Palindromic linearizations

• zr(σ, s) := (bd,bd−1, . . . ,bh+1,ebh,ebh−1,bh−2, . . . ,b1) if s6= 0 and kh−1 = 1, where eb_h = (ah−1 +kh−1+ 1 :a_h), i.e., eb_h = (s+ 1 :a_h) and

ebh−1 = (ah−2 +kh−2 :ah−1, ah−1+kh−1), i.e., ebh−1 = (bh−1, s).

• z_r(σ, s) := (b_d,bd−1, . . . ,b_h+1,eb_h, s,bh−1, . . . ,b₁) if s6= 0 and kh−1 >1, where eb_h = (ah−1+kh−1+ 1 : a_h), i.e., eb_h = (s+ 1 :a_h).

• z_r(σ, s) := (b_d,bd−1, . . . ,b₂,eb₁,eb₀) if s= 0, where eb₁ = (1 :a₁) and eb₀ = (0).

An index tuple of type-1 is defined as follows.

Definition 4.3.3 (Right and left index tuple of type-1). Let σ be a simple index tuple containing indices from {0 :m}. Let α := (s₁, . . . , s_k) and β be index tuples containing indices from σ. Then:

• α is said to be a right index tuple of type-1 relative to σ if, for i = 1 : k, si is a right index of type-1 relative to z_r(σ,(s₁, . . . , s_i−1)), where z_r(σ,(s₁, . . . , s_i−1)) :=

z_r z_r(σ,(s₁, . . . , s_i−2)), s_i−1

for i >2.

• β is said to be a left index tuple of type-1 relative to σ, if rev(β) is a right index tuple of type-1 relative to rev(σ). Moreover, if β is a left index tuple of type-1 relative to σ,we define z_l(β, σ) :=rev z_r(rev(σ), rev(β))

.

Example 4.3.4. Let σ = (5 : 7,1 : 2,0) and α = (5,6,5). Then σ is a simple index tuple containing indices from {0 : 7}. The simple tuple associated with (σ,5)is given by z_r(σ,5) = (6 : 7,5,1 : 2,0). Further, z_r(σ,(5,6)) = (7,5 : 6,1 : 2,0) and z_r(σ,(5,6,5)) = (7,6,5,1 : 2,0).

We now define P-admissible tuple which will play a central role in the construction of T-palindromic linearizations of P(λ).

Definition 4.3.5 (P-admissible tuple). Let P(λ) be a matrix polynomial of odd degree m ≥ 3. A sub-permutation σ of {0 : m} is said to be a P-admissible tuple relative to {0 :m} if the following hold: (a) 0∈σ (b) cardinality (i.e., the number of indices) of σ is (m+ 1)/2, and (c) if i∈σ then m−i /∈σ for i= 0 :m.

Recall that for a tuple X = (X₁, X₂, . . . , X_r) of n×n arbitrary matrices, we have

−X = (−X1,−X2, . . . ,−Xr),X^T = (X₁^T, X₂^T, . . . , X_r^T) andrev(X) = (Xr, . . . , X2, X1).

Lemma 4.3.6. Let σ be a P-admissible tuple relative to {0 : m}. Let σr and σl, respectively, be a right and a left index tuple of type-1 relative to σ. Then (σl, σ, σr) and (−m+rev(σ_r),−m+rev(σ),−m+rev(σ_l)) satisfy the SIP.

Proof. As σ is a sub-permutation of {0 : m}, it is enough to show that (σ, σ_r) and (σ_l, σ) satisfy the SIP. By induction on the number k of indices of σ_r we prove that (σ, σ_r) satisfies the SIP. Let k = 1, i.e., σ_r = (s₁). Since s₁ is a right index of type-1 relative to σ, there is a string (s₁ :b) in csf(σ) such that s₁ < b. Hence s₁+ 1 appears afters₁ in the indices ofσ which implies that (σ, s₁) satisfies the SIP, i.e., (σ, σ_r) satisfies the SIP. Now suppose that k > 0 and σ_r = (s₁, s₂. . . , sk−1, s_k). Then by the definition of right index tuple of type-1 we have eσ_r := (s₁. . . , sk−1) is a right index tuple of type- 1 relative to σ and sk is a right index of type-1 relative to zr(σ,(s1, . . . , sk−1)), i.e., zr(σ,eσr). By the induction hypothesis, (σ,eσr) satisfies the SIP. Now sincesk is a right index of type-1 relative toz_r(σ,eσ_r) there is a string (s_k:b) inz_r(σ,eσ_r) such thats_k < b.

By Definition 4.3.2 and Definition 4.3.3 it follows that s_k+ 1 appears to the right of the last index s_k in (σ,eσ_r). This which implies that (σ,eσ_r, s_k) satisfies the SIP, i.e., (σ, σ_r) satisfies the SIP.

Next we show that (σ_l, σ) satisfies the SIP. Note that an index tupleα satisfies the SIP if and only if rev(α) satisfies the SIP. Since σ_l is a left index tuple of type-1 relative to σ, rev(σ_l) is a right index tuple of type-1 relative to rev(σ). So by the above case, (rev(σ), rev(σ_l)) satisfies the SIP. Then (σ_l, σ) =rev(rev(σ), rev(σ_l)) satisfies the SIP.

Sinceβ := −m+rev(σ_r),−m+rev(σ),−m+rev(σ_l)

=−m+rev σ_l, σ, σ_r and σl, σ, σr

satisfies the SIP, we have β satisfies the SIP. This completes the proof.

Let σ be a P-admissible tuple relative to {0 : m}. Let σr and σl, respectively, be right and left index tuples of type-1 relative to σ with the condition that if an index j ∈ σ_r∪σ_l then j −1 ∈ σ for j = 1 : m. Let X and Y be any arbitrary nonsingular matrix assignments for σ_r and σ_l, respectively. Define L(λ) :=

M−m+rev(σr)(−rev(X^T))Mσ_l(Y) λM_−m+rev(σ)^P −M_σ^P

Mσr(X)M−m+rev(σ_l)(−rev(Y^T)).

(4.9) Then by Lemma 4.3.6,L(λ) is an EGFPR ofP(λ). We use pencilsL(λ) of the form (4.9) to construct T-palindromic linearizations of P(λ). Since arbitrary matrix assignments are used to define L(λ), the family of T-palindromic linearizations of P(λ) that we construct in this chapter contains infinite number of pencils.

Definition 4.3.7. [19] A matrix S ∈ C^mn is said to be a quasi-identity matrix if S = ₁I_n⊕· · ·⊕_mI_n, where_i ∈ {1,−1},i= 1 :m.We refer to(₁, . . . , _m)as the parameters of S.For i= 1 :m, we denote thei-th diagonal block ofS byS(i, i). Also, we denote the

quasi-identity matrix whose only negative parameter is i by Si. By default we denote by S0 and Sm+1 the identity matrix Imn, i.e., S0 :=Imn =:Sm+1.

For rest of this section, we defineR :=









I_n

· ·· I_n









∈C^mn.Note thatRR=I_mn, and SS =I_mn for any quasi-identity matrixS.

The following result will play a crucial role in constructing T-palindromic lineariza- tions of P(λ). For simplicity of notation, for the rest of this section we set M_i^T(X) :=

(M_i(X))^T for i∈ {±0,±1, . . . ,±m}, whereX ∈C^n×n is any arbitrary matrix.

Proposition 4.3.8. Let X ∈ C^n×n be an arbitrary matrix. Then the elementary ma- trices M±i(X), i= 0 :m, satisfy the following.

RM−i(X)R =S_i+1M_m−i^T (−X^T)S_i and

RMm−i(X)R=Sm+1−iM_−i^T (−X^T)Sm−i for i= 1 :m.

If X is invertible then both equalities hold for i= 0.

Proof. Easy to prove.

We need the following results for constructing T-palindromic linearizations of P(λ).

Theorem 4.3.9 ([27], Theorem 4.8). Let σ be a P-admissible tuple relative to {0 : m}. Define T(λ) := λM_−m+rev(σ)^P − M_σ^P. Then SRT(λ) is a T-palindromic strong linearization of P(λ), where S is the quasi-identity matrix given by

S(i, i) :=



















−I if











σ has an inversion at i−1,or σ has a consecution at m−i,or i∈σ and i−1∈/ σ

I otherwise.

Lemma 4.3.10 ([19], Lemma 3.1). Let P(λ) be a T-palindromic matrix polynomial. If T(λ) is a T-palindromic strong linearization of P(λ) and L(λ) = Q₁T(λ)Q₂ for some constant nonsingular matrices Q₁ and Q₂, then Q^T₂Q⁻¹₁ L(λ) is a T-palindromic strong linearization of P(λ).

By utilizing the EGFPRs, we now constructT-palindromic strong linearizations ofP(λ).

Lemma 4.3.11. Let σ be a P-admissible tuple relative to {0 : m} and let σ_r be a right index tuple of type-1 relative to σ with the condition that if an index j ∈σ_r then

j −1 ∈ σ for j = 1 : m. Let X be a nonsingular matrix assignment for σr. Define L(λ) := M−m+rev(σr)(−rev(X^T)) λM_−m+rev(σ)^P −M_σ^P

Mσr(X). Then SRL(λ) is a T- palindromic strong linearization of P(λ), where S is the quasi-identity matrix given by

S(i, i) :=



















−I if











σ has an inversion at i−1,or

z_r(σ, σ_r) has a consecution atm−i,or i∈σ and i−1∈/ σ

I otherwise.

(4.10)

Proof. We prove the result by induction on the number q of indices of σr. If q = 0, i.e., σ_r = ∅ then z_r(σ, σ_r) =σ, and hence the result follows from Theorem 4.3.9. Now suppose that q > 0. Assume that the result is true when σ_r contain q −1 indices.

Suppose that σ_r = (s₁, . . . , sq−1, s_q), where s_i denotes the i-th index in σ_r and let σe_r = (s₁, . . . , sq−1). Let X = (X₁, X₂, . . . , Xq−1, X_q). Set Xe = (X₁, X₂, . . . , Xq−1).

DefineL(λ) :=e M−m+rev(eσr)(−rev(Xe^T))

λM_−m+rev(σ)^P −M_σ^P

Meσr(X).e By the induction hypothesis, SRee L(λ) is a T-palindromic strong linearization of P(λ), where Se is the quasi-identity matrix given by

S(i, i) :=e



















−I if











σ has an inversion at i−1,or

z_r(σ,eσ_r) has a consecution atm−i,or i∈σ and i−1∈/ σ

I otherwise.

(4.11)

Note that

L(λ) =M−m+sq(−X_q^T)eL(λ)M_sq(X_q) =M−m+sq(−X_q^T)RSe SRe L(λ)e

M_sq(X_q) since RR=I and SeSe=I. Now, by Lemma 4.3.10,

M_s^T_q(Xq)

M−m+sq(−X_q^T)RSe −1

L(λ) = M_s^T_q(Xq)SRe

M−m+sq(−X_q^T) −1

L(λ) is aT-palindromic strong linearization ofP(λ). HenceSRL(λ) is aT-palindromic strong linearization of P(λ), where

S =M_s^T_q(X_q)SRe

M−m+s_q(−X_q^T)−1

R. (4.12)

Thus we only need to prove that S is the quasi-identity matrix satisfying (4.10).

Case-I: Assume that s_q > 0. For j = 1 : m−1, we have (M_j(Y))⁻¹ = M−j(−Y) for any matrix assignment Y. Thus by (4.12), S =M_s^T_q(X_q)SRMe m−sq(X_q^T)R. Now, by Proposition 4.3.8, we have

S =M_s^T_q(X_q)SSe m+1−sqM_−s^T _q(−X_q)Sm−sq. (4.13)

Now, since sq is a right index of type-1 relative to zr(σ,eσr), there exist a string (ai +ki : ai+1) in zr(σ,eσr) such that sq = ai+ki < ai+1. Next we show that M_s^T_q(Xq) and SSe _m+1−sq commute. Since

M_s^T_q(X_q) =

















I_{(m−sq−1)n}

X_q^T I_n I_n 0

I_(sq−1)n















 ,

it is enough to show that both (m−sq)-th and (m−sq+ 1)-th parameters of SSe m−sq+1

have the same sign. Recall from the definition of quasi-identity matrix that (m−s_q+ 1)- th parameter is the only negative parameter of Sm−s_q+1. Consequently, it is enough to show that both (m−s_q)-th and (m−s_q+ 1)-th parameters ofSehave the opposite sign.

Since s_q ∈ σ_r, we have s_q, s_q+ 1∈ σ. Again, recall the assumption that if j ∈σ_r then j−1∈σ forj = 1 :m−1.Since s_q ∈σ_r, by the assumption we haves_q−1∈σ. Thus s_q−1, s_q, s_q+1∈σ. Asσis aP-admissible tuple we havem−s_q−1, m−s_q, m−s_q+1∈/ σ.

Hence by (4.11), the parameter ofSeat the position (m−s_q)-th (resp., (m−s_q+ 1)-th) is −1 if and only if z_r(σ,σe_r) has a consecution at s_q (resp., s_q−1). Note that s_q is a right index of type-1 relative to zr(σ,eσr). Hence zr(σ,eσr) must be of the form

z_r(σ,eσ_r) =

(a_d−1+k_d−1 :a_d), . . . ,(s_q :a_i+1),(a_i−1+k_i−1 :a_i), . . . ,(a₀ :a₁) withs_q < a_i+1.Thusz_r(σ,eσ_r) has a consecution ats_q which implies that the parameters of Se at the position (m−s_q)-th is−1. Since s_q−1∈σ we have a_i =s_q−1. Hence

z_r(σ,σe_r) =

(ad−1+kd−1 :a_d), . . . ,(s_q :a_i+1),(ai−1+ki−1 :s_q−1), . . . ,(a₀ :a₁) . Thus zr(σ,eσr) has an inversion at sq−1 which implies that (m−sq+ 1)-th parameter of Seis 1. Hence (m−s_q)-th and (m−s_q+ 1)-th parameters of SSe _m−sq+1 are the same.

Thus M_s^T_q(X_q) andSSe m+1−s_q commute. Hence by (4.13), we have

S=SSe m+1−sqM_s^T_q(X_q)M_−s^T _q(−X_q)Sm−sq =SSe m−sq+1Sm−sq. (4.14) as M_s^T_q(X_q)M_−s^T _q(−X_q) = I_mn. Now we prove that S is the quasi-identity matrix satis- fying (4.10). Observe that by (4.14), the quasi-identity matricesS andSehave the same parameters except at the positions (m−sq) and (m−sq+ 1). So we only need to prove thatS and Sehave opposite parameters at the positions (m−sq) and (m−sq+ 1). Now by Definition 4.3.2, we have

z_r(σ, σ_r) =

(ad−1+kd−1 :a_d), . . . ,(s_q+ 1 :a_i+1),(ai−1 +ki−1 :s_q), . . . ,(a₀ :a₁) .

Thus the inversions and consecutions that occur in zr(σ,σer) and zr(σ, σr) are the same except

• at s_q wherez_r(σ,eσ_r) has a consecution andz_r(σ, σ_r) has an inversion, and

• at s_q−1 wherez_r(σ,σe_r) has an inversion and z_r(σ, σ_r) has a consecution.

Hence we have

• (m−s_q)-th parameters of Seand S are −1 and 1, respectively, and

• (m−s_q+ 1)-th parameters of Seand S are 1 and −1, respectively.

Thus the parameter ofS andSeat the positions (m−sq) and (m−sq+ 1) have opposite sign.

Case-II: Assume thats_q = 0. SinceX is a nonsingular matrix assignment forσ_r,X_q is a nonsingular matrix. This implies M₀(X_q) is invertible. By (4.12), we have

S =M₀^T(X_q)SRe

M_−m(−X_q^T)−1

R=M₀^T(X_q)SRMe _m(−X_q^T)R

asM_m(Y) = (M−m(Y))⁻¹ =M−m(Y⁻¹) for any nonsingularY ∈C^n×n.Now, by Propo- sition 4.3.8, S = M₀^T(X_q)SSe _m+1M₋₀^T (X_q)S_m = M₀^T(X_q)SMe ₋₀^T (X_q)S_m as S_m+1 = I_mn. Since M₀^T(X_q) =





I(m−1)n

X_q^T



, it trivially follows that M₀^T(X_q) and Se commute, which implies thatS =SMe ₀^T(X_q)M₋₀^T (X_q)S_m =SSe _m asM₀^T(X_q)M₋₀^T (X_q) =I_mn.Since s_q= 0 is a right index of type-1 relative to z_r(σ,eσ_r), we must have

z_r(σ,eσ_r) =

(ad−1+kd−1 :a_d), . . . ,(a₁+k₁ :a₂),(0 :a₁)

, wherea₁ >0, and

z_r(σ, σ_r) =

(ad−1+kd−1 :a_d), . . . ,(a₁+k₁ :a₂),(1 :a₁),0 .

By similar arguments as in Case-I, it follows thatSis the quasi-identity matrix satisfying (4.10). This completes the proof.

Theorem 4.3.12. Let σ be a P-admissible tuple relative to {0 : m}. Let σ_r and σ_l, respectively, be a right and a left index tuple of type-1 relative to σ with the condition that if an index j ∈σ_r∪σ_l then j −1∈σ for j = 1 :m. Let X and Y be any arbitrary nonsingular matrix assignments for σ_r and σ_l, respectively. Define L(λ) :=

M−m+rev(σr)(−rev(X^T))Mσ_l(Y) λM_−m+rev(σ)^P −M_σ^P

Mσr(X)M−m+rev(σ_l)(−rev(Y^T)).

Then SRL(λ) is a T-palindromic strong linearization of P(λ), where S is the quasi- identity matrix given by

S(i, i) :=



















−I if











z_l(σ_l, σ) has an inversion at i−1,or z_r(σ, σ_r) has a consecution at m−i,or i∈σ and i−1∈/σ

I otherwise.

(4.15)

Proof. We prove the result by induction on the number q of indices of σ_l.If q= 0 then the result follows from Lemma 4.3.11. Now suppose thatq >0. Assume that the result is true when σ_l contain q −1 indices. Suppose that σ_l = (s_q, sq−1, . . . , s₁), where s_i denotes an index in σ_l and let σe_l = (sq−1, . . . , s₁). Let Y = (Y_q, Yq−1, . . . , Y₂, Y₁). Set Ye = (Yq−1, . . . , Y₂, Y₁). Define L(λ) :=e

M−m+rev(σr)(−rev(X^T))M_σel(Ye)

λM_−m+rev(σ)^P −M_σ^P

M_σr(X)M−m+rev(eσl)(−rev(Ye^T)).

By the induction hypothesis, SRe L(λ) is ae T-palindromic strong linearization of P(λ), where Seis the quasi-identity matrix given by

S(i, i) =e



















−I if











z_l(σe_l, σ) has an inversion ati−1,or zr(σ, σr) has a consecution at m−i,or i∈σ and i−1∈/ σ

I otherwise.

(4.16)

Recall the assumption that if an index j ∈ σ_l∪σ_r then j −1 ∈ σ for j = 1 : m.

Let i ∈ σ_l ∪ σ_r. Then i − 1, i, i+ 1 ∈ σ. As σ is a P-admissible tuple we have m −i−1, m−i, m−i + 1 ∈/ σ. Hence m −i−1, m−i, m−i + 1 ∈/ σ_l ∪σ_r, i.e.,

−i−1,−i,−i+ 1 ∈ −m/ +rev(σ_l)∪ −m +rev(σ_r). So σ_l and −m +rev(σ_r) com- mute, and σ_r and −m+rev(σ_l) commute. Thus L(λ) =M_sq(Y_q)eL(λ)M−m+sq(−Y_q^T) = M_sq(Y_q)RSe SRe L(λ)e

M−m+sq(−Y_q^T) as RR=I and SeSe=I. Now, by Lemma 4.3.10, M_−m+s^T _q(−Y_q^T)

Msq(Yq)RSe −1

L(λ) = M_−m+s^T _q(−Y_q^T)SRe

Msq(Yq) −1

L(λ) is aT-palindromic strong linearization ofP(λ). HenceSRL(λ) is aT-palindromic strong linearization of P(λ), where

S=M_−m+s^T _q(−Y_q^T)SRe

M_sq(Y_q)−1

R. (4.17)

Thus we only need to prove that S is the quasi-identity matrix satisfying (4.15).

Case-I: Assume that s_q > 0. So

M_sq(Y_q)−1

= M−s_q(−Y_q) and hence by (4.17), S =M_−m+s^T _q(−Y_q^T)SRMe −sq(−Y_q)R. Now by Proposition 4.3.8, we have

S =M_−m+s^T _q(−Y_q^T)SSe _sq+1M_m−s^T _q(Y_q^T)S_sq. (4.18) Sinces_q is a left index of type-1 relative to z_l(eσ_l, σ), i.e., s_q is a right index of type-1 relative to rev(z_l(eσ_l, σ)), there exist a string (a_i+k_i : a_i+1) in rev(z_l(eσ_l, σ)) such that s_q=a_i+k_i < a_i+1. Next we show that M_m−s^T _q(Y_q^T) and SSe _sq+1 commute. Since

M_m−s^T _q(Y_q^T) =

















I_(sq−1)n

Y_q I_n I_n 0

I_{(m−sq−1)n}















 ,

it is enough to show that both (s_q)-th and (s_q+1)-th parameters ofSSe _sq+1have the same sign. Since (s_q+ 1)-th parameter is the only negative parameter of S_sq+1, it is enough to show that (s_q)-th and (s_q+ 1)-th parameters of Sehave the opposite sign. Note that for any index j ∈σ_l, we have j, j + 1∈ σ. Thus, since s_q ∈ σ_l, we have s_q, s_q+ 1 ∈σ.

Again, recall the assumption that if j ∈ σ_l then j −1 ∈ σ for j = 1 : m−1. Since s_q ∈ σ_l, we have s_q−1 ∈σ. Thus s_q−1, s_q, s_q + 1∈ σ. As σ is a P-admissible tuple we have m−(s_q−1), m−s_q, m−(s_q+ 1)∈/ σ. Hence by (4.16), the parameter of Seat the position (sq)-th (resp., (sq+ 1)-th) is −1 if and only ifzl(eσl, σ) has an inversion at sq−1 (resp., sq).

Note that s_q is a left index of type-1 relative to z_l(σe_l, σ), i.e., s_q is a right index of type-1 relative to rev(z_l(σe_l, σ)). Hencerev(z_l(σe_l, σ)) must be of the form

rev(z_l(σe_l, σ)) =

(a_d−1+k_d−1 :a_d), . . . ,(s_q :a_i+1),(a_i−1+k_i−1 :a_i), . . . ,(a₀ :a₁) , with s_q< a_i+1. Ass_q−1∈σ, we have a_i =s_q−1. Hencerev(z_l(σe_l, σ)) must be of the form

rev(z_l(σe_l, σ)) =

(a_d−1+k_d−1 :a_d), . . . ,(s_q :a_i+1),(a_i−1+k_i−1 :s_q−1), . . . ,(a₀ :a₁) . (4.19) Thusrev(z_l(eσ_l, σ)) has an inversion ats_q−1 and a consecution ats_qwhich imply that z_l(eσ_l, σ) has a consecution ats_q−1 and an inversion ats_q. Hence the parameters ofSeat the position (s_q)-th is 1 and at the position (s_q+ 1)-th is−1, i.e., (s_q)-th and (s_q+ 1)-th parameters of Se have the opposite sign. Thus M_m−s^T _q(Y_q^T) and SSe _sq+1 commute. By (4.18), we have

S =M_−m+s^T _q(−Y_q^T)M_m−s^T _q(Y_q^T)SSe _sq+1S_sq =SSe _sq+1S_sq (4.20)

as M_−m+s^T _q(−Y_q^T)M_m−s^T _q(Y_q^T) = Imn. Now we prove that S is the quasi-identity matrix satisfying (4.15). Note that by (4.20) the quasi-identity matricesS andSehave the same parameters except at the positions s_q and (s_q+ 1). So we only need to prove that the parameters of S and Seat the positions s_q and (s_q+ 1) have opposite sign.

By Definition 4.3.3, we have

rev(z_l(σ_l, σ)) = z_r(rev(σ), rev(σ_l)) =z_r(z_r(rev(σ), rev(eσ_l)), s_q)

= z_r(rev(z_l(eσ_l, σ)), s_q) (4.21) Hence by (4.19) and (4.21), we have

rev(z_l(σ_l, σ)) =

(ad−1+kd−1 :a_d), . . . ,(s_q+ 1 :a_i+1),(ai−1+ki−1 :s_q), . . . ,(a₀ :a₁) . Note that the cosecutions and inversions that occur inrev(zl(eσl, σ)) are the same as those that occur in rev(z_l(σ_l, σ)), except

• ats_q−1 whererev(z_l(eσ_l, σ)) has an inversion andrev(z_l(σ_l, σ)) has a consecution, i.e., z_l(eσ_l, σ) has a consecution at s_q−1 and z_l(σ_l, σ) has an inversion at s_q−1.

• at s_q where rev(z_l(eσ_l, σ)) has a consecution and rev(z_l(σ_l, σ)) has an inversion, i.e., z_l(eσ_l, σ) has an inversion at s_q and z_l(σ_l, σ) has a consecution at s_q.

Hence

• (s_q)-th parameters of Se and S are 1 and −1, respectively, and

• (s_q+ 1)-th parameters ofSeand S are −1 and 1, respectively.

Hence the parameters of S and Seat the positionss_q and s_q+ 1 have opposite sign.

Case II: Assume thats_q= 0.SinceY is a nonsingular matrix assignment forσ_l,Y_qis a nonsingular matrix. SoM₀(Y_q) is invertible. By (4.17),S =M_−m^T (−Y_q^T)SRe

M₀(Y_q)−1

R

=M_−m^T (−Y_q^T)SRMe −0(Yq)R.Now, by Proposition 4.3.8,S =M_−m^T (−Y_q^T)SSe 1M_m^T(−Y_q^T)S0

=M_−m^T (−Y_q^T)SSe ₁M_m^T(−Y_q^T) asS₀ =I_mn. SinceM_−m^T (−Y_q^T) =





−Yq

I(m−1)n



, it is easy to see that M_−m^T (−Y_q^T) and SSe ₁ commute, which implies that

S =SSe ₁M_−m^T (−Y_q^T)M_m^T(−Y_q^T) = SSe ₁ as M_−m^T (−Y_q^T)M_m^T(−Y_q^T) =I_mn.

Sinces_q = 0 is a left index of type-1 relative toz_l(σe_l, σ), i.e., 0 is a right index of type-1 relative to rev(z_l(σe_l, σ)), rev(z_l(σe_l, σ)) must be of the form

rev(z_l(σe_l, σ)) =

(ad−1+kd−1 :a_d), . . . ,(a₁+k₁ :a₂),(0 :a₁)

with a₁ >0.

Then we have

rev(z_l(σ_l, σ)) =

(a_d−1+k_d−1 :a_d), . . . ,(a₁+k₁ :a₂),(1 :a₁),0 .

By similar arguments as in Case-I, it follows thatSis the quasi-identity matrix satisfying (4.15). This completes the proof.

The following example construct aT-palindromic pencil of P(λ).

Example 4.3.13. Let P(λ) = P7

i=0λⁱA_i be T-palindromic. Consider σ = (4,0,1,2), σ_r = (0,1,0) and σ₂ = ∅. Then σ is a P-admissible tuple relative to {0 : 7} and σ_r is a right index tuple of type-1 relative to σ. Let (X, Y, Z) be a nonsingular matrix assignment for σ_r. Define

L(λ) :=M(−7,−6,−7)(−Z^T,−Y^T,−X^T) λM(−5,−6,−7,−3)^P −M_(4,0,1,2)^P

M_(0,1,0)(X, Y, Z).

Let S be the quasi-identity matrix given byS =₁I_n⊕ · · · ⊕₇I_n,where_i =−1fori= 4 andi = 1if i6= 4, i.e., parameters ofS are(1,1,1,−1,1,1,1). Then by Lemma 4.3.11, SRL(λ) is a T-palindromic strong linearization of P(λ) and is given by

λ



































0 0 0 0 0 X 0

0 0 0 0 0 Y Z

0 0 0 I_n A₃ 0 0

0 0 0 0 −I_n 0 0

A₇ A₆ A₅ 0 0 0 0

0 −X^T −Y^T 0 0 0 0

0 0 −Z^T 0 0 0 0

































 +



































0 0 0 0 A₀ 0 0

0 0 0 0 A₁ −X 0

0 0 0 0 A₂ −Y −Z

0 0 I_n 0 0 0 0

0 0 A₄ −I_n 0 0 0

X^T Y^T 0 0 0 0 0

0 Z^T 0 0 0 0 0

































 .

Observe that SRL(λ) is an anti-block penta-diagonal pencil.

Next, we consider an example where the matrix assignments X for σ_r and Y for σ_l in Theorem 4.3.12 are the trivial matrix assignment.

Example 4.3.14. Let P(λ) = P7

i=0λⁱA_i be T-palindromic. Consider σ = (5 : 6,4,0) andσ_r = 5. LetL(λ) :=M₋₂^P λM(−7,−3,−1,−2)^P −M_(5:6,4,0)^P

M₅^P. Then by Theorem 4.3.12,

SRL(λ) is a T-palindromic strong linearization of P(λ) and is given by

λ



































0 0 0 0 I_n A₂ A₁

0 0 0 In 0 A3 A2

0 0 0 0 0 0 −I_n

0 0 0 0 0 −In 0

0 −I_n 0 0 0 0 0

0 −A₅ I_n 0 0 0 0

A₇ 0 0 0 0 0 0

































 +



































0 0 0 0 0 0 A₀

0 0 0 0 −In −A2 0

0 0 0 0 0 I_n 0

0 I_n 0 0 0 0 0

I_n 0 0 0 0 0 0

A₅ A₄ 0 −I_n 0 0 0

A₆ A₅ −I_n 0 0 0 0

































 ,

where S = ₁I_n⊕ · · · ⊕₇I_n, with _i = 1 for i = 1,2,6,7, and _i = −1 for i = 3,4,5.

Observe that SRL(λ) is an anti block penta-diagonal pencil.

The following result is a corollary to Theorem 4.3.12.

Corollary 4.3.15. LetP(λ)be aT-palindromic matrix polynomial of odd degreem≥3.

Let σ be a P-admissible tuple relative to {0 : m}. Let σl be a left index tuple of type-1 relative to σ with the condition that if an index j ∈σ_l then j−1∈σ for j = 1 :m. Let Y be any arbitrary nonsingular matrix assignment for σ_l. Define

L(λ) := M_σl(Y) λM_−m+rev(σ)^P −M_σ^P

M−m+rev(σ_l)(−rev(Y^T)).

Then SRL(λ) is a T-palindromic strong linearization of P(λ), where S is the quasi- identity matrix given by

S(i, i) :=



















−I if











z_l(σ_l, σ) has an inversion at i−1,or σ has a consecution at m−i,or i∈σ and i−1∈/σ

I otherwise.

More generally, we have the following result.

Corollary 4.3.16. LetP(λ)be aT-palindromic matrix polynomial of odd degreem≥3.

Let σ be a P-admissible tuple relative to {0 :m}. Let σr and σl, respectively, be a right and a left index tuple of type-1 relative toσ with the condition that if an indexj ∈σ_r∪σ_l then j −1∈σ for j = 1 : m. Let X and Y be any arbitrary matrix assignments for σ_r and σ_l, respectively. Define L(λ) :=

M−m+rev(σr)(−rev(X^T))M_σl(Y) λM_−m+rev(σ)^P −M_σ^P

M_σr(X)M−m+rev(σ_l)(−rev(Y^T)).

Then L(λ)RS is a T-palindromic strong linearization of P(λ) where S is the quasi- identity matrix given by

S(i, i) :=



















−I if











z_l(σ_l, σ) has an inversion at m−i,or z_r(σ, σ_r) has a consecution at i−1,or i∈σ and i−1∈/ σ

I otherwise.

(4.22)

Proof. Note that P(λ)^T is aT-palindromic matrix polynomial. Also, σr (resp., σl) is a right (resp., left) index tuple of type-1 relative toσ if and only ifrev(σ_r) (resp.,rev(σ_l)) is a left (resp., right) index tuple of type-1 relative to rev(σ). This implies thatL(λ)^T is a strong linearization of P(λ)^T. Therefore, by Theorem 4.3.12,SRL(λ)^T = (L(λ)RS)^T is a T-palindromic strong linearization of P(λ)^T, where S is the quasi-identity matrix given by

S(i, i) :=



















−I if











z_l(rev(σ_r), rev(σ)) has an inversion ati−1,or z_r(rev(σ), rev(σ_l)) has a consecution atm−i,or i∈σ and i−1∈/ σ

I otherwise.

(4.23)

Since (L(λ)RS)^T is a T-palindromic strong linearization of P(λ)^T, L(λ)RS is a T- palindromic strong linearization of P(λ). Hence the desired result follows as S given in (4.23) and (4.22) are the same.

Next, we construct low bandwidth banded (anti-block tridiagonal and anti-block penta-diagonal)T-palindromic linearizations ofP(λ). The following result characterizes all anti-block penta-diagonalT-palindromic linearizations that can be obtained from the construction given in Theorem 4.3.12.

Theorem 4.3.17. Let P(λ), L(λ), S and R be as given in Theorem 4.3.12. Then SRL(λ) is an anti-block penta-diagonal T-palindromic strong linearization of P(λ) if and only if ct(σl, σ, σr)≤1 and it(σl, σ, σr)≤1 for any index 1≤t≤m−1.

Proof. Note that, for any index tuple α containing indices from {−m : m}, M_α(∗) is a block penta-diagonal matrix if and only if RM_α(∗) is an anti-block penta-diagonal matrix, where (∗) denotes any arbitrary matrix assignment for α. Since S is a quasi- identity matrix, it follows that SRL(λ) is anti-block penta-diagonal if and only if L(λ) is block penta-diagonal.

Recall the definition of an end index of a sub-permutation of {0 : m −1} (resp., {−m : −1}) from Definition 3.5.5. It follows from the condition j ∈ σr ∪σl =⇒ j −1 ∈ σ, j = 1 : m, that σ_r and σ_l do not contain the end points of σ. This implies that −m+rev(σ_r) and −m+rev(σ_l) do not contain the end points of −m+rev(σ).

Hence the desired result follows from Theorem 3.5.6 by considering σ₁ :=σ_l, σ₂ :=σ_r, τ :=−m+rev(σ), τ₁ :=−m+rev(σ_r) and τ₂ :=−m+rev(σ_l).

Note that, for any index tupleαcontaining indices from{−m:m},Mα(∗) is a block tridiagonal matrix if and only if RM_α(∗) is an anti-block tridiagonal matrix, where (∗) denotes any arbitrary matrix assignment for α. The following result characterizes all anti-block tridiagonal T-palindromic linearizations that can be obtained from the construction given in Theorem 4.3.12.

Theorem 4.3.18. Let P(λ), L(λ), S and R be as given in Theorem 4.3.12. Then SRL(λ) is an anti-block tridiagonal T-palindromic strong linearization of P(λ) if and only ifc_t(σ_l, σ, σ_r) = 0 and i_t(σ_l, σ, σ_r) = 0 for any index 1≤t≤m−1. Moreover, any anti-block tridiagonal SRL(λ) in Theorem 4.3.12 must be one of the following pencils









































A₀ λX

λI_n λA₂+A₁ −X

I_n 0 −λI_n

λI_n λA₄+A₃ −I_n I_n 0 −λI_n

· ·· · ·· · ·· λA_m λAm−1+Am−2 −I_n

X^T −λX^T









































(4.24)

for some nonsingular matrix X,









































−Y λY λI_n λA₂+A₁ A₀

In 0 −λIn

λI_n λA₄+A₃ −I_n In 0 −λIn

· ·· · ·· · ··

−λY^T λAm−1+Am−2 −In

Y^T λA_m









































(4.25)

for some nonsingular matrix Y and









































λI_n λA₁+A₀

I_n 0 −λI_n

λI_n λA₃+A₂ −I_n I_n 0 −λI_n

· ·· · ··

· ·· λI_n λAm−2+Am−3 −I_n

I_n 0 −λI_n

λA_m+Am−1 −I_n







































 .

(4.26) Proof. By Theorem 3.5.8 and similar arguments as in the proof of Theorem 4.3.17, it follows that SRL(λ) is an anti-block tridiagonal T-palindromic strong linearization of P(λ) if and only if c_t(σ_l, σ, σ_r) = 0 and i_t(σ_l, σ, σ_r) = 0 for any index 1 ≤t≤m−1.

We now show that any anti-block tridiagonal SRL(λ) must be one of the pencils given in (4.24), (4.25) and (4.26). Since c_t(σ_l, σ, σ_r) = 0 and i_t(σ_l, σ, σ_r) = 0 for all 1≤t≤m−1,σ must be a permutation of either {0 :₂ m−1}or{0,1 :₂ m−2}, where {a :₂ b} :={a+ 2j : j = 0 : `} are all equispaced points in [a, b] such that a+ 2` ≤ b.

Then we have the following choices for σl and σr.

Case-I: Suppose that σ is a permutation of {0 :2 m−1}. Then there are no choices for σ_l and σ_r (i.e., σ_l = ∅ = σ_r) and SRL(λ) is of the form (4.26), where the quasi identity matrix S is given by S(i, i) = −1 for i ∈ {2 :₂ m− 1} and S(i, i) = 1 for i /∈ {2 :₂ m−1}. In this case L(λ) is a GF pencil.

Case-II: Suppose thatσ is a permutation of {0,1 :₂ m−2}.

(a) Suppose that σ has a consecution at 0. Then σ_l = ∅, and σ_r = (0) or ∅. If σ_r = (0) then SRL(λ) is of the form (4.24) for some nonsingular matrix assignment X of σ_r, where S is given by S(i, i) = −1 for i ∈ {3 :₂ m −2} and S(i, i) = 1 for i /∈ {3 :₂ m−2}. In this case L(λ) is an EGFPR.

On the other hand, ifσ_r =∅, thenSRL(λ) is of the form (4.24) withX =I_n, where S is given by S(i, i) =−1 for i∈ {3 :2 m} and S(i, i) = 1 for i /∈ {3 :2 m}. In this case L(λ) is a GF pencil.

(b) Suppose thatσhas an inversion at 0. Then σ_r=∅, andσ_l = (0) or∅. Ifσ_l = (0) then SRL(λ) is of the form (4.25) for some nonsingular matrix assignment Y of σ_l, whereS is given byS(i, i) =−1 fori∈ {3 :₂ m−2}andS(i, i) = 1 fori /∈ {3 :₂ m−2}.

In this case L(λ) is an EGFPR.

On the other hand, ifσl =∅thenSRL(λ) is of the form (4.25) withY =−In, where S is given by S(i, i) =−1 for i ∈ {1 :2 m−2} and S(i, i) = 1 for i /∈ {1 :2 m−2}. In this caseL(λ) is a GF pencil.

Example 4.3.19. LetP(λ) =P7

i=0λⁱA_i beT-palindromic. Consider the pencilL(λ) :=

M−7(−X^T) λM(−6,−7,−4,−2)^P −M_(5,3,0:1)^P

M₀(X). Then by Lemma 4.3.11, SRL(λ) is a T-palindromic strong linearization of P(λ) and is given by



































0 0 0 0 0 A₀ λX

0 0 0 0 λI_n λA₂+A₁ −X

0 0 0 I_n 0 −λI_n 0

0 0 λI_n λA₄+A₃ −I_n 0 0

0 I_n 0 −λI_n 0 0 0

λA₇ λA₆+A₅ −I_n 0 0 0 0

X^T −λX^T 0 0 0 0 0

































 ,

where S = ₁I_n⊕ · · · ⊕₇I_n with _i = 1 for i = 1,2,4,6,7, and _i = −1 for i = 3,5.

Observe that SRL(λ) is an anti-block tridiagonal pencil.

Remark 4.3.20. It follows from Theorem 4.3.18 that SRL(λ) in Theorem 4.3.12 is not an anti-block tridiagonal pencil if σ is a permutation of {0 : 2}. Hence the construction given in [19] does not produce an anti-block tridiagonal T-palindromic linearization of a T-palindromic P(λ) having odd degree m ≥5.

Remark 4.3.21 (T-anti-palindromic linearizations). Let P(λ) be T-anti-palindromic.

Recall from Chapter 2 that P(λ) is T-anti-palindromic if and only if P(−λ) is T- palindromic [27]. Suppose that P(λ) is of odd degree m ≥ 3. Define Q(λ) := P(−λ).

Consider the EGFPR L(λ) :=

M−m+rev(σ_r)(−rev(X^T))M_σl(Y) λM_−m+rev(σ)^Q −M_σ^Q

M_σr(X)M−m+rev(σ_l)(−rev(Y^T)) of Q(λ) satisfying all conditions of Theorem 4.3.12. Then SRL(−λ) is a T-anti- palindromic strong linearization of P(λ), where S is the quasi-identity matrix as given in (4.15).

We mention that the number of T-palindromic strong linearizations obtained from EGFPRs as compare to FPRs increases substantially when the degree ofP(λ) increases.

We illustrate this by considering m= 7.