k2 andR2(z) :=k(A2−zI)⁻¹k2.Then the following holds.

Proposition 4.2.11. [24] Let a₁₂:=kA₁₂k₂, r_m(z) := min (R₁(z), R₂(z))andr_M(z) :=

max (R₁(z), R₂(z)). Then we have kR(z)k2 ≤

r

1 + a12rm(z)

2 ((a²₁₂rm(z)²+ 4)^1/2+a12rm(z)).

Then we have following pseudospectra inclusion. Recall that h_w,p(z) = k(1, z)k_w,p. Theorem 4.2.12. Let L∈L^p_w(C^n×n,k · k2) be given by L:=

·L1 L12

0 L₂

¸

, where Λ(L1)∩ Λ(L₂) =∅. Set g₁(²) :=²

r 1 + κ

² and κ:=|||L₁₂|||_w,p. Then for all ² >0 we have Λ_²(L)⊂Λ_g1(²)(L₁)∪Λ_g1(²)(L₂).

Proof: Without loss of generality, we consider L to be nonhomogeneous. So, let rM(λ) = max (kL1(λ)⁻¹k,kL2(λ)⁻¹k), rm(λ) = min (kL1(λ)⁻¹k,kL2(λ)⁻¹k). Then by Proposition 4.2.11,

kL(λ)⁻¹k ≤r_M(λ) r

1 + kL₁₂(λ)kr_m(λ) 2

³

(kL₁₂(λ)k²r_m(λ)²+ 4)^1/2+kL₁₂(λ)kr_m(λ)

´ . Now, (kL(λ)⁻¹k)≥ 1

²h_w⁻¹_,q(λ) gives 1

²²h_w⁻¹_,q(λ)² ≤r_M(λ)² µ

1 + kL₁₂(λ)kr_m(λ) 2

³

(kL₁₂(λ)k²r_m(λ)² + 4)^1/2 +kL₁₂(λ)kr_m(λ)

´¶

≤r_M(λ)² µ

1 + kL₁₂(λ)kr_M(λ) 2

³

(kL₁₂(λ)k²r_M(λ)²+ 4)^1/2+kL₁₂(λ)kr_M(λ)

´¶

≤r_M(λ)²+ κh_w⁻¹_,q(λ)r_M(λ)³

2 (κ²h_w⁻¹_,q(λ)²r_M(λ)²+ 4)^1/2+ κ²h_w⁻¹_,q(λ)²r_M(λ)⁴

2 . Thus we have 1

²²h_w⁻¹_,q(λ)² −r_M(λ)² µ

1 + κ²h_w⁻¹_,q(λ)²r_M(λ)² 2

¶

≤ κh_w⁻¹_,q(λ)r_M(λ)³ 2

¡κ²h_w⁻¹_,q(λ)²r_M(λ)² + 4¢_1/2 .

For simplicity of notations, set d := h_w⁻¹_,q(λ), r_M := r_M(λ) and recall that κ =

|||∆L₁₂|||_w,p. Then we have 1

²²d² −r_M² µ

1 + κ²d²r²_M 2

¶

≤ κdr_M³ 2

µq

κ²d²r²_M + 4

¶

. (4.6)

Now two choices arise, namely.

(a) 1

²²d² −r²_M µ

1 + κ²d²r_M² 2

¶

≥0 and (b) 1

²²d² −r_M² µ

1 + κ²d²r²_M 2

¶

≤0.

Now under the case (a) squaring both sides of (4.6) we get 1

²⁴d⁴+r⁴_M µ

1 + κ²d²r_M² 2

¶₂

− 2

²²d²r_M² µ

1 + κ²d²r²_M 2

¶

≤ κ²d²r⁶_M

4 (κ²d²r²_M + 4)⇒ 1

²⁴d⁴ +r_M⁴ − 2

²²d²r_M² − κ²

²²r_M⁴ ≤0⇒ µ

r²_M − 1

²²d²

¶₂

≤ r⁴_Mκ²

²² ⇒

·µ

r_M² − 1

²²d²

¶

−r²_Mκ

²

¸ ·µ

r²_M − 1

²²d²

¶

+r_M² κ

²

¸

≤ 0. Hence

we have ·

r_M²

³ 1− κ

²

´

− 1

²²d²

¸ · r²_M

³ 1 + κ

²

´

− 1

²²d²

¸

≤0. (4.7)

Case-I: Letκ≤². Then either r²_M

³ 1−κ

²

´

≤ 1

²²d² and r²_M

³ 1 + κ

²

´

≥ 1

²²d² or

r_M²

³ 1− κ

²

´

≥ 1

²²d² and r_M²

³ 1 + κ

²

´

≤ 1

²²d². In the first case we have 1

²p

1 +κ/² ≤ r_Md ≤ 1

²p

1−κ/² and in the second case, we

have 1

²p

1−κ/² ≤r_Md≤ 1

²p

1 +κ/², which is not possible.

Case-II: Letκ≥². Then

· r_M²

³ 1− κ

²

´

− 1

²²d

¸

≤0. Hence from (4.7) we have

· r_M²

³ 1 + κ

²

´

− 1

²²d

¸

≥0⇒r_Md≥ 1

²p

1 +κ/². This shows thatr_Md≥ 1

²p

1 +κ/² for all ².

Next consider the case (b). In this case, we have 1

²²d² −r_M² µ

1 + κ²d²r²_M 2

¶

≤ 0

which gives r_M(z)² ≥

−1 + r

1 + 2κ²

²²

κ²d² .We now show that for all ²≥0,we have

−1 + r

1 + 2κ²

²²

κ²d² ≥ 1

²²d²(1 +κ/²).

Indeed, −1 + r

1 + 2κ²

²² ≥ κ²

²²(1 +κ/²) implies that 1 ≥

·

1 + κ²

²²(1 + ^κ_²)

¸₂

−2κ²/²² ⇒

−2κ³

²²(κ+²) + κ⁴

²²(²+κ)² ≤0⇒κ+ 2²≥0 which is obviously true. Hence we have

r_M(z)² ≥

−1 + r

1 + 2κ²

²²

κ²d² ≥ 1

²²d²(1 +κ/²) for all ²≥0.

This shows thatr_Md≥ 1

²p

1 +κ/² = 1

g₁(²). Consequently, we have Λ_²(L)⊂Λ_g1(²)(L₁)∪Λ_g1(²)(L₂) for all ². ¥

ForL ∈L^p_w(C^n×n,k · k), we have the following pseudospectra inclusion.

Theorem 4.2.13. Let L∈L^p_w(C^n×n,k · k2) be given by L=

·L1 L12

0 L₂

¸

, where Λ(L1)∩ Λ(L2) = ∅. Let (Z1, Z2) be the solution of the generalized Sylvester equation L1Z1 − Z₂L₂ = −L₁₂. Set φ(²) := ²(1 +kZ₁k+kZ₂k) and f(²) := ²+√

²²+ 4² κ

2 , where κ :=

|||L₁₂|||_w,p. Then we have

(a)Λ²(L)⊆Λ_φ(²)(L1)∪Λ_φ(²)(L2), (b)Λ²(L)⊆Λ_f(²)(L1)∪Λ_f(²)(L2).

Proof: Without loss of generality, assume that L is nonhomogeneous. For λ ∈ C, set d(λ) := max (kL₁(λ)⁻¹k, kL₂(λ)⁻¹k). Then

L⁻¹(λ) =

·L₁(λ)⁻¹ −L₁(λ)⁻¹L₁₂(λ)L₂(λ)⁻¹

0 L₂(λ)⁻¹

¸ . Consequently, we have

kL⁻¹(λ)k ≤max¡

kL₁(λ)⁻¹k,kL₂(λ)⁻¹k¢ +°

°L₁(λ)⁻¹ L₁₂(λ)L₂(λ)⁻¹°

°. SinceL₁(λ)Z₁−Z₂L₂(λ) =−L₁₂(λ), we have

°°L1(λ)⁻¹ L12(λ)L2(λ)⁻¹°

° = °

°L1(λ)⁻¹(Z2L2(λ)−L1(λ)Z1)L2(λ)⁻¹°

°

≤ d(λ) (kZ₁k+kZ₂k)

and kL(λ)⁻¹k ≤d(λ) + d(λ) (kZ₁k+kZ₂k) = d(λ) (1 +kZ₁k+kZ₂k).

Now, kL(λ)⁻¹k ≥(h_w⁻¹_,q(λ)²)⁻¹ gives (h_w⁻¹_,q(λ)²)⁻¹ ≤ kL(λ)⁻¹k ≤d(λ)(1 +kZ₁k+ kZ2k) = d(λ)φ(²) ⇒ d(λ)h_w⁻¹_,q(λ) ≥ 1

²(1 +kZ₁k+kZ₂k) = 1

φ(²). Hence Λ²(L) ⊆ Λ_φ(²)(L₁)∪Λ_φ(²)(L₂).

Next, we have kL(λ)⁻¹k ≤ d(λ) + d(λ)²κ h_w⁻¹_,q(λ) ⇒ ²⁻¹ ≤ d(λ)h_w⁻¹_,q(λ) + (d(λ)h_w⁻¹_,q(λ))²κ. Now, set d⁰(λ) = d(λ)h_w⁻¹_,q(λ). Then ²d⁰(λ) +² κd⁰(λ)² −1 ≥ 0 which implies that d⁰(λ)≥ −²+√

²²+ 4² κ

2² κ = 2

²+√

²²+ 4² κ ⇒d(λ)h_w⁻¹_,q(λ)≥ 1 f(²). Thus Λ_²(L)⊆Λ_f(²)(L₁)∪Λ_f(²)(L₂). ¥