Frobenius norm and structured backward errors

Consider the pencil 4L(z) =4A+z(4A)^∗.Then4L∈S andL(λ)x+4L(λ)x= 0.

Proof: The proof is computational and is easy to check. ¥.

We also consider general classes of linearly structured pencils whose coefficient matrices are elements of certain Jordan and/or Lie algebras and show that for these pencils struc- tured backward perturbation analysis ultimately reduces to that of one of the ten classes of structured pencils discussed above.

This givesa11+λb11=x^Tranda1+λb1=Q^T₁rwhose minimum norm solutions are

(a1 b1) =Q^T₁r à 1

λ

!_†

⇒a1= 1

1 +|λ|²Q^T₁r, b1= λ 1 +|λ|²Q^T₁r

and (a₁₁ b₁₁) =x^Tr à 1

λ

!_†

⇒a₁₁= 1

1 +|λ|²x^Tr, b₁₁= λ

1 +|λ|²x^Tr. Hence we have

4Ag= Ã 1

1+|λ|²x^Tr _1+|λ|¹ 2(Q^T₁r)^T

1

1+|λ|²Q^T₁r A1

!

, 4Bg = Ã λ

1+|λ|²x^Tr _1+|λ|^λ 2(Q^T₁r)^T

λ

1+|λ|²Q^T₁r B1

! .

This shows that the Frobenius norms of4Agand4Bgare minimized whenA1= 0 andB1= 0.

Hence k4Ak²_F =k4Akg ²_F =|a11|²+ 2ka1k²₂ and k4Bk²_F =k4Bkg ²_F =|b11|²+ 2kb1k²₂. Note that QQ^H =I⇒Q1Q^H₁ =I−xx^H ⇒Q₁Q^T₁ =I−xx^T.Consequently, we have

|||4L|||= (k4Ak²_F+k4Bk²_F)^1/2=

p|x^Tr|²+ 2k(I−xx^T)rk²₂ k(1, λ)k2 =

p2krk²₂− |x^Tr|² k(1, λ)k2 . Next, we have

4A = Q4AQg ^H= 1

1 +|λ|²xx^Trx^H+ 1

1 +|λ|²[xr^TQ₁Q^H₁ +Q₁Q^T₁rx^H] +Q₁A₁Q^H₁

= 1

1 +|λ|²[xr^T +rx^H−(r^Tx)xx^H] +Q₁A1Q^H₁, 4B = Q4BQg ^H = λ

1 +|λ|²xx^Trx^H+ 1

1 +|λ|²[λxr^TQ1Q^H₁ +λQ₁Q^T₁rx^H] +Q₁B1Q^H₁

= λ

1 +|λ|²[xr^T +rx^H−(r^Tx)xx^H] +Q₁B1Q^H₁

from which we obtain the desired pencil by settingA1= 0 and B1 = 0.This completes the proof. ¥

Observe that if Y is symmetric and Y x = 0 then Y = (I −xx^H)^TZ(I −xx^H) for some symmetric matrix Z.Consequently, we have Q1A1Q^H₁ = (I−xx^H)^TZ1(I−xx^H) and Q1B1Q^H₁ = (I−xx^H)^TZ2(I−xx^H) for some symmetric matricesZ1 andZ2.Hence from the proof of Theorem 3.3.1 we have following.

Corollary 3.3.2. LetL be aT-symmetric pencil and(λ, x)∈C×Cⁿ. Setr:=−L(λ)x.Let K be a T-symmetric pencil. Then L(λ)x+ K(λ)x= 0 if and only if K(z) = 4L(z) + (I− xx^H)^TN(z)(I−xx^H)for some T-symmetric pencil N, where 4L is the T-symmetric pencil given in Theorem 3.3.1.

Next, we considerT-skew-symmetric pencils.

Theorem 3.3.3. Let S be the space of T-skew-symmetric pencils and let L ∈S be given by L(z) =A+zB.Let(λ, x)∈C×Cⁿ andr:=−L(λ)x.Thenη^S(λ, x,L) =√

2krk2/k(1, λ)k2=

√2η(λ, x,L). Further, for theT-skew-symmetric pencil4Lgiven in Theorem 3.2.2, we have L(λ)x+4L(λ)x= 0and|||4L|||=η^S(λ, x,L).

Proof: AsAandB are skew-symmetric, from the proof of Theorem 3.3.1, we have 4Ag =Q^T4AQ=

à 0 a^T₁

−a1 A1

!

, 4Bg=Q^T4BQ=

à 0 b^T₁

−b1 B1

!

, Q^Tr= Ã x^Tr

Q^T₁r

! ,

where A1 and B1 are skew-symmetric matrices of size n−1. Consequently, as before, we have (4Ag+λ4B)Qg ^Hx =

à x^Tr Q^T₁r

!

which gives

à 0

−a1−λb1

!

=

à x^Tr Q^T₁r

!

. Note that x^Tr= 0 and the smallest norm solution of−a1−λb1=Q^T₁ris given by

(a1 b1) =Q^T₁r Ã

−1

−λ

!_†

⇒a1=− 1

1 +|λ|²Q^T₁r, b1=− λ

1 +|λ|²Q^T₁r.

Hence we have 4A=Q

à 0 −_1+|λ|¹ 2(Q^T₁r)^T

1

1+|λ|²Q^T₁r A1

!

Q^H, 4B=Q

à 0 −_1+|λ|^λ 2(Q^T₁r)^T

λ

1+|λ|²Q^T₁r B1

! Q^H.

Setting A1= 0 andB1= 0 we obtain4L such that |||4L|||=η^S(λ, x,L) =√

2krk2/k(1, λ)k2. SinceQ₁Q^T₁ =I−xx^T,we have

4A=− 1

1 +|λ|²[xr^T−rx^H] +Q₁A1Q^H₁ and4B=− λ

1 +|λ|²[xr^T −rx^H] +Q₁B1Q^H₁. Setting A1=B1= 0 we obtain theT-skew-symmetric pencil4L given in Theorem 3.2.2. ¥

Using the fact that ifY is skew-symmetric and Y x= 0 thenY = (I−xx^H)^TZ(I−xx^H) for some skew-symmetric matrix Z, we obtain an analogue of Corollary 3.3.2 for T-skew- symmetric pencils.

Next, we derive structured backward errors forT-even andT-odd pencils.

Theorem 3.3.4. Let S ∈ {T-even, T-odd} and L ∈ S be given by L(z) = A+zB. Let (λ, x)∈C×Cⁿ andr:=−L(λ)x.Then we have

η^S(λ, x,L) = s

|x^TAx|²+2krk²₂−2|x^Tr|² 1 +|λ|² =

p2krk²₂+ (|λ|²−1)|x^Tr|² k(1, λ)k2

when LisT-even and

η^S(λ, x,L) =





 q

|x^TBx|²+^2krk_1+|λ|^22−2|x2^Tr|2 =

p2krk²₂+ (|λ|⁻²−1)|x^Tr|²

k(1, λ)k₂ , ifλ6= 0.

√2η(λ, x,L), ifλ= 0,

when L is T-odd. The pencil 4L∈S given in Theorem 3.2.2 satisfies L(λ)x+4L(λ)x= 0 and|||4L|||=η^S(λ, x,L).

Proof: First, assume that L isT-even. Then noting that A=A^T andB =−B^T,the proof follows from similar arguments as those employed for T-symmetric and T-skew-symmetric

pencils. Indeed, considering a unitary matrixQ:= [x, Q1],we have 4Ag:=Q^T4AQ=

à a11 a^T₁ a1 A1

!

, 4Bg:=Q^T4BQ=

à 0 b^T₁

−b1 B1

!

, Q^Tr= Ã x^Tr

Q^T₁r

! ,

whereA1=A^T₁ andB1=−B^T₁ are of sizen−1.Consequently, we have (4Ag+λ4B)Qg ^Hx= Ã

x^Tr Q^T₁r

!

⇒ Ã

a11

a1−λb1

!

= Ã

x^Tr Q^T₁r

!

. This gives a11 = −x^TAx. The smallest norm solution ofa1+λb1=Q^T₁ris given by

(a1 b1) =Q^T₁r à 1

−λ

!_†

⇒a1= 1

1 +|λ|²Q^T₁r, b1=− λ

1 +|λ|²Q^T₁r.

Consequently, we have 4A=Q

à −x^TAx (_1+|λ|¹ 2Q^T₁r)^T

1

1+|λ|²Q^T₁r A1

!

Q^H, 4B=Q

à 0 (−_1+|λ|^λ 2Q^T₁r)^T

λ

1+|λ|²Q^T₁r B1

! Q^H.

Setting A1=B1= 0 and using the fact thatQ₁Q^T₁ =I−xx^T,we obtain the pencil4L such that

|||4L|||=η^S(λ, x,L) = s

|x^TAx|²+2krk²₂−2|x^Tr|² 1 +|λ|² . Now simplifying expressions for 4Aand4B,we obtain

4A = −xx^TAxx^H+ 1

1 +|λ|²[xr^T+rx^H−2(x^Tr)xx^H] +Q₁A1Q^H₁,

4B = − λ

1 +|λ|²[xr^T −rx^H] +Q₁B1Q^H₁.

Setting A1=B1= 0 we obtain theT-even pencil4L given in Theorem 3.2.2.

When L isT-odd, the results follow by interchanging the role ofAandB. ¥ It follows from Theorem 3.3.4 that for aT-even pencil, we haveη^S(λ, x,L)≤√

2η(λ, x,L) when|λ| ≤1 andη^S(λ, x,L)≤ k(1, λ)k2η(λ, x,L) when|λ|>1.Similarly, for aT-odd pencil, we have η^S(λ, x,L)≤√

2η(λ, x,L) when|λ| ≥1 andη^S(λ, x,L)≤ k(1, λ⁻¹)k2η(λ, x,L) when λ6= 0 and |λ|<1.

We mention that an analogue of Corollary 3.3.2 holds for T-even and T-odd pencils as well. Now, we consider aT-palindromic pencil L(z) =A+zA^T.

Theorem 3.3.5. Let S be the space of T-palindromic pencils andL ∈S be given by L(z) = A+zA^T. Let(λ, x)∈C×Cⁿ andr:=−L(λ)x.Then we have

η^S(λ, x,L) =







√2 q

|x^TAx|²+^krk_1+|λ|^22−|xT2^r|2 =√ 2

pkrk²₂−2reλ|x^TAx|²

k(1, λ)k2 , if λ6=−1,

√2η(λ, x,L), if λ=−1,

In particular, we haveη^S(λ, x,L) =√

2 η(λ, x,L), if λ∈iR.

Now define 4A=

( 1

1+|λ|²[λxr^T(I−xx^H) + (I−xx^T)rx^H], if λ=−1,

−xx^TAxx^H+_1+|λ|¹ 2[λxr^T(I−xx^H) + (I−xx^T)rx^H], if λ6=−1,

and consider the pencil 4L(z) = 4A+z(4A)^T. Then L(λ)x+4L(λ)x = 0 and |||4L||| = η^S(λ, x,L).

Proof: By Theorem 3.2.3, there exists a T-palindromic pencil 4L(z) =4A+z4A^T such that (L(λ) +4L(λ))x= 0.LetQ1∈C^n×(n−1)be such that Q:= [x Q1] is unitary. Then

4Ag:=Q^T4AQ=

à a11 a^T₁ b1 A1

!

, Q^Tr= Ã x^Tr

Q^T₁r

! .

Now, if λ6=−1 then by Theorem 3.2.1, we havex^T(4A+A)x= 0⇒x^T4Ax =−x^TAx.

Hence we have a11 =−x^TAx. Whenλ = −1, we haveλa^T₁₁+a11 =x^Tr = 0 for any a11. Since the aim is to minimize the Frobenius norm of4A, we seta11= 0.

Next, the minimum norm solution ofa1λ+b1=Q^T₁ris given by

³ a1 b1

´

=Q^T₁r Ã

λ 1

!_†

⇒a1= λQ^T₁r

1 +|λ|², b1= Q^T₁r 1 +|λ|².

Therefore when λ=−1, we have 4A =Q



 0 (_1+|λ|^λQT1r2)^T

Q^T₁r

1+|λ|² A1



Q^H. Setting A₁ = 0, we obtainη^S(λ, x,L) =|||4L|||=√

2krk2/p

1 +|λ|² =√

2η(λ, x,L). SinceQ1Q^H₁ =I−xx^H ⇒ Q₁Q^T₁ =I−xx^T,simplifying the expression for4A,we obtain

4A= 1

1 +|λ|²[λxr^T(I−xx^H) + (I−xx^T)rx^H] +Q₁A₁Q^H₁.

Whenλ6=−1,we have4A=Q



 −x^TAx (_1+|λ|^λQT1r2)^T

Q^T₁r

1+|λ|² A1



Q^H.SettingA1= 0,we obtain

η^S(λ, x,L) =|||4L|||= s

2|x^TAx|²+2k[I−xx^T]rk²₂ 1 +|λ|² =√

2 s

|x^TAx|²+krk²₂− |x^Tr|² 1 +|λ|² from which the result follows. Since |x^Tr|² = |x^TAx|²(1 +|λ|²) when λ ∈ iR, we have η^S(λ, x,L) = √

2krk2/k(1, λ)k2, for λ ∈ iR. Again, simplifying the expression for 4A, we obtain

4A=−xx^TAxx^H+ 1

1 +|λ|²[λxr^T(I−xx^H) + (I−xx^T)rx^H] +Q₁A1Q^H₁. This completes the proof. ¥

Observe that by Theorem 3.3.5 we have η^S(λ, x,L) ≤ √

2η(λ, x,L) when reλ > 0 and η^S(λ, x,L)≤ k(1,p

|reλ|/|1 +λ|)k2η(λ, x,L) when λ6=−1 andreλ <0.

Note that if Y ∈ C^n×n is such that Y x= 0 andY^Tx= 0 thenY = (I−xx^H)^TZ(I− xx^H) for some matrix Z.Hence from the proof of Theorem 3.3.5, we obtain an analogue of Corollary 3.3.2 for T-palindromic pencil. Indeed, if K is a T-palindromic pencil such that L(λ)x+ K(λ)x= 0 then K(z) =4L(z) + (I−xx^H)^TN(z)(I−xx^H) for someT-palindromic pencil N,where 4L is given in Theorem 3.3.5.

Now we turn to H-Hermitian, H-skew-Hermitian, H-even, H-odd and H-palindromic pencils.

Theorem 3.3.6. Let S∈ {H-Hermitian,H-skew-Hermitian} andL∈S be given byL(z) = A+zB.For(λ, x)∈C×Cⁿ,set r:=−L(λ)x.Then we have

η^S(λ, x,L) =











p2krk²₂− |x^Hr|² k(1, λ)k2

≤√

2η(λ, x,L) ifλ∈R, s

|x^HAx|²+|x^HBx|²+2krk²₂−2|x^Hr|²

1 +|λ|² , ifλ∈C\R.

In particular, we haveη^S(λ, x,L) =krk2=√

2η(λ, x,L),if λ=±i.

Whenλ∈R,define 4A :=

( 1

1+λ²[xr^H+rx^H−(r^Hx)xx^H], if A=A^H

1

1+λ²[rx^H−xr^H+ (r^Hx)xx^H], if A=−A^H 4B :=

( λ

1+λ²[xr^H+rx^H−(r^Hx)xx^H], if B=B^H

λ

1+λ²[rx^H−xr^H+ (r^Hx)xx^H], if B=−B^H

and consider the pencil 4L(z) = 4A+z4B. Then 4L ∈ S, L(λ)x+4L(λ)x = 0 and

|||4L|||=η^S(λ, x,L).

Whenλ∈C\R,the H-Hermitian/H-skew-Hermitian pencil 4L given in Theorem 3.2.2 satisfies L(λ)x+4L(λ)x= 0and|||4L|||=η^S(λ, x,L).

Proof: Suppose that L(z) = A+zB is H-Hermitian so that A = A^H and B = B^H. By Theorem 3.2.2 there existsH-Hermitian pencil4L(z) =4A+z4Bsuch that (4A+λ4B)x= r. Again, choosing a unitary matrixQ:= [x, Q1],we have

4Ag:=Q^H4AQ=

à a11 a^H₁ a1 A1

!

, 4Bg :=Q^H4BQ=

à b11 b^H₁ b1 B1

!

, Q^Hr=

à x^Hr Q^H₁r

! ,

whereA1=A^H₁ andB1=B₁^H are of sizen−1.This gives (4Ag+λ4B)Qg ^Hx=

à x^Hr Q^H₁ r

!

⇒ Ã a₁₁+λb₁₁

a1+λb1

!

=

à x^Hr Q^H₁r

!

.The minimum norm solution ofa1+λb1=Q^H₁ris given by

(a1 b1) =Q^H₁ r à 1

λ

!_†

⇒a1= 1

1 +|λ|²Q^H₁ r, b1= λ

1 +|λ|²Q^H₁r.

For the equationa11+λb11=x^Hr, two cases arise.

Case-I: Whenλ∈R,the minimum norm solution is given by

(a11 b11) =x^Hr Ã

1 λ

!_†

⇒a11= 1

1 +λ²x^Hr∈R, b11= λ

1 +λ²x^Hr∈R.

Hence we have 4A=Q

à 1

1+λ²x^Hr _1+λ¹2(Q^H₁r)^H

1

1+λ²Q^H₁ r A1

!

Q^H,4B=Q Ã λ

1+λ²x^Hr (_1+λ^λ2Q^H₁r)^H

λ

1+λ²Q^H₁ r B1

! Q^H.

Setting A1=B1= 0 and using the fact thatQ1Q^H₁ =I−xx^H,we have η^S(x, λ,L) =|||4L|||=

p2krk²₂− |x^Hr|² k(1, λ)k2 . Now simplifying the expressions for4Aand4B, we have

4A = 1

1 +λ²xx^Hrx^H+ 1

1 +λ²[xr^HQ1Q^H₁ +Q1Q^H₁ rx^H] +Q1A1Q^H₁

= 1

1 +λ²[xr^H+rx^H−(r^Hx)xx^H] +Q1A1Q^H₁,

4B = λ

1 +λ²xx^Hrx^H+ λ

1 +λ²[xr^HQ1Q^H₁ +Q1Q^H₁ rx^H] +Q1B1Q^H₁

= λ

1 +λ²[xr^H+rx^H−(r^Hx)xx^H] +Q1B1Q^H₁. Hence the results follow.

Case-II: Suppose that λ∈C\R. Then by Theorem 3.2.1, we have x^H(A+4A)x= 0 and x^H(B+4B)x= 0. Hence we havea11=−x^HAxandb11=−x^HBx.Consequently,

4A=Q

à −x^HAx (_1+|λ|¹ 2Q^H₁r)^H

1

1+|λ|²Q^H₁r A₁

!

Q^H, 4B=Q

à −x^HBx (_1+|λ|^λ 2Q^H₁r)^H

λ

1+|λ|²Q^H₁r B₁

! Q^H.

Setting A1=B1= 0,we obtain

η^S(λ, x,L) =|||4L|||= s

|x^HAx|²+|x^HBx|²+2k(I−xx^H)rk²₂ 1 +|λ|² . Hence the result follows.

Now simplifying the expressions for4Aand4B,we have 4A = −xx^HAxx^H+ 1

1 +|λ|²[xr^H(I−xx^H) + (I−xx^H)rx^H] +Q1A1Q^H₁ , 4B = −xx^HBxx^H+ 1

1 +|λ|²[λxr^H(I−xx^H) +λ(I−xx^H)rx^H] +Q1B1Q^H₁. Setting A1=B1= 0,we obtain the H-Hermitian pencil4L given in Theorem 3.2.2.

The proof is similar for the case when L isH-skew-Hermitian. ¥

Needless to mention that an analogue of Corollary 3.3.2 holds forH-Hermitian/H-skew- Hermitian pencils.

Theorem 3.3.7. Let S ∈ {H-even,H-odd} and L ∈ S be given by L(z) = A+zB. For (λ, x)∈C×Cⁿ, setr:=−L(λ)x. Then we have

η^S(λ, x,L) =











p2krk²₂− |x^Hr|² k(1, λ)k2 ≤√

2η(λ, x,L) if λ∈iR, s

|x^HAx|²+|x^HBx|²+2krk²₂−2|x^Hr|²

1 +|λ|² , if λ∈C\iR.

In particular, we haveη^S(λ, x,L) =krk2=√

2η(λ, x,L),if λ=±1.

Whenλ∈iR,define 4A :=

( 1

1+|λ|²[xr^H+rx^H−(r^Hx)xx^H], ifA=A^H

1

1+|λ|²[rx^H−xr^H+ (r^Hx)xx^H], ifA=−A^H 4B :=

( −λ

1+|λ|²[rx^H−xr^H+ (r^Hx)xx^H], ifB=B^H

−λ

1+|λ|²[rx^H+xr^H−(r^Hx)xx^H], ifB =−B^H

and consider the pencil 4L(z) = 4A+z4B. Then 4L ∈ S, L(λ)x+4L(λ)x = 0 and

|||4L|||=η^S(λ, x,L).

Whenλ∈C\iR, theH-even/H-odd pencil4L given in Theorem 3.2.2 satisfiesL(λ)x+ 4L(λ)x= 0 and|||4L|||=η^S(λ, x,L).

Proof: First, suppose that L(z) = A+zB is H even. ThenA =A^H and B = −B^H. By Theorem 3.2.2 there exists H-even pencil4L(z) =4A+z4B such that 4L(λ)x=r. Now choosing a unitary matrixQ:= [x, Q1] and noting that4A=4A^H, 4B=−4B^H,we have

4A:=Q

à a11 a^H₁ a₁ A₁

!

Q^H and4B=Q

à b11 b^H₁

−b₁ B₁

! Q^H,

where A1 = A^H₁ and B1 = −B₁^H are matrices of size n−1. Then 4L(λ)x = r gives à a11+λb11

a1−λb1

!

=

à x^Hr Q^H₁r

!

. The minimum norm solution of a1−λb1 = Q^H₁r is given by

(a1 b1) =Q^H₁r Ã

1

−λ

!_†

⇒a1= 1

1 +|λ|²Q^H₁ r, b1=− λ

1 +|λ|²Q^H₁ r.

For the solution of a11+λb11 = x^Hr two cases arise. When λ ∈ iR, the minimum norm solution is given by

(a11 b11) =x^Hr Ã

1 λ

!_†

⇒a11= 1

1 +|λ|²x^Hr∈R, b11= λ

1 +|λ|²x^Hr∈iR.

Whenλ∈C\iR,by Theorem 3.2.1, x^H(A+4A)x= 0 =x^H(B+4B)x⇒a11 =−x^HAx andb11=−x^HBx.Consequently, we have

4A=Q Ã 1

1+|λ|²x^Hr (_1+|λ|¹ 2Q^H₁r)^H

1

1+|λ|²Q^H₁r A1

!

Q^H,4B=Q Ã λ

1+|λ|²x^Hr (−_1+|λ|^λ 2Q^H₁ r)^H

λ

1+|λ|²Q^H₁r B1

! Q^H

whenλ∈iRand 4A=Q

à −x^HAx (_1+|λ|¹ 2Q^H₁r)^H

1

1+|λ|²Q^H₁r A1

!

Q^H,4B=Q

à −x^HBx (−_1+|λ|^λ 2Q^H₁ r)^H

λ

1+|λ|²Q^H₁r B1

! Q^H

when λ∈C\iR.Hence the desired results follow. Finally, reversing the role ofA andB we obtain the results for the case when L(z) =A+zB isH-odd. ¥

We have the following result forH-palindromic pencils.

Theorem 3.3.8. Let S be the space of H-palindromic pencils andL∈S be given byL(z) = A+zA^H. Let(λ, x)∈C×Cⁿ andr:=−L(λ)x.Then we have

η^S(λ, x,L) =









√2 s

|x^HAx|²+krk²₂− |x^Hr|²

1 +|λ|² if |λ| 6= 1, q

krk²₂−¹₂|x^Hr|², if |λ|= 1.

Now define 4A:=

( 1

1+|λ|²[rx^H+λxr^H(I−xx^H)], if|λ|= 1,

−xx^HAxx^H+_1+|λ|¹ 2[λxr^H(I−xx^H) + (I−xx^H)rx^H], if|λ| 6= 1, and consider4L(z) :=4A+z(4A)^H. ThenL(λ)x+4L(λ)x= 0and|||4L|||=η^S(λ, x,L).

Proof: LetQ:= [x, Q1] be unitary. Then 4Ag :=Q^H4AQ=

à a₁₁ a^H₁ b1 A1

!

andQ^Hr= Ã

x^Hr Q^H₁r

!

. Hence 4L(λ)x = r gives Ã

λa^H₁₁+a11

λa1+b1

!

= Ã

x^Hr Q^H₁r

!

. If |λ| 6= 1 then by Theorem 3.2.1, we have x^H(4A+A)x = 0 ⇒ x^H4Ax = −x^HAx. Hence we have a11 =

−x^HAx.On the other hand, when|λ|= 1, the minimum norm solution is given by

(a11 a11) =x^Hr à λ

1

!_†

= ( λx^Hr 1 +|λ|²

x^Hr 1 +|λ|²).

Note that when|λ|= 1 we havex^Hr=λx^Hr.Next, the minimum solution ofa1λ+b1=Q^H₁r is given by (a1, b1) =Q^H₁ r

à λ 1

!_†

=

³ λQ^H₁r 1+|λ|²

Q^H₁r 1+|λ|²

´

.Consequently, when|λ| 6= 1,we

have 4A=Q



 −x^HAx (_1+|λ|^λQH1r2)^H

Q^H₁r

1+|λ|² A1



Q^H.Setting A₁= 0,we obtain

η^S(λ, x,L) =|||4L|||= s

2|x^HAx|²+2k[I−xx^H]rk²₂ 1 +|λ|² =√

2 s

|x^HAx|²+krk²₂− |r^Hx|² 1 +|λ|² . Using the fact thatQ₁Q^H₁ =I−xx^H,we have

4A=−xx^HAxx^H+ 1

1 +|λ|²[λxr^H(I−xx^H) + (I−xx^H)rx^H] +Q1A1Q^H₁.

Setting A1= 0,the result follows.

For the case when |λ|= 1, we have 4A =Q



 ^x

Hr

1+|λ|² (_1+|λ|^λQH1r2)^H

Q^H₁r

1+|λ|² A1



Q^H. Again setting A1= 0 we obtain

η^S(λ, x,L) =|||4L|||= r

krk²₂−1 2|x^Hr|². SinceQ1Q^H₁ = (I−xx^H),simplifying the expression for4A,we obtain

4A:= 1

1 +|λ|²[rx^H+λxr^H(I−xx^H)] +Q1A1Q^H₁. Hence the proof. ¥

Remark 3.3.9. Let(λ, x)∈C×Cⁿwithx^Hx= 1andS∈ {T-symmetric, T-skew-symmetric, T-odd, T-even, T-palindromic, H-Hermitian, H-skew-Hermitian, H-odd, H-even, H-palindromic}.For L∈S, consider the set

S(λ, x,L) :={K∈S: L(λ)x+ K(λ)x= 0}.

ThenS(λ, x,L)6=∅ and there exists a unique4L∈S(λ, x,L)such that min{|||K|||: K∈S(λ, x,L)}=|||4L|||=η^S(λ, x,L).

Further, each pencil in S(λ, x,L) is of the form4L + (I−xx^H)^∗Z(I−xx^H)for someZ∈S, where∗is either the transpose or the conjugate transpose depending upon the structure defined by S. In other words, we haveS(λ, x,L) =4L + (I−xx^H)^∗S(I−xx^H).

We mention that the results obtained above are easily extended to the case of pencils having more general structures. Indeed, let M be a unitary matrix such that M^T =M or M^T = −M. Consider the Jordan algebra J := {A ∈ C^n×n : M⁻¹A^TM = A} and the Lie algebra L := {A ∈ C^n×n : M⁻¹A^TM = −A} associated with the scalar product (x, y) 7→

y^TM x. Consider a pencil L(z) = A+zB, where A and B are in J and/or in L. Then the pencilML given byML(z) =M A+zM B is eitherT-symmetric,T-skew-symmetric,T-even orT-odd. Hence replacingA, BandrbyM A, M BandM r,respectively, in the above results, we obtain corresponding results for the pencil L.

Similarly, whenM is unitary andM =M^HorM =−M^H,we consider the Jordan algebra J:={A∈C^n×n:M⁻¹A^HM =A}and the Lie algebraL:={A∈C^n×n :M⁻¹A^HM =−A}

associated with the scalar product (x, y) 7→ y^HM x. Now, let L(z) = A+zB be a pencil where A and B are in J and/or in L. Then the pencil ML(z) = M A+zM B is either H- Hermitian, H-skew-Hermitian, H-even or H-odd. Hence replacing A, B and r byM A, M B and M r,respectively, in the above results, we obtain corresponding results for the pencil L.

In particular, whenM :=J,whereJ:=

Ã

0 I

−I 0

!

∈C^2n×2n,the Jordan algebraJconsists of skew-Hamiltonian matrices and the Lie algebra L consists of Hamiltonian matrices. So, for example, considering the pencil L(z) :=A+zB,whereAis Hamiltonian and B is skew- Hamiltonian, we see that the pencil JL(z) = JA+zJB is H-even. Hence extending the results obtained forH-even pencil to the case of L, we have the following.

Theorem 3.3.10. Let S be the space of pencils of the form L(z) = A+zB, where A is Hamiltonian and B is skew-Hamiltonian. For (λ, x) ∈ C×Cⁿ, set r :=−L(λ)x.Then we have

η^S(λ, x,L) =











p2krk²₂− |x^HJr|² k(1, λ)k2 ≤√

2η(λ, x,L) ifλ∈iR, s

|x^HJAx|²+|x^HJBx|²+2krk²₂−2|x^HJr|²

1 +|λ|² , ifλ∈C\iR.

We mention that Remark 3.3.9 remains valid for structured pencils inSwhose coefficient matrices are element of Jordan and/or Lie algebras associated with a scalar product considered above. In such a case the ∗in (I−xx^H)^∗ is the adjoint induced by the scalar product that defines the Jordan and Lie algebras.