Spectral norm and structured backward errors

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.

Proof: Suppose that L isT-symmetric. Then from the proof of Theorem 3.3.1, we have 4A=Q

à x^Tr

1+|λ|² 1

1+|λ|²(Q^T₁r)^T

1

1+|λ|²(Q^T₁r) A1

!

Q^H,4B=Q

à λ x^Tr

1+|λ|² λ

1+|λ|²(Q^T₁r)^T

λ

1+|λ|²(Q^T₁r) B1

! Q^H,

such that4L(λ)x+L(λ)x= 0.Now, forµ4A:=_1+|λ|^krk22 andµ4B:= ^{|λ| krk}_1+|λ|2²,by the DKW The- orem 1.2.5, we haveA1=− x^Tr(Q^T₁r)(Q^T₁r)^T

(1 +|λ|²) (krk²₂− |x^Tr|²) andB1=− λ x^Tr(Q^T₁r)(Q^T₁r)^T (1 +|λ|²) (krk²₂− |x^Tr|²). This gives η^S(λ, x,L) = (k4Ak²₂+k4Bk²₂)^1/2= krk2

k(1, λ)k2

. Simplifying expressions for 4A and4B,we obtain the desired results.

When L isT-skew-symmetric, from the proof of Theorem 3.3.3, we have

4A=Q

à 0 −^(Q_1+|λ|^T1r)T2

1

1+|λ|²Q^T₁r A1

!

Q^H,4B=Q

à 0 −^λ(Q_1+|λ|^T1r)2^T

λ

1+|λ|²Q^T₁r B1

! Q^H,

such that 4L(λ)x+ L(λ)x= 0.Now, for µ4A := _1+|λ|^krk22 and µ4B := ^{|λ| krk}_1+|λ|2², by the DKW Theorem 1.2.5, we obtain A1 = 0 = B1. Consequently, we have η^S(λ, x,L) = (k4Ak²₂ + k4Bk²₂)^1/2 = krk2/k(1, λ)k2. Simplifying the expressions for 4A and 4B, we obtain the desired results. ¥

Remark 3.4.2. If|x^Tr|=krk2,thenkQ^T₁rk2= 0. In such a case, consideringA1= 0 =B1

we obtain the desired results.

Next, we considerT-even andT-odd pencils. Recall that forz∈C,sign(z) :=z/|z|when z6= 0 and sign(z) := 1 whenz= 0.

Theorem 3.4.3. 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+krk²₂− |x^Tr|² 1 +|λ|² =

pkrk²₂+|λ|²|x^Tr|² k(1, λ)k2

, ifL isT-even, q

|x^TBx|²+^krk_1+|λ|^22−|xT2^r|2 =

pkrk²₂+|λ|⁻²|x^Tr|²

k(1, λ)k2 , ifL isT-odd,λ6= 0,

η(λ, x,L), ifL isT-odd, λ= 0.

Now, define

4A :=









−xx^TAxx^H+_1+|λ|¹ 2[xr^T +rx^H−2(x^Tr)xx^H] + ^xTAx(I_krk^−xx2^{T)rrT(I−xxH)} 2−|x^Tr|² ,

ifA=A^T,

1

1+|λ|²[rx^H−xr^T], ifA=−A^T. 4B :=









−xx^TBxx^H+_1+|λ|^λ 2[xr^T+rx^H−2(x^Tr)xx^H]−^{sign(λ)2xTBx}_krk^(I−xx2 ^{T)rrT(I−xxH)}

2−|x^Tr|² ,

if B=B^T.

−_1+|λ|^λ 2[xr^T −rx^H], if B=−B^T,

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).

Proof: Suppose that L isT-even. Then from the proof of Theorem 3.3.4, we have

4A=Q



 −x^TAx ^(Q_1+|λ|^T1r)T2

Q^T₁r

1+|λ|² A1



Q^H, 4B =Q

à 0 −_1+|λ|^λ 2(Q^T₁r)^T

λ

1+|λ|²(Q^T₁r) B1

! Q^H,

such that4L(λ)x+ L(λ)x= 0.Now, for

µ4A:=

s

|x^TAx|²+krk²₂− |x^Tr|²

(1 +|λ|²)² andµ4B:=

s

|λ|²(krk²₂− |x^Tr|²) (1 +|λ|²)²

by the DKW Theorem 1.2.5, we have A1= x^TAx

krk²₂− |x^Tr|²(Q^T₁r)(Q^T₁r)^T and B1 = 0. This gives

η^S(λ, x,L) = s

|x^TAx|²+krk²₂− |x^Tr|² 1 +|λ|² =

pkrk²₂+|λ|²|x^Tr|² k(1, λ)k2 .

Simplifying the expressions for4Aand4B,we obtain the desired results. When L isT-odd the results follow by interchanging the role ofA andB. ¥

It follows that for aT-even pencil we haveη^S(λ, x,L)≤ k(1, λ)k₂η(λ, x,L) whereas for a T-odd pencil we haveη^S(λ, x,L)≤ k(1, λ⁻¹)k2η(λ, x,L) whenλ6= 0.

Theorem 3.4.4. Let S∈ {H-Hermitian,H-skew-Hermitian} andL∈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) =







η(λ, x,L), if λ∈R, s

|x^HAx|²+|x^HBx|²+krk²₂− |x^Hr|²

1 +|λ|² , ifλ∈C\R When λ∈R, define

4A :=





1

1+λ²[xr^H+rx^H−(r^Hx)xx^H−^xHr(I−xx_krk2^{H)rrH(I−xxH)}

2−|x^Hr|² ], ifA=A^H,

1

1+λ²[rx^H−xr^H+ (r^Hx)xx^H+^rHx(I−xx_krk2^{H)rrH(I−xxH)}

2−|x^Hr|² ], if A=−A^H. 4B :=





λ

1+λ²[xr^H+rx^H−(r^Hx)xx^H−^xHr(I−xx_krk2^{H)rrH(I−xxH)}

2−|x^Hr|² ], ifB =B^H,

λ

1+λ²[rx^H−xr^H+ (r^Hx)xx^H+^rHx(I−xx_krk2^{H)rrH(I−xxH)}

2−|x^Hr|² ], if B=−B^H.

Whenλ∈C\R,define

4A :=













−xx^HAxx^H+_1+|λ|¹ 2[xr^H(I−xx^H) + (I−xx^H)rx^H] + ^xHAx(I−xx_krk2^{H)rrH(I−xxH)} 2−|x^Hr|² ,

ifA=A^H,

−xx^HAxx^H+_1+|λ|¹ 2[(I−xx^H)rx^H−xr^H(I−xx^H)] +^xHAx(I_krk^−xx2^{H)rrH(I−xxH)} 2−|x^Hr|² ,

if A=−A^H.

4B :=













−xx^HBxx^H+_1+|λ|¹ 2[λxr^H(I−xx^H) +λ(I−xx^H)rx^H] +^{xHBx (I−xx}_krk2 ^{H)rrH(I−xxH)} 2−|x^Hr|² ,

ifB=B^H,

−xx^HBxx^H−_1+|λ|^λ 2xr^H(I−xx^H) +_1+|λ|^λ 2(I−xx^H)rx^H+^xHBx(I−xx_krk2^{H)rrH(I−xxH)} 2−|x^Hr|² ,

if B=−B^H. Consider4L(z) :=4A+z4B.Then4L∈S, L(λ)x+4L(λ)x= 0and|||4L|||=η^S(λ, x,L).

Proof: First, suppose that L isH-Hermitian. Assume thatλ∈R.Thenx^Hr∈R.Now from the proof of Theorem 3.3.6, we have

4A=Q Ã 1

1+λ²x^Hr _1+λ¹2(Q^H₁r)^H

1

1+λ²Q^H₁r A1

!

Q^H and4B =Q Ã λ

1+λ²x^Hr _1+λ^λ2(Q^H₁ r)^H

λ

1+λ²Q^H₁r B1

! Q^H

such that 4L(λ)x+ L(λ)x= 0. For µ4A := _1+λ^krk22 and µ4B := ^{|λ| krk}_1+λ2² by the DKW The- orem 1.2.5, we have A1=− x^Hr(Q^H₁r)(Q^H₁r)^H

(1 +λ²) (krk²₂− |x^Hr|²), B1=− λ x^Hr(Q^H₁r)(Q^H₁r)^H (1 +λ²) (krk²₂− |x^Hr|²).This gives

η^S(λ, x,L) = (k4Ak²₂+k4Bk²₂)^1/2= krk2

k(1, λ)k2.

Now simplifying the expressions for4Aand4B,we obtain the desired results.

Next, suppose thatλ∈C\R.Then again from the proof of Theorem 3.3.6, we have

4A=Q

à −x^HAx _1+|λ|¹ 2(Q^H₁r)^H

1

1+|λ|²Q^H₁r A1

!

Q^H,4B:=Q

à −x^HBx _1+|λ|^λ 2(Q^H₁r)^H

λ

1+|λ|²Q^H₁r B1

! Q^H.

For, µ4A :=

q

|x^HAx|²+^krk_(1+|λ|^22−|xH2)^r|22, µ4B :=

q

|x^HBx|²+^|λ|2(krk_(1+|λ|^22−|x2)^H2r|2), by the DKW Theorem 1.2.5, we have

A1= x^HAx

krk²₂− |x^Hr|²(Q^H₁r)(Q^H₁r)^H andB1= x^HBx

krk²₂− |x^Tr|²(Q^H₁r)(Q^H₁r)^H.

Hence we haveη^S(λ, x,L) = s

|x^HAx|²+|x^HBx|²+krk²₂− |x^Hr|²

1 +|λ|² .Now, simplifying the ex- pressions for 4A and 4B, we obtain the desired results. The proof is similar for the case when L is H-skew-Hermitian. ¥

We mention that whenQ^H₁ r= 0,the desired results follow by considering A1= 0 =B1. Theorem 3.4.5. Let S ∈ {H-even, H-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) =







η(λ, x,L), if λ∈iR, s

|x^HAx|²+|x^HBx|²+krk²₂− |x^Hr|²

1 +|λ|² , if λ∈C\iR.

When λ∈iR,define

4A :=





1

1+|λ|²[xr^H+rx^H−(r^Hx)xx^H−^xHr(I−xx_krk2^{H)rrH(I−xxH)}

2−|x^Hr|² ], ifA=A^H,

1

1+|λ|²[rx^H−xr^H+ (r^Hx)xx^H+^rHx(I−xx_krk2^{H)rrH(I−xxH)}

2−|x^Hr|² ], if A=−A^H. 4B :=









1

1+|λ|²[λrx^H+λxr^H−λ(r^Hx)xx^H+^{λxHr (I−xx}_krk2^{H)rrH(I−xxH)}

2−|x^Hr|² ], ifB =B^H,

1

1+|λ|²[λxx^Hrx^H−λxr^H(I−xx^H) +λ(I−xx^H)rx^H+^λrHx(I_krk^−xx2^{H)rrH(I−xxH)} 2−|x^Hr|² ],

if B=−B^H. Whenλ∈C\iR,define

4A :=













−xx^HAxx^H+_1+|λ|¹ 2[xr^H(I−xx^H) + (I−xx^H)rx^H] + ^xHAx(I−xx_krk2^{H)rrH(I−xxH)} 2−|x^Hr|² ,

ifA=A^H,

−xx^HAxx^H+_1+|λ|¹ 2[(I−xx^H)rx^H−xr^H(I−xx^H)] +^xHAx(I_krk^−xx2^{H)rrH(I−xxH)} 2−|x^Hr|² ,

if A=−A^H.

4B :=













−xx^HBxx^H+_1+|λ|¹ 2[λxr^H(I−xx^H) +λ(I−xx^H)rx^H] +^{xHBx (I−xx}_krk2 ^{H)rrH(I−xxH)} 2−|x^Hr|² ,

if B=B^H,

−xx^HBxx^H−^λxrH_1+|λ|^(I−xx2 ^H)+^λ(I−xx_1+|λ|^H)rx2 ^H +^xHBx(I_krk^−xx2^{H)rrH(I−xxH)} 2−|x^Hr|² ,

if B=−B^H. Consider4L(z) :=4A+z4B.Then4L∈S, L(λ)x+4L(λ)x= 0and|||4L|||=η^S(λ, x,L).

Proof: First, suppose that L is H-even. Next, assume that λ∈ iR. Then it follows that x^Hr∈R. Now from the proof of Theorem 3.3.7, we have

4A=Q Ã 1

1+|λ|²x^Hr _1+|λ|¹ 2(Q^H₁r)^H

1

1+|λ|²Q^H₁r A1

!

Q^H,4B:=Q Ã λ

1+|λ|²x^Hr −_1+|λ|^λ 2(Q^H₁r)^H

λ

1+|λ|²Q^H₁r B1

! Q^H

such that 4L(λ)x+ L(λ)x = 0. For µ4A := _1+|λ|^krk22, µ4B := ^{|λ| krk}_1+|λ|2², by the DKW Theo- rem 1.2.5 we have

A1=− x^Hr(Q^H₁ r)(Q^H₁ r)^H

(1 +|λ|²) (krk²₂− |x^Hr|²) and B1= λ x^Hr(Q^H₁r)(Q^H₁r)^H (1 +|λ|²) (krk²₂− |x^Hr|²). This gives η^S(λ, x,L) = (k4Ak²₂+k4Bk²₂)^1/2= krk2

k(1, λ)k2. Simplifying expressions for 4A and4B,we obtain the desired result.

Now suppose thatλ∈C\iR.The again from the proof of Theorem 3.3.7, we have

4A=Q

à −x^HAx _1+|λ|¹ 2(Q^H₁r)^H

1

1+|λ|²Q^H₁r A1

!

Q^H,4B:=Q

à −x^HBx −_1+|λ|^λ 2(Q^H₁r)^H

λ

1+|λ|²Q^H₁r B1

! Q^H.

In this case we have,µ4A= q

|x^HAx|²+^krk_(1+|λ|^22−|x2^H)^r|22, µ4B= q

|x^HBx|²+^|λ|2(krk_(1+|λ|^22−|x2)^H2r|2). By the DKW Theorem 1.2.5, we have

A1= x^HAx

krk²₂− |x^Hr|²(Q^H₁r)(Q^H₁r)^H andB1= x^HBx

krk²₂− |x^Tr|²(Q^H₁r)(Q^H₁r)^H.

Consequently, we haveη^S(λ, x,L) = s

|x^HAx|²+|x^HBx|²+krk²₂− |x^Hr|²

1 +|λ|² .Now, simplifying the expressions for4Aand4B,we obtain the desired results.

When L isH-odd, the desired results follow by interchanging the role of AandB.¥

As before, the above results are easily extended to the case of general structured pencils where the coefficient matrices are elements of Jordan and/or Lie algebras. In particular, for the pencil L(z) :=A+zB,whereAis Hamiltonian andB is skew-Hamiltonian, we have the following result.

Theorem 3.4.6. Let S be the space of pencils of the form L(z) = A+zB, where A is Hamiltonian and B is skew-Hamiltonian. Let L∈S and (λ, x)∈C×Cⁿ.Set r :=−L(λ)x.

Then we have

η^S(λ, x,L) =







η(λ, x,L), ifλ∈iR s

|x^HJAx|²+|x^HJBx|²+krk²₂− |x^HJr|²

1 +|λ|² , ifλ∈C\iR.

Now we consider palindromic pencils.

Theorem 3.4.7. Let S be the space ofT-palindromic pencils and L∈S be given byL(z) :=

A+zA^T. Let(λ, x)∈C×Cⁿ andr:=−L(λ)x.Then we have

η^S(λ, x,L) =

















√2 s

|x^TAx|²+|λ|² (krk²₂− |x^Tr|²)

(1 +|λ|²)² , if|λ|>1,

√2 s

|x^TAx|²+krk²₂− |x^Tr|²

(1 +|λ|²)² , if|λ| ≤1 andλ6=±1,

η(λ, x,L), ifλ=±1.

Now define

4A:=



















−xx^TAxx^H+_1+|λ|¹ 2[λxr^T(I−xx^H) + (I−xx^T)rx^H] +^{λ xTAx}_|λ|2^(I−xx(krk²₂^T−|x^)rrTTr|^(I−xx2) ^H), if |λ|>1,

−xx^TAxx^H+_1+|λ|¹ 2[λxr^T(I−xx^H) + (I−xx^T)rx^H] +^{λ xTAx}_krk^(I−xx2 ^{T)rrT(I−xxH)} 2−|x^Tr|² , if |λ| ≤1 andλ6=−1,

1

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

Consider the pencil 4L(z) := 4A+z(4A)^T. Then 4L ∈ S, L(λ)x+4L(λ)x = 0 and

|||4L|||=η^S(λ, x,L).

Proof: Suppose thatλ6=−1.Then from the proof of Theorem 3.3.5, we have

4A = Q

à −x^TAx _1+|λ|^λ 2(Q^T₁r)^T

1

1+|λ|²Q^T₁r A1

! Q^H

such that4L(λ)x+ L(λ)x= 0.Now for

µ_4A :=



 q

|x^TAx|²+^|λ|2(krk_(1+|λ|^22−|x2)^T2r|2), if|λ|>1, q

|x^TAx|²+^krk_(1+|λ|^22−|x2^T)^r|22, if|λ| ≤1, by the DKW Theorem 1.2.5, we have

A1 =





λ x^TAx

|λ|² (krk²₂−|x^Tr|²)Q^T₁r(Q^T₁r)^T, if|λ|>1,

λ x^TAx

krk²₂−|x^Tr|²Q^T₁r(Q^T₁r)^T, if|λ| ≤1.

Consequently, we have

η^S(λ, x,L) =





√2 q

|x^TAx|²+^{|λ|2 (krk}_(1+|λ|^22−|x2)^2Tr|2), if|λ|>1,

√2 q

|x^TAx|²+^krk_(1+|λ|^22−|x2^T)^r|22, if|λ| ≤1.

Now simplifying the expression for 4A,we obtain the desired results.

Next, suppose thatλ=−1.Then again from the proof of Theorem 3.3.5, we have

4A = Q

à 0 _1+|λ|^λ 2(Q^T₁r)^T

1

1+|λ|²Q^T₁r A₁

!

Q^H. Forµ4A:= krk2

1 +|λ|², by the DKW Theorem 1.2.5, we have A1 = 0. Hence η^S(λ, x,L) = 1

√2krk2. Simplifying the expression for4A,we obtain the desired result. ¥

ForH-palindromic pencils we have the following.

Theorem 3.4.8. Let S be the space ofH-palindromic pencils andL∈S be given by L(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+|λ|² (krk²₂− |x^Hr|²)

(1 +|λ|²)² , if |λ|>1,

√2 s

|x^HAx|²+krk²₂− |x^Hr|²

(1 +|λ|²)² , if |λ|<1,

η(λ, x,L), if |λ|= 1.

Now define

4A:=



















−xx^HAxx^H+_1+|λ|¹ 2[λxr^H(I−xx^H) + (I−xx^H)rx^H] +^xHAx_λ^(I−xx_(krk2^{H)rrH(I−xxH)}

2−|x^Hr|²) , if|λ|>1,

−xx^HAxx^H+_1+|λ|¹ 2[λxr^H(I−xx^H) + (I−xx^H)rx^H] +^{λ xHAx}_krk^(I−xx2 ^{H)rrH(I−xxH)} 2−|x^Hr|² ,

if|λ|<1,

1

1+|λ|²[rx^H+λxr^H(I−xx^H)−^xHr(I−xx_(krk2^{H)rrH(I−xxH)}

2−|x^Hr|²) ], if |λ|= 1,

and consider the pencil 4L(z) :=4A+z(4A)^H. Then4L ∈S, L(λ)x+4L(λ)x= 0 and

|||4L|||=η^S(λ, x,L).

Proof: First, suppose that|λ| 6= 1.Then from the proof of Theorem 3.3.8, we have

4A = Q

à −x^HAx _1+|λ|^λ 2(Q^H₁ r)^H

1

1+|λ|²Q^H₁r A1

! Q^H

such that4L(λ)x+ L(λ)x= 0.In this case, we have

µ4A =



 q

|x^HAx|²+^|λ|2(krk_(1+|λ|^22−|x2)^H2r|2), if|λ|>1, q

|x^HAx|²+^krk_(1+|λ|^22−|x2^H)^r|22, if|λ|<1.

Hence by the DKW Theorem 1.2.5, we have

A1 =





λ x^HAx

|λ|²(krk²₂−|x^Hr|²)Q^H₁r(Q^H₁r)^H, if|λ|>1,

λ x^HAx

krk²₂−|x^Hr|²Q^H₁ r(Q^H₁r)^H, if|λ|<1.

This gives

η^S(λ, x,L) =





√2 q

|x^HAx|²+^{|λ|2 (krk}_(1+|λ|^22−|x2)^2Hr|2), if|λ|>1,

√2 q

|x^HAx|²+^krk_(1+|λ|^22−|xH2)^r|22, if|λ|<1.

Simplifying the expression for4A, we obtain the desired result.

When|λ|= 1,again from the proof of Theorem 3.3.8, we have

4A = Q

à x^Hr

1+|λ|² λ

1+|λ|²(Q^H₁ r)^H

1

1+|λ|²Q^H₁r A1

! Q^H

Now, we haveµ4A= krk2

1 +|λ|². Hence by the DKW Theorem 1.2.5, we have A1=−x^Hr (I−xx^H)rr^H(I−xx^H)

(1 +|λ|²)(krk²₂− |x^Hr|²) . Consequently, we haveη^S(λ, x,L) = krk2

√2 .Simplifying the expression for 4A,we obtain the desired result. ¥

Remark 3.4.9. Let(λ, x)∈C×Cⁿ withx^Hx= 1andS∈ {T-symmetric, T-skew-symmetric,

T-odd, T-even, T-palindromic, H-Hermitian, H-skew-Hermitian, H-odd, H-even, H-palindromic}.

ForL∈S, consider the set

S(λ, x,L) :={K∈S: L(λ)x+ K(λ)x= 0}.

ThenS(λ, x,L)6=∅ andmin{|||K|||: K∈S(λ, x,L)}=η^S(λ, x,L). Further, Sopt(λ, x,L) :={4L∈S(x, λ,L) :|||4L|||=η^S(λ, x,L)}

is an infinite set and is characterized by the DKW Theorem 1.2.5 by taking into account the nonzero contractions. Let 4L ∈Sopt(λ, x,L). Then each pencil in S(λ, x,L) is of the form 4L + (I−xx^H)^∗Z(I−xx^H)for some Z∈S,where∗is either the transpose or the conjugate transpose depending upon the structure defined byS.In other words, we have

S(λ, x,L) =4L + (I−xx^H)^∗S(I−xx^H).

Needless to mention that Remark 3.4.9 remains valid for structured pencils in S whose coefficient matrices are element of Jordan and/or Lie algebras associated with a scalar product considered in the previous section. In such a case the ∗in (I−xx^H)^∗ is the adjoint induced by the scalar product that defines the Jordan and Lie algebras.

We now illustrate various structured and unstructured backward errors by numerical ex- amples. We use matlab.7.0 for our computation. We generate AandB as follows:

>> randn(‘state’,15), A = randn(50)+ i*randn(50); A = A ± A^∗;

>> randn(‘state’,25), B = randn(50)+i*randn(50); B = B ± B^∗;

ForT-palindromic/H-palindromic pencils, we generateAandB by

>> randn(‘state’,15), A = randn(50)+ i*randn(50); B = A^∗; Here A^∗=A^T or A^∗=A^H.Finally, we compute (λ, x) by

>> [V,D] = eig(A,B); λ = -D(2,2); x = V(:,2)/norm(V(:,2));

We denote byη_F^S(λ, x,L) andη^S₂(λ, x,L) the backward errorη^S(λ, x,L) whenC^n×nis equipped with the Frobenius norm and the spectral norm, respectively. Note thatη(λ, x,L) is the same for the spectral and the Frobenius norms. Table 3.2 gives the computed result.

Note that structured backward errors are bigger than or equal to unstructured backward errors but they are marginally so. On the other hand, structured condition numbers are less than or equal to unstructured condition numbers [16, 54]. Consequently, structured backward errors when combined with structured condition numbers provide almost the same approx- imate upper bounds on the errors in the computed eigenelements as do their unstructured counterparts. We mention that thematlabeigcommand does not ensure spectral symmetry in the computed eigenvalues.