Structured backward error of approximate eigenpair

In this section we derive structured backward error of an approximate eigenpair (λ, x) of P∈S. The backward error of (λ, x) is defined as the smallest, in norm, perturbation4P of P such that (λ, x) is an eigenpair of P +4P, that is, (P(λ) +4P(λ))x= 0. We make the convention throughout the chapter that,4P∈Pm(C^n×n) is of the form4P =P_m

j=0z^j4Aj. Recall that the Frobenius and the spectral norm defined on Pm(C^n×n) are given by

|||P|||F :=



 Xm

j=0

kAjk²_F





1/2

and|||P|||2:=



 Xm

j=0

kAjk²₂





1/2

respectively where P(z) = P_m

i=0z^jAj. Then it follows that, for any λ ∈ C,kP(λ)k ≤

|||P|||F,2 kΛmk2,where Λm= [1, λ, . . . , λ^m]^T.

By convention, if (λ, x)∈C×Cⁿ,thenxis assumed to be nonzero, that is,x6= 0.Treating (λ, x) as an approximate eigenpair of the polynomial P∈Pm(C^n×n),we define the backward error of (λ, x) by

ηF(λ, x,P) := min

4P∈Pm(C^n×n){|||4P|||F : P(λ)x+4P(λ)x= 0}, η2(λ, x,P) := min

4P∈Pm(C^n×n){|||4P|||2: P(λ)x+4P(λ)x= 0}.

Setting r:=−P(λ)x,we haveηF(λ, x,P) =krk2/kxk2kΛmk2=η2(λ, x,P),see (4.1). Hence- forth, we denote ηF(λ, x,P) andη2(λ, x,P) byη(λ, x,P).

Next assume that P∈S.Then we define the structured backward error of (λ, x) by η_F^S(λ, x,P) := min

4P∈S{|||4P|||F : P(λ)x+4P(λ)x= 0}

η₂^S(λ, x,P) := min

4P∈S{|||4P|||2: P(λ)x+4P(λ)x= 0}.

Unless otherwise stated, we denoteη^S(λ, x,P) for bothη^S_F(λ, x,P) andη₂^S(λ, x,P).

By Theorem 5.2.1 and Theorem 5.2.2 it is obvious to see that η^S(λ, x,P) < ∞ and η(λ, x,P)≤η^S(λ, x,P).

To make the presentation simple we define Πi:C^m→C^m,by

Πi([x0, x1, x1, . . . , xm−1]^T) := [x0, x1, x2, . . . , xi−1,0,0, . . . ,0]^T.

We consider flip operatorR=





1 . .. 1



∈R^m×m.

Theorem 5.3.1. Let S be a space of T-palindromic polynomials and P ∈ S. Assume that (λ, x)is an approximate eigen-pair of Pand setr:=−P(λ)x.Then we have the following

1. mis odd:

η_F^S(λ, x,P) =







√2η(λ, x,P), ifλ=−1

√2 r

|x^Tr|²

kΠ_(m+1)/2(Λm+RΛm)k²₂ +^krk_kΛ^22−|xTr|2

mk²₂ ,ifλ6=±1

η₂^S(λ, x,P) =















η(λ, x,P)if λ=±1

√2 r

|x^Tr|²

kΠ_(m+1)/2(Λm+RΛm)k²₂ +^{kΠ(m+1)/2RΛ}_kΛ^{mk22 (krk22−|xTr|2)}

mk⁴₂ ,if |λ|>1

√2 r

|x^Tr|²

kΠ_(m+1)/2(Λm+RΛm)k²₂ +^{kΠ(m+1)/2Λm}_kΛ^{k2 (krk22−|xTr|2)}

mk⁴₂ , if |λ| ≤1 LetEj:= _kΠ ^λj+λm−j

(m+1)/2(Λm+RΛm)k²₂(x^Tr)xx^H, Fj:= _kΛ^λj

mk²₂P_x^Trx^H+_kΛ^λm−j

mk²₂xr^TPx. Forj = 0 : (m−1)/2, define

4Aj:=

( Fj, λ=−1 Ej+Fj, λ6=−1 and 4Am−j = 4A^T_j. Then 4P(z) = P_m

j=0z^j4Aj is a unique polynomial such that 4P∈S,P(λ)x+4P(λ)x= 0 and|||4P|||F =η^S_F(λ, x,P).Next, define

4Aj :=

















Fj, ifλ=−1

Ej+Fj− (|λ^j|²λ^m−j+|λ^m−j|²λ^j) x^TrP_x^Trr^TPx

|λ^m−j|² kΠ_(m+1)/2(Λm+RΛm)k²₂ (krk²₂− |x^Tr|²), if|λ|>1 Ej+Fj− (|λ^j|²λ^m−j+|λ^m−j|²λ^j) x^TrP_x^Trr^TPx

|λ^j|² kΠ_(m+1)/2(Λm+RΛm)k²₂ (krk²₂− |x^Tr|²),if |λ| ≤1, λ6=−1 forj= 0 : (m−1)/2and4Am−j =4A^T_j.Then4P(z) =P_m

j=0z^j4Aj is a polynomial such that4P∈S,P(λ)x+4P(λ)x= 0and|||4P|||2=η^S₂(λ, x,P).

2. mis even:

η_F^S(λ, x,P) =







√1 m+1

p2krk²₂− |x^Tr|²≤√

2η(λ, x,P), if λ=±1

r

(2kΠ_(m/2)+1Λm+Π_m/2RΛmk²₂−|λ^m/2|²)|x^Tr|²

kΠ_(m/2)+1Λm+Π_m/2RΛmk⁴₂ + 2^krk_kΛ^22−|xTr|2

mk²₂ ,if λ6=±1

η₂^S(λ, x,P) =



























η(λ, x,P), ifλ=±1

r

kΛm+RΛmk²₂−3|λ^m/2|²

kΠ_(m/2)+1Λm+Π_m/2RΛmk⁴₂|x^Tr|²+^{(2kΠm/2RΛmk22+|λ}_kΛ^{m/2|2) (krk22−|xTr|2)}

mk⁴₂ ,

if|λ|>1;

r

kΛm+RΛmk²₂−3|λ^m/2|²

kΠ_(m/2)+1Λm+Π_m/2RΛmk⁴₂|x^Tr|²+^{(2kΠm/2Λmk22+|λ}_kΛ^{m/2|2) (krk22−|xTr|2)}

mk⁴₂ ,

if|λ| ≤1.

Set

Gj := λ^j+λ^m−j

kΠ(m/2)+1Λm+ Πm/2RΛmk²₂xx^Trx^H+ λ^j

kΛmk²₂P_x^Trx^H+ λ^m−j kΛmk²₂xr^TPx

H_m/2 := λ^m/2

kΠ(m/2)+1Λm+ Πm/2RΛmk²₂xx^Trx^H+ λ^m/2

kΛmk²₂P_x^Trx^H+ λ^m/2

kΛmk²₂xr^TPx. Now forj= 0 : (m−2)/2 define

4Aj :=

( λ^j

kΛ_mk²₂(x^Tr)xx^H+_kΛ¹

mk²₂[λ^jP_x^Trx^H+λ^m−jxr^TPx], ifλ=±1

Gj, ifλ6=±1

4A_m/2 :=

( λ^m/2

kΛmk²₂ [(x^Tr)xx^H+P_x^Trx^H+xr^TPx], ifλ=±1

H_m/2, ifλ6=±1

and 4Am−j = 4A^T_j. Then 4P(z) = P_m

j=0z^j4Aj is a unique polynomial such that 4P∈S,P(λ)x+4P(λ)x= 0 and|||4P|||F =η^S_F(λ, x,P).

Further, forj= 0 : (m−2)/2 define

4Aj :=































λ^j

kΛmk²₂(x^Tr)xx^H+_kΛ¹

mk²₂[λ^jP_x^Trx^H+λ^m−jxr^TPx]−_kΛ^{λm−j xTrPxTrrTPx}

mk²₂ (krk²₂−|x^Tr|²), ifλ=±1 Gj− (|λ^j|²λ^m−j+|λ^m−j|²λ^j) x^Tr P_x^Trr^TPx

kΠ_(m/2)+1Λm+ Π_m/2RΛmk²₂ |λ^m−j|² (krk²₂− |x^Tr|²), if|λ|>1

Gj− (|λ^j|²λ^m−j+|λ^m−j|²λ^j)x^Tr P_x^Trr^TPx

kΠ(m/2)+1Λm+ Πm/2RΛmk²₂ |λ^j|² (krk²₂− |x^Tr|²),if|λ| ≤1, λ6=±1

4A_m/2 :=









 λ^m/2

kΛmk²₂ [(x^Tr)xx^H+P_x^Trx^H+xr^TPx]− λ^m/2x^Tr P_x^Trr^TPx

kΛmk²₂ (krk²₂− |x^Tr|²),ifλ=±1 H_m/2− λ^m/2 x^Tr P_x^Trr^TPx

kΠ(m/2)+1Λm+ Πm/2RΛmk²₂ (krk²₂− |x^Tr|²), ifλ6=±1, and 4Am−j =4A^T_j. Then 4P(z) = P_m

j=0z^j4Aj is a polynomial such that a 4P∈ S,P(λ)x+4P(λ)x= 0and|||4P|||2=η^S₂(λ, x,P).

Proof: First suppose that mis odd. By Theorem 5.2.1 there exists a polynomial 4P ∈S such that4P(λ)x+ P(λ)x= 0.Forj = 0 : (m−1)/2,consider

4Ag_j:=Q^T4AQ=

à a_jj a^T_j bj Xj

!

and 4A^T_j =4Am−j, where Q:= [x, Q1] is a unitary matrix. Now since4P(λ)x+ P(λ)x= 0,we have,

à P_m

j=0λ^jajj

P_(m−1)/2

j=0 λ^jbj+P_(m−1)/2

j=0 λ^m−jaj

!

= Ã x^Tr

Q^T₁r

! .

The minimum norm solution of P_(m−1)/2

j=0 λ^jbj+P_(m−1)/2

j=0 λ^m−jaj =Q^T₁r is given by bj =

λ^j

kΛmk²₂Q^T₁r andaj = _kΛ^λm−j

mk²₂Q^T₁r.

Forλ=−1, we havex^Tr= 0.Hence the minimum norm solution ofP_m

j=0λ^jajj =x^Tr, isajj = 0.So we have4Aj forj= 0 :m, as follows:

4Aj=Q



 0 (_kΛ^λm−j

mk²₂Q^T₁r)^T

λ^j

kΛmk²₂Q^T₁r Xj



Q^H, 4A^T_j =4Am−j, j= 0 : (m−1)/2.

Setting Xj= 0, j= 0 :m,we obtainη^S_F(λ, x,P) =√

2^√krk_m+1² =√

2η(λ, x,P).Simplifying the expressions of4Aj, j= 0 :m,we obtain the desired result.

Now letλ 6=−1. Notice that ajj = am−j,m−j, j = 0 : (m−1)/2. Hence the minimum norm solution ofP_m

j=0λ^jajj =x^Trwhich implies P_(m−1)/2

j=0 (λ^j+λ^m−j)ajj =x^Tr, is ajj = λ^j+λ^m−j

kΠ_(m+1)/2(Λm+RΛm)k²₂x^Tr.

Thus we obtain

4Aj = Q







λ^j+λ^m−j

kΠ(m+1)/2(Λm+RΛm)k²₂x^Tr (_kΛ^λm−j

mk²₂Q^T₁r)^T

λ^j

kΛmk²₂Q^T₁r Xj





Q^H, j= 0 : (m−1)/2 4Am−j = A^T_j, j= 0 : (m−1)/2.

Now setting Xj= 0, j= 0 : (m−1)/2,we have

η^S_F(λ, x,P) = √ 2

s

|x^Tr|²

kΠ(m+1)/2(Λm+RΛm)k²₂ +krk²₂− |x^Tr|² kΛmk²₂ . Simplifying the expression of4Aj, i= 0 :mwe obtain the desired result.

In particular, whenλ= 1,we getkΛmk²₂=m+1 andkΠ(m+1)/2(Λm+RΛm)k²₂= 2(m+1), which gives,

η^S_F(λ, x,P) = 1

√m+ 1 q

2krk²₂− |x^Tr|²≤√

2η(λ, x,P).

Now we consider spectral norm. For λ =−1, by DKW Theorem 1.2.5 applied to4Aj

givesµ4Aj = _kΛ^krk2

mk²₂ andXj = 0, j= 0 : (m−1)/2.Hence η^S₂(λ, x,P) = ^√krk_m+1² =η(λ, x,P).

Thus we obtain the same4Aj that we have obtained for Frobenius norm.

Next letλ6=−1.Forj= 0 : (m−1)/2,applying DKW Theorem 1.2.5 to4Aj,we have

µ4Aj =







 r

|λ^j+λ^m−j|²

kΠ_(m+1)/2(Λm+RΛm)k⁴₂|x^Tr|²+^|λm−j|2_kΛ^{(krk22−|xTr|2)}

mk⁴₂ , if|λ|>1 r

|λ^j+λ^m−j|²

kΠ_(m+1)/2(Λm+RΛm)k⁴₂|x^Tr|²+^{|λj|2 (krk}_kΛ^22−|xTr|2)

mk⁴₂ , if|λ| ≤1,

and

Xj =











− (|λ^j|²λ^m−j+|λ^m−j|²λ^j)x^TrQ^T₁r(Q^T₁r)^T

|λ^m−j|² kΠ_(m+1)/2(Λm+RΛm)k²₂(krk²₂− |x^Tr|²), if|λ|>1

− (|λ^j|²λ^m−j+|λ^m−j|²λ^j)x^TrQ^T₁r(Q^T₁r)^T

|λ^j|² kΠ_(m+1)/2(Λm+RΛm)k²₂(krk²₂− |x^Tr|²), if|λ| ≤1.

This gives,

η₂^S(λ, x,P) =









√2 r

|x^Tr|²

kΠ(m+1)/2(Λm+RΛm)k²₂ +^{kΠ(m+1)/2RΛ}_kΛ^{mk22 (krk22−|xTr|2)}

mk⁴₂ , if|λ|>1

√2 r

|x^Tr|²

kΠ_(m+1)/2(Λm+RΛm)k²₂ +^{kΠ(m+1)/2Λm}_kΛ^{k2 (krk22−|xTr|2)}

mk⁴₂ , if|λ| ≤1.

Simplifying the expression of4Aj, j= 0 : (m−1)/2 we obtain the desired result.

Now ifλ= 1, we havekΛmk²₂ =m+ 1 andkΠ_(m+1)/2(Λm+RΛm)k²₂= 2(m+ 1),which gives,

η₂^S(λ, x,P) = krk2

√m+ 1 =η(λ, x,P).

Next, suppose thatmis even. Note that,A_m/2=A^T_m/2.Then we have

à P_m

j=0λ^jajj

P_(m−2)/2

j=0 λ^jbj+P_(m−2)/2

j=0 λ^m−jaj+λ^m/2am/2

!

= Ã x^Tr

Q^T₁r

! .

The minimum norm solution ofP_(m−2)/2

j=0 λ^jbj+P_(m−2)/2

j=0 λ^m−jaj+λ^m/2am/2=Q^T₁ris given by

bj = λ^j

kΛmk²₂Q^T₁r, aj= λ^m−j

kΛmk²₂Q^T₁r, a_m/2= λ^m/2 kΛmk²₂Q^T₁r.

Forλ=±1,the minimum norm solution ofP_m

j=0λ^jajj =x^Tris ajj =_kΛ^λj

mk²₂x^Tr. Hence we have,

4Aj=Q







λ^j

kΛmk²₂x^Tr _kΛ^λm−j

mk²₂(Q^T₁r)^T

λ^j

kΛmk²₂Q^T₁r Xj





Q^H, 4A_m/2=Q







λ^m/2

kΛmk²₂x^Tr _kΛ^λm/2

mk²₂(Q^T₁r)^T

λ^m/2

kΛmk²₂Q^T₁r X_m/2





Q^H,

and 4Am−j =4A^T_j, j = 0 : (m−2)/2.Setting Xj = 0, j = 0 :m, we have η_F^S(λ, x,P) =

√1 m+1

p2krk²₂− |x^Tr|².Simplifying4Ajs forj= 0 :mwe obtain the desired result.

Again, ifλ6=±1,the minimum norm solution ofP_m

j=0λ^jajj =x^Tris ajj = λ^j+λ^m−j

kΠ_(m+2)/2Λm+ Π_m/2RΛmk²₂x^Tr, am/2,m/2= λ^m/2

kΠ_(m+2)/2Λm+ Π_m/2RΛmk²₂x^Tr.

Hence we have

4Aj = Q







λ^j+λ^m−j

kΠ_(m+2)/2Λm+Π_m/2RΛmk²₂x^Tr _kΛ^λm−j

mk²₂(Q^T₁r)^T

λ^j

kΛmk²₂Q^T₁r Xj





Q^H, j= 0 : (m−2)/2,

4A_m/2 = Q







λ^m/2

kΠ_(m+2)/2Λm+Π_m/2RΛmk²₂x^Tr _kΛ^λm/2

mk²₂(Q^T₁r)^T

λ^m/2

kΛmk²₂Q^T₁r Xm/2





Q^H

4Am−j = 4A^T_j, j= 0 : (m−2)/2.

Setting Xj = 0, j= 0 :m, we have

η_F^S(λ, x,P) = s

(2kΠ_(m+2)/2Λm+ Π_m/2RΛmk²₂− |λ^m/2|²)|x^Tr|²

kΠ(m+2)/2Λm+ Πm/2RΛmk⁴₂ + 2krk²₂− |x^Tr|² kΛmk²₂ . which gives the desired result. Simplifying the expressions of4Aj the desired result follows.

Next we consider spectral norm. If λ= ±1, applying DKW Theorem 1.2.5 to4Aj we have, µ4Aj = _kΛ^krk2

mk²₂ for j = 0 : m and η₂^S(λ, x,P) = ^√krk_m+1² = η(λ, x,P). Now by DKW Theorem 1.2.5, we have

Xj=− λ^m−j Q^T₁r(Q^T₁r)^T

kΛmk²₂ (krk²₂− |x^Tr|²), j= 0 : (m−2)/2, Xm/2=− λ^m/2Q^T₁r(Q^T₁r)^T kΛmk²₂(krk²₂− |x^Tr|²). Therefore, we have

4Aj = λ^j

kΛmk²₂rx^H+ λ^m−j

kΛmk²₂xr^TPx− λ^m−j P_x^Trr^TPx

kΛmk²₂ (krk²₂− |x^Tr|²), 4A_m/2 = λ^m/2

kΛ_mk²₂ [rx^H+xr^TPx]− λ^m/2 P_x^Trr^TPx

kΛ_mk²₂ (krk²₂− |x^Tr|²). Again for λ6=±1,we have

µ4Aj =







 r

|λ^j+λ^m−j|² |x^Tr|²

kΠ(m+2)/2Λm+Πm/2RΛmk⁴₂ +^|λm−j|2_kΛ^{(krk22−|xTr|2)}

mk⁴₂ , if|λ|>1 r

|λ^j+λ^m−j|² |x^Tr|²

kΠ(m+2)/2Λm+Πm/2RΛmk⁴₂ +^{|λj|2 (krk}_kΛ^22−|xTr|2)

mk⁴₂ , if|λ| ≤1 µ4Am/2 =

s

|λ^m/2|²|x^Tr|²

kΠ_(m+2)/2Λm+ Π_m/2RΛmk⁴₂ +|λ^m/2|² (krk²₂− |x^Tr|²) kΛmk⁴₂ . Consequently by DKW Theorem 1.2.5 we have

Xj =











− (|λ^j|²λ^m−j+|λ^m−j|²λ^j)x^Tr Q^T₁r(Q^T₁r)^T

kΠ(m+2)/2Λm+ Πm/2RΛmk²₂ |λ^m−j|² (krk²₂− |x^Tr|²), if|λ|>1

− (|λ^j|²λ^m−j+|λ^2m−j|²λ^j)x^Tr Q^T₁r(Q^T₁r)^T

kΠ(m+2)/2Λm+ Πm/2RΛmk²₂ |λ^j|² (krk²₂− |x^Tr|²), if|λ| ≤1;

X_m/2 = − λ^m/2x^Tr Q^T₁r(Q^T₁r)^T

kΠ_(m+2)/2Λm+ Π_m/2RΛmk²₂(krk²₂− |x^Tr|²).

Hence we obtain

η^S₂(λ, x,P) =



















 r

kΛm+RΛmk²₂−3|λ^m/2|²

kΠ(m+2)/2Λm+Πm/2RΛmk⁴₂|x^Tr|²+^{2kΠm/2RΛmk22+|λ}_kΛ^{m/2|2 (krk22−|xTr|2)}

mk⁴₂ ,

if|λ|>1 r

kΛm+RΛmk²₂−3|λ^m/2|²

kΠ(m+2)/2Λm+Πm/2RΛmk⁴₂|x^Tr|²+^{2kΠm/2Λmk22+|λ}_kΛ^{m/2|2 (krk22−|xTr|2)}

mk⁴₂ ,

if|λ| ≤1.

Simplifying the expressions of4Ajs we obtain the desired result.¥

Note that ifY ∈C^n×n is such thatY x= 0 andY^Tx= 0 thenY = (I−xx^H)^TZ(I−xx^H) for some matrix Z. Hence from the proof of Theorem 5.3.1, we obtain that if K is a T- palindromic polynomial such that P(λ)x+K(λ)x= 0 then K(z) =4P(z)+(I−xx^H)^TN(z)(I−

xx^H) for someT-palindromic polynomial N,where 4P is given in Theorem 5.3.1.

Remark 5.3.2. If |x^Tr|=krk2,then kQ^T₁rk2= 0. In such a case, considering Xj = 0, j = 0 :m, we obtain the desired results.

Theorem 5.3.3. Let S be the space of T-anti-palindromic polynomials and P∈S. Assume that (λ, x)∈C×Cⁿ and setr:=−P(λ)x.Then we have the following.

1. mis odd: we have

η^S_F(λ, x,P) =





√2 η(λ, x,P), if λ= 1

√2 r

|x^Tr|²

kΠ_(m+1)/2(Λm−RΛm)k²₂ +^krk_kΛ^22−|xTr|2

mk²₂ , if λ6= 1.

η^S₂(λ, x,P) =























η2(λ, x,P), ifλ= 1

√2 r

|x^Tr|²

kΠ(m+1)/2(Λm−RΛm)k²₂ +^{kΠ(m+1)/2RΛ}_kΛ^{mk22(krk22−|xTr|2)}

mk⁴₂ , if|λ|>1,

√2 r

|x^Tr|²

kΠ(m+1)/2(Λm−RΛm)k²₂ +^kΠ(m+1)/2Λ_kΛ^{mk22(krk22−|xTr|2)}

mk⁴₂ , if|λ| ≤1.

Set

Ej := λ^j−λ^m−j

kΠ_(m+1)/2(Λm−RΛm)k²₂(x^Tr)xx^H− 1

kΛmk²₂[λ^m−jxr^TPx−λ^jP_x^Trx^H] Fj := λ^j−λ^m−j

kΠ(m+1)/2(Λm−RΛm)k²₂(x^Tr)xx^H− 1

kΛmk²₂[λ^m−jxr^TPx−λ^jP_x^Trx^H].

Forj= 0 : (m−1)/2,define 4Aj:=

( 1

kΛmk²₂[rx^H−xr^T], if λ= 1

Ej, λ6= 1

and4Am−j=−A^T_j.Then4P(z) =P_m

j=0z^j4Aj is a unique polynomial4P∈S such

thatP(λ)x+4P(λ)x= 0 and|||4P|||F =η_F^S(λ, x,P).Next, define

4Aj:=





















 1

kΛmk²₂[rx^H−xr^T], if λ= 1

Ej+ (|λ^j|²λ^m−j− |λ^m−j|²λ^j)x^TrP_x^Trr^TPx

|λ^m−j|²kΠ_(m+1)/2(Λm−RΛm)k²₂(krk²₂− |x^Tr|²), if |λ|>1

Ej+ (|λ^j|²λ^m−j− |λ^m−j|²λ^j)x^TrP_x^Trr^TPx

|λ^j|²kΠ(m+1)/2(Λm−RΛm)k²₂(krk²₂− |x^Tr|²), if |λ| ≤1, λ6= 1 and 4Am−j = −A^T_j, j = 0 : (m−1)/2.Then 4P(z) = P_m

j=0z^j4Aj is a polynomial such that4P∈S,P(λ)x+4P(λ)x= 0and|||4P|||2=η^S₂(λ, x,P).

2. mis even:

η_F^S(λ, x,P) =





√2 η(λ, x,P), if λ= 1

√2 r

|x^Tr|²

kΠ_m/2(Λm−RΛm)k²₂ +^krk_kΛ^22−|xTr|2

mk²₂ , if λ6= 1.

η₂^S(λ, x,P) =















η(λ, x,P), if λ= 1,

r

2|x^Tr|²

kΠm/2(Λm−RΛm)k²₂ +^{(2kΠm/2RΛmk2+|λ}_kΛ^{m/2|2)(krk22−|xTr|2)}

mk⁴₂ ,if|λ|>1, r

2|x^Tr|²

kΠ_m/2(Λm−RΛm)k²₂ +^{(2kΠm/2Λmk2+|λ}_kΛ^{m/2|2)(krk22−|xTr|2)}

mk⁴₂ , if|λ| ≤1.

Forj= 0 : (m−2)/2,consider

4A_j :=







1

kΛmk²₂[rx^H−xr^T], ifλ= 1

Fj, ifλ6= 1

, 4A_m/2:=









1

kΛmk²₂[rx^H−xr^T], ifλ= 1

λ^m/2

kΛmk²₂[rx^H−xr^T], ifλ6= 1 and 4Am−j = −A^T_j, j = 0 : (m−2)/2. Then 4P(z) = P_m

j=0z^j4Aj is a unique polynomial such that 4P ∈ S,P(λ)x+4P(λ)x = 0 and |||4P|||F = η_F^S(λ, x,P). Next, define

4Aj :=





















 1

kΛmk²₂[rx^H−xr^T], ifλ= 1 Fj+ (|λ^j|²λ^m−j− |λ^m−j|²λ^j)x^TrP_x^Trr^TPx

|λ^m−j|²kΠm/2(Λm−RΛm)k²₂(krk²₂− |x^Tr|²), if|λ|>1

Fj+ (|λ^j|²λ^m−j− |λ^m−j|²λ^j)x^TrP_x^Trr^TPx

|λ^j|²kΠm/2(Λm−RΛm)k²₂(krk²₂− |x^Tr|²), if|λ| ≤1, λ6= 1

4A_m/2 :=







 1

kΛmk²₂[rx^H−xr^T], ifλ= 1 λ^m/2

kΛmk²₂[rx^H−xr^T], ifλ6= 1

and4Am−j =−A^T_j, j = 0 : (m−2)/2. Then 4P(z) = P_m

j=0z^j4Aj is a polynomial such that4P∈S,4P∈Ssuch that P(λ)x+4P(λ)x= 0and|||4P|||2=η^S₂(λ, x,P).

Proof: First assume that P∈S andmis odd. By Theorem 5.2.2 there exists 4P∈Ssuch that 4P(λ)x+ P(λ)x= 0.Forj= 0 : (m−1)/2,consider

4Ag_j :=Q^T4AQ=

à ajj a^T_j bj Xj

!

and 4Am−j =−4A^T_j, j= 0 : (m−1)/2, where Q= [x, Q1] is a unitary matrix. Since4P(λ)x+ P(λ)x= 0,we have,





P_(m−1)/2

j=0 λ^ja_jj−P_(m−1)/2

j=0 λ^m−ja_jj P_(m−1)/2

j=0 λ^jbj−P_(m−1)/2

j=0 λ^m−jaj



= Ã x^Tr

Q^T₁r

! .

The minimum norm solution of P_(m−1)/2

j=0 λ^jbj−P_(m−1)/2

j=0 λ^m−jaj =Q^T₁r is given by bj =

λ^j

kΛmk²₂Q^T₁r andaj =−_kΛ^λm−j

mk²₂Q^T₁r, j= 0 : (m−1)/2.

Forλ= 1,we havex^Tr= 0. Hence the minimum norm solution ofP_m

j=0λ^jajj =x^Tr, is ajj = 0, j= 0 :m.Thus we have

4Aj=Q



 0 −^λm−j_kΛ^(QT1r)T

mk²₂ λ^jQ^T₁r

kΛmk²₂ Xj



Q^H, Am−j=A^T_j j= 0 : (m−1)/2.

Setting Xj = 0, j = 0 : mwe have the minimum Frobenius norm of 4Aj. Consequently we have,

η_F^S(λ, x,P) =√ 2 krk2

kΛmk2

=√

2η(λ, x,P).

Simplifying the expression of4Aj, j= 0 :mwe obtain the desired result.

Next suppose thatλ6= 1.Then the minimum norm solution of

(m−1)/2X

j=0

λ^jajj−

(m−1)/2X

j=0

λ^m−jajj =x^Tr

is given by

ajj = λ^j−λ^m−j

kΠ_(m+1)/2(Λm−RΛm)k²₂x^Tr, j= 0 : (m−1)/2.

Therefore we have

4Aj=Q







λ^j−λ^m−j

kΠ(m+1)/2(Λm−RΛm)k²₂x^Tr −^λm−j_kΛ^(QT1r)T

mk²₂

λ^jQ^T₁r

kΛmk²₂ Xj





Q^H, j= 0 : (m−1)/2.

TakingXj= 0,we obtain the minimum Frobenius norm of4Aj and consequently we have

η^S_F(λ, x,P) = √ 2

vu

utP_(m−1)/2

j=0 |λ^j−λ^m−j|²|x^Tr|² kΠ_(m+1)/2(Λm−RΛm)k⁴₂ +

P_(m−1)/2

j=0 (|λ^m−j|²+|λ^j|²)(krk²₂− |x^Tr|²) kΛmk⁴₂

= √

2 s

|x^Tr|²

kΠ_(m+1)/2(Λm−RΛm)k²₂+krk²₂− |x^Tr|² kΛmk²₂ .

Simplifying the expression of4Aj for allj = 0 : (m−1)/2 we obtain the desired result.

Next we consider the spectral norm. Assume thatλ= 1.Applying DKW Theorem 1.2.5 to 4Aj we have µ4Aj =_kΛ^krk2

mk2, j = 0 : (m−1)/2 andXj = 0.Therefore we have η₂^S(λ, x,P) =

krk2

kΛmk2 =η(λ, x,P).Simplifying the expression of4Ajwe obtain4Aj =−_kΛ¹

mk²₂[xr^T−rx^H].

Hence the result follows.

Next, suppose thatλ6= 1.In this case we have

µ4Aj =







 r

|λ^j−λ^m−j|²|x^Tr|²

kΠ(m+1)/2(Λm−RΛm)k⁴₂ +^|λm−j|2_kΛ^{(krk22−|xTr|2)}

mk⁴₂ , if |λ|>1 r

|λ^j−λ^m−j|²|x^Tr|²

kΠ(m+1)/2(Λm−RΛm)k⁴₂ +^|λj|2(krk_kΛ^22−|xTr|2)

mk⁴₂ , if |λ| ≤1 forj= 0 : (m−1)/2.Therefore applying DKW Theorem 1.2.5 to4Aj we have

Xj =











(|λ^j|²λ^m−j− |λ^m−j|²λ^j)x^Tr(Q^T₁r)(Q^T₁r)^T

|λ^m−j|²kΠ_(m+1)/2(Λm−RΛm)k²₂(krk²₂− |x^Tr|²), if|λ|>1, (|λ^j|²λ^m−j− |λ^m−j|²λ^j)x^Tr(Q^T₁r)(Q^T₁r)^T

|λ^j|²kΠ_(m+1)/2(Λm−RΛm)k²₂(krk²₂− |x^Tr|²), if|λ| ≤1, forj= 0 : (m−1)/2.

Hence we have

η^S₂(λ, x,P) =









√2 r

|x^Tr|²

kΠ_(m+1)/2(Λm−RΛm)k²₂ +^{kΠ(m+1)/2RΛ}_kΛ^{mk22(krk22−|xTr|2)}

mk⁴₂ , if |λ|>1,

√2 r

|x^Tr|²

kΠ(m+1)/2(Λm−RΛm)k²₂ +^kΠ(m+1)/2Λ_kΛ^{mk22(krk22−|xTr|2)}

mk⁴₂ , if |λ| ≤1.

Simplifying the expression of4Aj forj= 0 : (m−1)/2 we obtain the desired result.

Next, assume that m is even. Note that in this case A^T_m/2 =−A_m/2. Consequently we

have 



P_(m−2)/2

j=0 λ^jajj−P_(m−2)/2

j=0 λ^m−jajj

P_(m−2)/2

j=0 λ^jbj+am/2λ^m/2−P_(m−2)/2

j=0 λ^m−jaj



= Ã x^Tr

Q^T₁r

! .

Note thatam/2,m/2= 0,sinceAm/2is a skew-symmetric matrix. The minimum norm solution ofP_(m−2)/2

j=0 λ^jbj+am/2λ^m/2−P_(m−2)/2

j=0 λ^m−jaj=Q^T₁ris given by bj =λ^jQ^T₁r

kΛmk²₂, aj=−λ^m−jQ^T₁r

kΛmk²₂ , am/2=λ^m/2Q^T₁r

kΛmk²₂ , j= 0 : (m−2)/2.

Forλ= 1, x^Tr= 0.Hence the minimum norm solution ofP_(m−2)/2

j=0 λ^jajj−P_(m−2)/2

j=0 λ^m−jajj = 0 is given by ajj = 0.Consequently, we have,

4Aj =Q



 0 −^(Q_kΛ^T1r)T

mk²₂ Q^T₁r

kΛ_mk²₂ Xj



Q^H, j = 0 : (m−2)/2.

Setting Xj = 0, j = 0 : (m−2)/2 we obtain the minimum norm of4Aj and consequently we haveη^S_F(λ, x,P) =√

2η(λ, x,P).Simplifying the expression of4Aj we obtain the desired

result.

Next, suppose thatλ6= 1.Then the minimum norm solution of

(m−2)/2X

j=0

λ^jajj−

(m−2)/2X

j=0

λ^m−jajj =x^Tr

is given by

ajj = λ^j−λ^m−j

kΠm/2(Λm−RΛm)k²₂x^Tr andam/2,m/2= 0. Consequently, forj= 0 : (m−2)/2,we obtain

4Aj = Q







λ^j−λ^m−j

kΠ_m/2(Λm−RΛm)k²₂x^Tr −^λm−j_kΛ^(QT1r)T

mk²₂

λ^jQ^T₁r

kΛmk²₂ Xj





Q^H,

4Am/2 = Q



 0 −^λm/2_kΛ^(QT1r)T

mk²₂ λ^m/2Q^T₁r

kΛmk²₂ Xm/2



Q^H.

SettingXj= 0 =X_m/2,we get the minimum Frobenius norm of4Ajand4A_m/2respectively and these give,

η_F^S(λ, x,P) = √ 2

s

|x^Tr|²

kΠ_m/2(Λm−RΛm)k²₂ +krk²₂− |x^Tr|² kΛmk²₂ . Simplifying the expressions of4Aj forj= 0 :m/2 we obtain the desired result.

Next consider spectral norm. Suppose thatλ= 1.Then considerµ4Aj = _kΛ^krk2

mk2 and by DKW Theorem 1.2.5, we have Xj = 0.Therefore we have η^S₂(λ, x,P) = _kΛ^krk2

mk2 =η(λ, x,P).

Simplifying the expression of 4Aj we obtain 4Aj = −_kΛ¹

mk²₂[xr^T −rx^H] for j = 0 :m/2.

Hence the result follows.

Next, suppose λ 6= 1. Consider µ4Am/2 =

q|λ^m|²(krk²₂−|x^Tr|²)

kΛmk⁴₂ . Then again by DKW Theorem 1.2.5 we have X_m/2= 0.Now forj= 0 : (m−2)/2 consider

µ4Aj =







 r

|λ^j−λ^m−j|²|x^Tr|²

kΠ_m/2(Λm−RΛm)k⁴₂ +^|λm−j|2_kΛ^{(krk22−|xTr|2)}

mk⁴₂ , if |λ|>1, r

|λ^j−λ^m−j|²|x^Tr|²

kΠ_m/2(Λm−RΛm)k⁴₂ +^|λj|2(krk_kΛ^22−|xTr|2)

mk⁴₂ , if |λ| ≤1.

Then by DKW Theorem 1.2.5 applied to4Aj we have

Xj=











(|λ^j|²λ^m−j− |λ^m−j|²λ^j)x^Tr(Q^T₁r)(Q^T₁r)^T

|λ^m−j|²kΠm/2(Λm−RΛm)k²₂(krk²₂− |x^Tr|²), if |λ|>1, (|λ^j|²λ^m−j− |λ^m−j|²λ^j)x^Tr(Q^T₁r)(Q^T₁r)^T

|λ^j|²kΠm/2(Λm−RΛm)k²₂(krk²₂− |x^Tr|²), if |λ| ≤1,

forj= 0 : (m−2)/2.Therefore we have

η^S₂(λ, x,P) =







 r

2|x^Tr|²

kΠ_m/2(Λm−RΛm)k²₂ +^{(2kΠm/2RΛmk2+|λ}_kΛ^{m/2|2)(krk22−|xTr|2)}

mk⁴₂ , if |λ|>1, r

2|x^Tr|²

kΠm/2(Λm−RΛm)k²₂ +^{(2kΠm/2Λmk2+|λ}_kΛ^{m/2|2)(krk22−|xTr|2)}

mk⁴₂ , if |λ| ≤1.

Simplifying the expressions of4Aj forj= 0 :m/2 we obtain the desired result. ¥

Analogous to theT-palindromic polynomial we conclude that if K is aT-anti-palindromic polynomial such that P(λ)x+ K(λ)x= 0 then K(z) =4P(z) + (I−xx^H)^TN(z)(I−xx^H) for some T-anti-palindromic polynomial N,where4P is given in Theorem 5.3.3.

Remark 5.3.4. If |x^Tr|=krk2,then kQ^T₁rk2= 0. In such a case, considering Xj = 0, j = 0 :m, we obtain the desired results.

Next we consider H-palindromic/H-anti-palindromic polynomials. To make the presen- tation simple we proceed as follows.

Letz=a+i b∈C.Define a map vec :C→R²by vec(z) =

à re(z)

im(z)

!

.Further define a map M:C→R^2×2 byM(z) =

Ã

re(z) −im(z) im(z) re(z)

!

. Then the following hold:

• vec(z) = Σ vec(z),where Σ =

à 1 0 0 −1

! .

• vec(z1z2) =M(z1)vec(z2), z1, z2∈C.

• M(z) =M(z)^T.

Assume that λ∈C, andajj ∈C.Then if we haveP_m

j=0λⁱajj =x^Hrthen applying the map vec both sides we obtain

vec(

Xm

j=0

λⁱajj) = vec(x^Hr) ⇒ Xm

j=0

M(λⁱ)vec(ajj) = vec(x^Hr). (5.1)

Theorem 5.3.5. Let S be the space ofH-palindromic polynomials. Assume that P∈S and (λ, x)∈C×Cⁿ. Settingr:=−P(λ)x,we have the following.

1. Let mbe odd. Then

η_F^S(λ, x,P) =





√1 m+1

p2krk²₂− |r^Hx|²≤√

2η(λ, x,P), if|λ|= 1

√2 q

kbrk²₂+^krk_kΛ^22−|xHr|2

mk²₂ , if|λ| 6= 1.

η₂^S(λ, x,P) =

















η(λ, x,P), if|λ|= 1

√2 r

kbrk²₂+^{kΠ(m+1)/2RΛ}_kΛ^{mk2 (krk22−|xHr|2)}

mk⁴₂ , if|λ|>1

√2 r

kbrk²₂+^{kΠ(m+1)/2Λm}_kΛ^{k2 (krk22−|xHr|2)}

mk⁴₂ , if|λ|<1.

wherebr= h

H0 H1 . . . H_(m−1)/2 i_†

vec(x^Hr) and

Hj=





re(λ^j) +re(λ^m−j) −im(λ^j) +im(λ^m−j) im(λ^j) +im(λ^m−j) re(λ^j)−re(λ^m−j)



, j= 0 : (m−1)/2.

Let Ej:= _kΛ¹

mk²₂[λ^m−jxr^HPx+λ^jPxrx^H].Forj= 0 : (m−1)/2 define 4Aj :=





λ^j

kΛmk²₂xr^Hxx^H+Ej, if |λ|= 1 e^T_jbrxx^H+Ej, if |λ| 6= 1 and 4Am−j = 4A^H_j . Then 4P(z) = P_m

j=0z^j4Aj is a unique polynomial such that 4P∈S,P(λ)x+4P(λ)x= 0 and|||4P|||F =η^S_F(λ, x,P).Next define

4Aj:=



















 λ^j

kΛmk²₂xr^Hxx^H+Ej− x^Hr λ^m−jPxrr^HPx

kΛmk²₂ (krk²₂− |x^Hr|²), if |λ|= 1 e^T_jrxxb ^H+Ej− e^T_jr λb ^j λ^m−j Pxrr^HPx

|λ^m−j|² (krk²₂− |x^Hr|²), if |λ|>1 e^T_jrxxb ^H+Ej−e^T_jbr λ^j λ^m−j Pxrr^HPx

|λⁱ|² (krk²₂− |x^Hr|²) , if |λ|<1 forj= 0 : (m−1)/2with4Am−j=4A^H_j .Then4P(z) =P_m

j=0z^j4Ajis a polynomial such that4P∈S,P(λ)x+4P(λ)x= 0and|||4P|||2=η^S₂(λ, x,P).

2. Let mbe even. Then

η_F^S(λ, x,P) =





√1 m+1

p2krk²₂− |r^Hx|²≤√

2η(λ, x,P), if|λ|= 1 q

(2kbrk²₂− |e^T_m/2br|²) + 2^krk_kΛ^22−|xHr|2

mk²₂ , if|λ| 6= 1.

η₂^S(λ, x,P) =













η(λ, x,P) if|λ|= 1

r

(2kbrk²₂− |e^T_m/2br|²) +^{(2kΠm/2RΛmk22+|λ}_kΛ^{m/2|2)(krk22−|xHr|2)}

mk⁴₂ , if|λ|>1 r

(2kbrk²₂− |e^T_m/2br|²) +^{(2kΠm/2Λmk22+|λ}_kΛ^{m/2|2)(krk22−|xHr|2)}

mk⁴₂ , if|λ|<1.

wherebr=h

H0 H1 . . . H_(m−2)/2 H_m/2 i_†

vec(x^Hr)and

Hj = Ã

re(λ^j) +re(λ^m−j) −im(λ^j) +im(λ^m−j im(λ^j) +im(λ^m−j) re(λ^j)−re(λ^m−j)

!

, j = 0 : (m−2)/2,

Hm/2 =

à re(λ^m/2) im(λ^m/2)

! .

LetFj:= _kΛ¹

mk²₂[λ^m−jxr^HPx+λ^jPxrx^H]andGm/2:=_kΛ¹

mk²₂[λ^m/2xr^HPx+λ^m/2Pxrx^H].

Then for j= 0 : (m−2)/2,define

4Aj:=







λ^j

kΛmk²₂xr^Hxx^H+Fj, if |λ|= 1 e^T_jbrxx^H+Fj, if |λ| 6= 1

,4A_m/2:=

















λ^m/2

kΛmk²₂xr^Hxx^H+Gm/2, if|λ|= 1 e^T_m/2brxx^H+Gm/2,

if|λ| 6= 1 and 4Am−j = 4A^H_j . Then 4P(z) = P_m

j=0z^j4Aj is a unique polynomial such that 4P∈S,P(λ)x+4P(λ)x= 0 and|||4P|||F =η^S_F(λ, x,P).Next define

4Aj :=



















 λ^j

kΛmk²₂xx^Hrx^H+Fj− λ^m−j r^Hx Pxrr^HPx

kΛmk²₂ (krk²₂− |r^Hx|²), if |λ|= 1 e^T_jbrxx^H+Fj− e^T_jbr λ^j λ^m−j Pxrr^HPx

|λ^m−j|² (krk²₂− |x^Hr|²), if |λ|>1 e^T_jbrxx^H+Fj−e^T_jbr λ^j λ^m−j Pxrr^HPx

|λⁱ|² (krk²₂− |x^Hr|²) , if |λ|<1

4A_m/2 :=









 λ^m/2

kΛmk²₂xx^Hrx^H+Gm− λ^m/2 r^Hx Pxrr^HPx

kΛmk²₂ (krk²₂− |r^Hx|²), if|λ|= 1 e^T_m/2brxx^H+G_m/2−e^T_m/2r Pb xrr^HPx

(krk²₂− |x^Hr|²), if|λ| 6= 1.

and 4Am−j = 4A^H_j . Then 4P(z) = P_m

j=0z^j4Aj is a polynomial such that 4P ∈ S,P(λ)x+4P(λ)x= 0and|||4P|||2=η^S₂(λ, x,P).

Proof: Let P∈Sgiven by P(z) =P_m

j=0z^jAj.First suppose thatmis odd. By Theorem 5.2.2 there exists4P∈S,such that4P(λ)x+ P(λ)x= 0.Define

4A_j:=Q

à ajj a^H_j bj Xj

!

Q^H, X_m−j =X_j^H, j= 0 : (m−1)/2,

where Q= [x, Q1] is a unitary matrix. Since4P(λ)x+ P(λ)x= 0,we have

à P_m

j=0λ^jajj

P_(m−1)/2

j=0 λ^jbj+P_(m−1)/2

j=0 λ^m−jaj

!

=

à x^Hr Q^H₁ r

! .

Consequently, the minimum norm solution of P_(m−1)/2

j=0 λ^jbj+P_(m−1)/2

j=0 λ^m−jaj =Q^H₁r is given by

bj= λ^j

kΛmk²₂Q^H₁r, aj = λ^m−j

kΛmk²₂Q^H₁ r, j= 0 : (m−1)/2.

Note that for |λ| = 1, we have x^Hr = λ^mx^Hr. Hence the minimum norm solution of

P_m

j=0λ^jajj =x^Hrisajj =_kΛ^λj

mk²₂x^Hr, j= 0 :m.Therefore we have

4Aj = Q







λ^j

kΛmk²₂x^Hr _kΛ^λm−j

mk²₂(Q^H₁ r)^H

λ^j

kΛmk²₂Q^H₁ r Xj





Q^H,

4Am−j = 4A^H_j , j= 0 : (m−1)/2.

Setting Xj = 0 givesη^S_F(λ, x,P) = ^√_m+1¹ p

2krk²₂− |r^Hx|². Simplifying the expressions for 4Ajs we obtain the desired result.

Next, let|λ| 6= 1. The value ofajj for the equationP_m

j=0λ^jajj =x^Hr, j= 0 : (m−1)/2 is achieved after employing the condition ajj =am−j,m−jj= 0 : (m−1)/2 in equation (5.1) which becomes

(m−1)/2X

j=0

(M(λ^j) +M(λ^m−j)Σ)vec(ajj) = vec(x^Hr) and the solution is given by ajj =e^T_jbrwhere

b r=h

H0 H1 . . . H(m−1)/2

i†

vec(x^Hr) and

H_j=

à re(λ^j) +re(λ^m−j) −im(λ^j) +im(λ^m−j) im(λ^j) +im(λ^m−j) re(λ^j)−re(λ^m−j)

! . Thus we obtain,

4Aj=Q







e^T_jbr λ^m−j(Q_kΛ^H1r)H

mk²₂

λ^jQ^H₁r

kΛmk²₂ Xj





Q^H, j= 0 : (m−1)/2.

SettingXj= 0,we obtainη_F^S(λ, x,P) =√ 2

q

kbrk²₂+^krk_kΛ^22−|xHr|2

mk²₂ .Simplifying the expressions of4Ajs we obtain the desired result.

Next we consider spectral norm. If|λ|= 1 applying DKW Theorem 1.2.5 to4Ajwe have µ4Aj = _kΛ^krk2

mk²₂ and

Xj=−x^Hr λ^m−jQ^H₁ r(Q^H₁r)^H

kΛmk²₂ (krk²₂− |x^Hr|²), j= 0 : (m−1)/2.

This gives, η₂^S(λ, x,P) = ^√krk_m+1² =η(λ, x,P).Simplifying the expressions for 4Ajs we obtain the desired result.

Further, for|λ| 6= 1,we consider,

µ4Aj =





 q

|e^T_jr|b²+^|λm−j|2_kΛ^{(krk22−|xHr|2)}

mk⁴₂ , if|λ|>1 q

|e^T_jr|b²+^{|λi|2 (krk}_kΛ^22−|xHr|2)

mk⁴₂ , if|λ|<1

forj= 0 : (m−1)/2.Then by DKW Theorem 1.2.5 we have

Xj=







−^eTj_|λ^{br λ}m−j^{j λ}|^2m−j(krk^Q2₂^H1−|x^r(QHH1r|^2r))^H, if|λ|>1

−^{eTjbr λ}

j λ^m−j Q^H₁r(Q^H₁r)^H

|λⁱ|² (krk²₂−|x^Hr|²) , if|λ|<1.

This gives

η₂^S(λ, x,P) =









√2 r

kbrk²+^{kΠ(m+1)/2RΛ}_kΛ^{mk2 (krk22−|xHr|2)}

mk⁴₂ , if|λ|>1

√2 r

kbrk²+^{kΠ(m+1)/2Λm}_kΛ^{k2(krk22−|xHr|2)}

mk⁴₂ , if|λ|<1.

Simplifying the expressions of4Aj for allj = 0 : (m−1)/2 we obtain the desired result.

Now suppose that P∈ S and m is even. Notice that Am/2 =A^H_m/2. By Theorem 5.2.2 there exists a polynomial 4P∈S such that4P(λ)x+ P(λ)x= 0.Define

4Aj := Q Ã

ajj a^H_j bj Xj

!

Q^H, Xm−j =X_j^H, j= 0 : (m−2)/2,

4Am/2 := Q

à am/2,m/2 a^H_m/2 a_m/2 X_m/2

!

Q^H, Xm/2=X_m/2^H . Since4P(λ)x+ P(λ)x= 0, we have

à P_m

j=0λ^jajj

P_(m−2)/2

j=0 λ^jbj+P_(m−2)/2

j=0 λ^m−jaj+λ^m/2am/2

!

=

à x^Hr Q^H₁ r

! .

The minimum norm solution of P_(m−2)/2

j=0 λ^jbj+P_(m−2)/2

j=0 λ^m−jaj+λ^m/2am/2 = Q^H₁r is given by

bj = λ^j

kΛmk²₂, aj= λ^m−j

kΛmk²₂, am/2= λ^m/2 kΛmk²₂.

Note that for |λ| = 1, we have x^Hr = λ^mx^Hr. Hence the minimum norm solution of P_m

j=0λ^jajj =x^Hris given byajj = _kΛ^λj

mk²₂x^Hr, j= 0 :m.Therefore we have

4Aj = Q







λ^j

kΛmk²₂x^Hr _kΛ^λm−j

mk²₂(Q^H₁ r)^H

λ^j

kΛmk²₂Q^H₁ r Xj





Q^H,

4Am/2 = Q







λ^m/2

kΛmk²₂x^Hr _kΛ^λm/2

mk²₂(Q^H₁ r)^H

λ^m/2

kΛmk²₂Q^H₁ r Xm/2





Q^H,

4Am−j = 4A^H_j , j= 0 : (m−2)/2, which gives, η^S_F(λ, x,P) = ^√_m+1¹ p

2krk²₂− |r^Hx|². Simplifying the expressions of4Aj for j = 0 :m/2 we obtain the desired result.

Now let |λ| 6= 1. Then the value of a_jj from the equation P_m

j=0λ^ja_jj = x^Hr, j = 0 :

(m−2)/2 is achieved after employing the conditionajj =am−j,m−j, j = 0 : (m−2)/2 and a_m/2,m/2∈Rin equation (5.1) which becomes

(m−2)/2X

j=0

(M(λ^j) +M (λ^m−j)Σ)vec(ajj) +M(λ^m/2)vec(a_m/2,m/2) = vec(x^Hr).

The solution is then given byajj =e^T_jbr, j= 0 :m/2 where b

r= h

H0 H1 . . . H_(m−2)/2 H_m/2 i_†

vec(x^Hr) and

Hj = Ã

re(λ^j) +re(λ^m−j) −im(λ^j) +im(λ^m−j) im(λ^j) +im(λ^m−j) re(λ^j)−re(λ^m−j)

!

, j= 0 : (m−2)/2, and

Hm/2 =

à reλ^m/2 imλ^m/2

! .

Therefore we obtain,

4A_j = Q







e^T_jbr λ^m−j(Q_kΛ^H1r)H

mk²₂

λ^jQ^H₁r

kΛmk²₂ Xj





Q^H,

4A_m/2 = Q







e^T_m/2rb λ^{m/2 (}_kΛ^QH1r)H

mk²₂

λ^m/2Q^H₁r

kΛmk²₂ X_m/2





Q^H,

4A_m−j = 4A^H_j , j= 0 : (m−2)/2.

Setting Xj = 0 =Xm/2, j= 0 : (m−2)/2 we obtain

η_F^S(λ, x,P) = s

(2kbrk²₂− |e^T_m/2r|b²) + 2krk²₂− |x^Hr|² kΛmk²₂ . Simplifying the expressions for 4Ajs we obtain the desired result.

Next we consider spectral norm. Note that for |λ| = 1, we have x^Hr =λ^mx^Hr. Conse- quently, λ^j r^Hx =λ^m−j x^Hr. Consider µ4Aj = _kΛ^krk2

mk²₂. Applying DKW Theorem 1.2.5 to 4A_j,we have

Xj=−λ^m−j r^Hx Q^H₁ r(Q^H₁ r)^H

kΛmk²₂ (krk²₂− |r^Hx|²), Xm/2=−λ^m/2 r^Hx Q^H₁r(Q^H₁r)^H kΛmk²₂ (krk²₂− |r^Hx|²)

which givesη^S₂(λ, x,P) = ^√krk_m+1² .Simplifying the expressions for 4Ajs we obtain the desired result.

If|λ| 6= 1 consider

µ4Aj =





 q

|e^T_jr|b²+^|λm−j|2_kΛ^{(krk22−|xHr|2)}

mk⁴₂ , if|λ|>1 q

|e^T_jr|b²+^{|λi|2 (krk}_kΛ^22−|xHr|2)

mk⁴₂ , if|λ|<1 µ4Am/2 =

s

|e^T_m/2br|²+|λ^m|² (krk²₂− |x^Hr|²) kΛmk⁴₂

forj= 0 : (m−2)/2.Then again by DKW Theorem 1.2.5 we have

Xj =







−^eTj_|λ^{br λ}m−j^{j λ}|^2m−j(krk^Q2₂^H1−|x^r(QHH1r|^2r))^H, if|λ|>1

−^{eTjbr λ}_|λi^j|^2λ(krk^m−j2₂^Q−|x^H1r(QHr|^H12)^r)H, if|λ|<1, X_m/2 = −e^T_m/2br Q^H₁r(Q^H₁r)^H

(krk²₂− |x^Hr|²) forj= 0 : (m−2)/2.This gives

η^S₂(λ, x,P) =







 r

(2kbrk²₂− |e^T_m/2br|²) +^{(2kΠm/2RΛmk22+|λ}_kΛ^{m/2|2)(krk22−|xHr|2)}

mk⁴₂ , if|λ|>1 r

(2kbrk²₂− |e^T_m/2br|²) +^{(2kΠm/2Λmk22+|λ}_kΛ^{m/2|2)(krk22−|xHr|2)}

mk⁴₂ , if|λ|<1.

Simplifying the expressions of4Ajs we obtain the desired result.¥

It is evident from the proof of Theorem 5.3.5 that, if K is a H-palindromic polynomial such that P(λ)x+ K(λ)x = 0 then K(z) = 4P(z) + (I−xx^H)^TN(z)(I−xx^H) for some H-palindromic polynomial N,where4P is given in Theorem 5.3.5.

Remark 5.3.6. If |x^Hr|=krk2,thenkQ^H₁ rk2= 0. In such a case, consideringXj= 0, j = 0 :m, we obtain the desired results.

Theorem 5.3.7. Let S be the space of H-anti-palindromic polynomials. Assume that P∈S and(λ, x)∈C×Cⁿ. Settingr:=−P(λ)x, we have the following.

1. mbe odd:

η_F^S(λ, x,P) =





√1 m+1

p2krk²₂− |r^Hx|²≤√

2η(λ, x,P), if|λ|= 1

√2 q

kbrk²₂+^krk_kΛ^22−|xHr|2

mk²₂ , if|λ| 6= 1.

η₂^S(λ, x,P) =













η(λ, x,P), if|λ|= 1

√2 r

kbrk²₂+^{kΠ(m+1)/2RΛ}_kΛ^{mk2 (krk22−|xHr|2)}

mk⁴₂ , if|λ|>1

√2 r

kbrk²₂+^{kΠ(m+1)/2Λm}_kΛ^{k2 (krk22−|xHr|2)}

mk⁴₂ , if|λ|<1.

wherebr= h

H0 H1 . . . H_(m−1)/2 i_†

vec(x^Hr) and

Hj= Ã

re(λ^j)−re(λ^m−j) −im(λ^j)−im(λ^m−j) im(λ^j)−im(λ^m−j) re(λ^j) +re(λ^m−j)

! .

Let Ej:= _kΛ¹

mk²₂[λ^jPxrx^H−λ^m−jxr^HPx].Then for j= 0 : (m−1)/2 define

4Aj:=





 λ^j

kΛmk²₂xr^Hxx^H+Ej, if|λ|= 1 e^T_jbrxx^H+E_j, if|λ| 6= 1 and4Am−j =−4A^H_j . Then 4P(z) = P_m

j=0z^j4Aj is a unique polynomial such that 4P∈S,P(λ)x+4P(λ)x= 0 and|||4P|||F =η^S_F(λ, x,P).Next define

4Aj:=



















 λ^j

kΛmk²₂xr^Hxx^H+Ej+ x^Hr λ^m−jPxrr^HPx

kΛmk²₂ (krk²₂− |x^Hr|²), if |λ|= 1 e^T_jrxxb ^H+Ej+ e^T_jr λb ^j λ^m−j Pxrr^HPx

|λ^m−j|² (krk²₂− |x^Hr|²), if |λ|>1 e^T_jrxxb ^H+Ej+e^T_jbr λ^j λ^m−j Pxrr^HPx

|λⁱ|² (krk²₂− |x^Hr|²) , if |λ|<1 for j = 0 : (m−1)/2 with 4Am−j = −4A^H_j . Then 4P(z) = P_m

j=0z^j4Aj is a polynomial such that 4P∈S,P(λ)x+4P(λ)x= 0 and|||4P|||2=η₂^S(λ, x,P).

2. mis even:

η_F^S(λ, x,P) =





√1 m+1

p2krk²₂− |r^Hx|²≤√

2η(λ, x,P), if|λ|= 1 q

(2kbrk²₂− |e^T_m/2br|²) + 2^krk_kΛ^22−|xHr|2

mk²₂ , if|λ| 6= 1.

η₂^S(λ, x,P) =













η(λ, x,P) if|λ|= 1

r

(2kbrk²₂− |e^T_m/2br|²) +^{(2kΠm/2RΛmk22+|λ}_kΛ^{m/2|2)(krk22−|xHr|2)}

mk⁴₂ , if|λ|>1 r

(2kbrk²₂− |e^T_m/2br|²) +^{(2kΠm/2Λmk22+|λ}_kΛ^{m/2|2)(krk22−|xHr|2)}

mk⁴₂ , if|λ|<1.

wherebr=h

H0 H1 . . . H(m−2)/2 Hm/2

i†

vec(x^Hr),

H_j =

à re(λ^j) +re(λ^m−j) −im(λ^j) +im(λ^m−j) im(λ^j) +im(λ^m−j) re(λ^j)−re(λ^m−j)

!

, j= 0 : (m−2)/2, and

H_m/2 =

à −im(λ^m/2) re(λ^m/2)

! .

LetFj:= _kΛ¹

mk²₂[λ^jPxrx^H−λ^m−jxr^HPx]andGm/2:=_kΛ¹

mk²₂[λ^m/2Pxrx^H−λ^m/2xr^HPx].

Then for j= 0 : (m−2)/2,define

4Aj :=





 λ^j

kΛmk²₂xr^Hxx^H+Fj, if |λ|= 1 e^T_jbrxx^H+Fj, if |λ| 6= 1

,4A_m/2:=













 λ^m/2

kΛ_mk²₂xr^Hxx^H+G_m/2, if|λ|= 1 i e^T_m/2brxx^H+G_m/2,

if|λ| 6= 1 and4Am−j =−4A^H_j . Then 4P(z) = P_m

j=0z^j4Aj is a unique polynomial such that 4P∈S,P(λ)x+4P(λ)x= 0 and|||4P|||F =η^S_F(λ, x,P).Next define

4Aj :=



















 λ^j

kΛmk²₂xx^Hrx^H+Fj+ λ^m−j r^Hx Pxrr^HPx

kΛmk²₂ (krk²₂− |r^Hx|²), if |λ|= 1 e^T_jbrxx^H+Fj+ e^T_jbr λ^j λ^m−j Pxrr^HPx

|λ^m−j|² (krk²₂− |x^Hr|²), if |λ|>1 e^T_jbrxx^H+Fj+e^T_jbr λ^j λ^m−j Pxrr^HPx

|λⁱ|² (krk²₂− |x^Hr|²) , if |λ|<1

4A_m/2 :=









 λ^m/2

kΛmk²₂xx^Hrx^H+G_m/2+ λ^m/2 r^Hx Pxrr^HPx

kΛmk²₂ (krk²₂− |r^Hx|²), if|λ|= 1 i e^T_m/2rxxb ^H+Gm/2+ie^T_m/2br Pxrr^HPx

(krk²₂− |x^Hr|²), if|λ| 6= 1.

and4Am−j =−4A^H_j forj= 0 : (m−2)/2.Then4P(z) =P_m

j=0z^j4Aj is a polyno- mial such that4P∈S,P(λ)x+4P(λ)x= 0 and|||4P|||2=η₂^S(λ, x,P).

Proof: First assume thatm is odd. By Theorem 5.2.2 we know that there exists4P ∈S, such that4P(z)x+ P(z)x= 0.Define

4Aj:=Q

à ajj a^H_j bj Xj

!

Q^H, Am−j =−A^H_j , j= 0 : (m−1)/2.

Since4P(λ)x+ P(λ)x= 0, we have

à P_m

j=0λ^jajj

P_(m−1)/2

j=0 λ^jbj−P_(m−1)/2

j=0 λ^jaj

!

=

à x^Hr Q^H₁ r

! .

The minimum norm solution of P_(m−1)/2

j=0 λ^jbj−P_(m−1)/2

j=0 λ^m−jaj =Q^H₁ ris given by bj =

λ^j

kΛ_mk²₂ andaj =−_kΛ^λm−j

mk²₂.

Note that for |λ| = 1, we havex^Hr =−λ^mx^Hr. Hence the minimum norm solution of P_m

j=0λ^jajj =x^Hrisajj =_kΛ^λj

mk²₂x^Hr, j= 0 : (m−1)/2.Thus we have

4Aj = Q







λ^j

kΛmk²₂x^Hr −_kΛ^λm−j

mk²₂(Q^H₁r)^H

λ^j

kΛmk²₂Q^H₁r Xj





Q^H, 4Am−j = −4A^H_j j= 0 : (m−1)/2.

SettingXj= 0 we haveη^S(λ, x,P) =^√_m+1¹ p

2krk²₂− |r^Hx|².Simplifying the expressions for 4Aj forj= 0 : (m−1)/2 we obtain the desired result.

Now suppose that|λ| 6= 1.The the value ofajj fromP_m

j=0λ^jajj =x^Hr, is achieved after employing the conditionajj =−am−j,m−j j= 0 : (m−1)/2.applying vec both sides by (5.1) we have

(m−1)/2X

j=0

(M(λ^j)−M(λ^m−j)Σ)vec(ajj) = vec(x^Hr)

which givesajj =e^T_jr, jb = 0 : (m−1)/2 wherebr= h

H0 H1 . . . H_(m−1)/2 i_†

vec(x^Hr) and

Hj=

à re(λ^j)−re(λ^m−j) −im(λ^j)−im(λ^m−j) im(λ^j)−im(λ^m−j) re(λ^j) +re(λ^m−j)

! .

Consequently we have

4Aj=Q







e^T_jbr −_kΛ^λm−j

mk²₂(Q^H₁r)^H

λ^j

kΛmk²₂Q^H₁r X_j





Q^H,4Am−j =−4A^H_j , j= 0 : (m−1)/2.

Setting X_j = 0 we obtain the minimum Frobenius norm of4A_j and hence we have

η^S_F(λ, x,P) =√ 2

s

kbrk²₂+krk²₂− |x^Hr|² kΛmk²₂ . Simplifying the expression of4Aj we obtain the desired result.

Next consider spectral norm. From the construction of 4Aj it is easily seen that the η^S₂(λ, x,P) will be same as that given in Theorem 5.3.5 and hence desired results follow.

Now suppose that mis even. By Theorem 5.2.2 we know that there exists 4P∈Ssuch that 4P(λ)x+ P(λ)x= 0.Forj= 0 : (m−2)/2 define

4Aj :=Q Ã

ajj a^H_j bj Xj

!

Q^H,4Am/2=Q

à a_m/2,m/2 −a^H_m/2 am/2 Xm/2

! Q^H.

Since4P(λ)x+ P(λ)x= 0, we have

à P_m

j=0λ^jajj

P_(m−2)/2

j=0 λ^jbj−P_(m−2)/2

j=0 λ^m−jaj+λ^m/2am/2

!

=

à x^Hr Q^H₁ r

! .

The minimum norm solution of P_(m−2)/2

j=0 λ^jbj−P_(m−2)/2

j=0 λ^m−jaj+λ^m/2am/2=Q^H₁ris bj= λ^j

kΛmk²₂Q^H₁r, aj =− λ^m−j

kΛmk²₂Q^H₁r, a_m/2= λ^m/2 kΛmk²₂Q^H₁r.

Note that for |λ| = 1, we havex^Hr =−λ^mx^Hr. Hence the minimum norm solution of P_m

j=0λ^jajj =x^Hrisajj = _kΛ^λj

mk²₂x^Hr, j= 0 : (m−2)/2.Note thatam/2,m/2∈iR.Thus we

have

4Aj = Q







λ^j

kΛmk²₂x^Hr −_kΛ^λm−j

mk²₂(Q^H₁r)^H

λ^j

kΛmk²₂Q^H₁ r Xj





Q^H,

4A_m/2 = Q







λ^m/2

kΛmk²₂x^Hr −_kΛ^λm/2

mk²₂(Q^H₁r)^H

λ^m/2

kΛmk²₂Q^H₁ r Xm/2





Q^H,

4Aj = −4A^H_m−j, j= 0 : (m−2)/2.

Setting Xj = 0 = X_m/2 givesη^S_F(λ, x,P) = ^√_m+1¹ p

2krk²₂− |r^Hx|² if|λ|= 1.Simplifying the expressions for4Aj we obtain the desired result.

Now suppose that|λ| 6= 1.The value of ajj from the equation P_m

j=0λ^jajj =x^Hr, j= 0 : (m−2)/2 is obtained after employing the conditionajj =−am−j,m−j j= 0 : (m−2)/2 and a_m/2,m/2∈iR.Then applying vec operator both sides, by (5.1) we haveajj =e^T_jvec(br), j = 0 : (m−2)/2 andam/2,m/2=i e^T_m/2vec(br) where

b r = h

H0 H1 . . . H(m−2)/2 Hm/2

i_†

vec(x^Hr), Hj =

à re(λ^j) +re(λ^m−j) −im(λ^j) +im(λ^m−j) im(λ^j) +im(λ^m−j) re(λ^j)−re(λ^m−j)

!

, j= 0 : (m−2)/2, and

Hm/2 = Ã

−im(λ^m/2) re(λ^m/2)

! .

Consequently we have

4Aj = Q







e^T_jrb −_kΛ^λm−j

mk²₂(Q^H₁r)^H

λ^j

kΛmk²₂Q^H₁ r Xj





Q^H,

4A_m/2 = Q







i e^T_m/2br −_kΛ^λm/2

mk²₂(Q^H₁r)^H

λ^m/2

kΛmk²₂Q^H₁ r Xm/2





Q^H,

4Aj = −4A^H_m−j, j= 0 : (m−2)/2.

Setting Xj = 0 =X_m/2 we obtain

η_F^S(λ, x,P) = s

(2kbrk²₂− |e^T_m/2r|b²) + 2krk²₂− |x^Hr|² kΛmk²₂ . Simplifying the expressions for 4Aj’s we obtain the desired result.

Next consider spectral norm. From the construction of4Ajit is easily seen thatη^S₂(λ, x,P) will be same as that given in Theorem 5.3.5 and hence the proof is similar.¥

It follows from the proof of Theorem 5.3.5 that, if K is aH-anti-palindromic polynomial

such that P(λ)x+ K(λ)x = 0 then K(z) = 4P(z) + (I−xx^H)^TN(z)(I−xx^H) for some H-anti-palindromic polynomial N,where4P is given in Theorem 5.3.7.

Remark 5.3.8. If |x^Hr|=krk2,thenkQ^H₁ rk2= 0. In such a case, consideringXj= 0, j = 0 :m, we obtain the desired results.

5.4 Structured backward error and palindromic lineariza-