Eigenvalue at infinity and recovery of eigenvectors

1.2 Polynomial and rational matrices

2.1.2 Eigenvalue at infinity and recovery of eigenvectors

thatc0(σ, σ2) = 2. Thus, by Corollary 2.1.13,(e^T₄₋₂⊗In)u=u2 ∈ Nr(P(λ)). It is easily verified that u2 6= 0and P(λ)u2 = 0 showing that u2 is indeed an eigenvector ofP(λ).

We now present an example to illustrate a case when our recovery rule coincides with that in Theorem 2.1.18.

Example 2.1.20. Let P(λ) := P7

i=0λⁱA_i. Consider σ = (4 : 6,2 : 3,0 : 1), σ₁ = (2,1,3), σ₂ = (2,4,5,3,4), τ = (−7) and τ₁ = τ₂ = ∅. Clearly, rev(σ₁) and σ₂ are right index tuples of type-1 relative to rev(σ) and σ, respectively. Consider the FPR L(λ) := M_(2,1,3)^P λM₋₇^P −M(4:6,2:3,0:1)^P

M(4,5,2,3,4)^P . Let u ∈ N_r(L(λ)) and v ∈ N_l(L(λ)).

Define u_i := (e^T_i ⊗I_n)u and v_i := (e^T_i ⊗I_n)v for i= 1 : 7.

We have c0(σ, σ2) = 4 and i0(σ1, σ) = 2. Hence, by Corollary 2.1.13, (e^T₇₋₄⊗In)u= u3 ∈ Nr(P(λ)) and (e^T₇₋₂ ⊗In)v =v5 ∈ Nl(P(λ)).

On the other hand,σ₂ is a right index tuple of type-1 relative toσ and the simple tuple associated with (σ, σ₂) is given byz_r(σ, σ₂) = (6,5,0 : 4). Thus c₀(z_r(σ, σ₂)) = 4. Hence by Theorem 2.1.18, (e^T₇₋₄ ⊗I_n)u = u₃ ∈ N_r(P(λ)). Similarly, rev(σ₁) is a right index tuple of type-1 relative to rev(σ) and the simple tuple associated with (rev(σ), rev(σ₁)) is z_r(rev(σ), rev(σ₁)) = (6,5,4,3,0 : 2). Thus c₀(z_r(rev(σ), rev(σ₁))) = 2. Hence by Theorem 2.1.18, (e^T₇₋₂⊗I_n)v =v₅ ∈ N_l(P(λ)).

We remark that our recovery rules for right eigenvectors and right minimal bases of P(λ) from those of GFPRs (and hence of FPRs) can be read off fromc₀(σ, σ₂) whereas the rules for left eigenvectors and left minimal bases can be read off fromi0(σ1, σ). Most importantly,c₀(σ, σ₂) and i₀(σ₁, σ) can be easily read off by looking at the index tuples σ, σ₂ and σ₁. By contrast, the recovery formulae for right (resp., left) eigenvectors and right (resp., left) minimal bases from those of the type-1 FPRs in Theorem 2.1.18 require the number of consecutions of the simple index tuplez_r(σ, σ₂) (resp.,z_r(rev(σ), rev(σ₁))) at 0. The simple index tuples z_r(σ, σ₂) and z_r(rev(σ), rev(σ₁)) are defined recursively and hence cannot be read off readily from the index tuplesσ, σ₂ andσ₁. Therefore, even though for certain type-1 FPRs our recovery rules coincide with those in Theorem 2.1.18, our recovery rules are automatic and can be easily read off from the index tuples σ, σ₁ and σ₂.

Lemma 2.1.21. Let α be an index tuple containing indices from {−m:−1} such that α satisfies the SIP. Then we have the following.

(a) If −s∈α and c−s(α) = p, then α∼ α^L,−s,−(s−1), . . . ,−(s−p), α^R

for some index tuples α^L and α^R such that −s /∈α^L and −(s−p),−(s−p−1)∈/ α^R. (b) if −t ∈ α and i−t(α) =q, then α ∼ α^L,−(t−q), . . . ,−(t−1),−t, α^R

for some index tuples α^L and α^R such that −t /∈α^R and −(t−q),−(t−q−1)∈/ α^L. Proof. Since α satisfies the SIP, we can write α in the column standard form. Let the column standard form of α be given by

csf(α) =

(

(−a1 :−1), . . . ,(−ak−1 :−(k−1)),(−ak:−k), . . . ,(−am :−m)

)

, where −k is the largest integer such that −s ∈ (−a_k : −k). Then (−a_k : −k) = (−a_k,−a_k+ 1, . . . ,−s−1,−s, . . . ,−k). Now, set

α^L:=

(

(−a₁ :−1), . . . ,(−a_k−1 :−(k−1)),−a_k,−a_k+ 1, . . . ,−s−1

)

and α^R:=

(

(−a_k+1 :−(k+ 1)), . . . ,(−a_m :−m)

)

. Then csf(α) = α^L,−s,−(s−1), . . . ,−k, α^R

, where −s /∈α^L and −k,−(k−1)∈/ α^R. Thus (−s,−(s− 1), . . . ,−k) is a subtuple of α and (−s,−(s−1), . . . ,−k,−(k−1)) is not a subtuple of α. So s− k = c−s(α), i.e., s −k = p. Thus k = s − p and

−(s−p),−(s−p−1)∈/ α^R. This proves (a) as α∼csf(α).

(b) Let the row standard form of α be given by

rsf(α) =

(

rev(−b_m :−m), . . . , rev(−b_k:−k), rev(−bk−1 :−(k−1)), . . . , rev(−b₁ :−1)

)

, where −k is the largest integer such that −t ∈ rev(−b_k : −k). Then rev(−b_k : −k) = (−k, . . . ,−t,−t−1, . . . ,−b_k+ 1,−b_k). Now, set

α^L:=

(

rev(−b_m :−m), . . . , rev(−b_k+1 :−(k+ 1))

)

and α^R:=

(

−t−1, . . . ,−b_k+ 1,−b_k, rev(−bk−1 :−(k−1)), . . . , rev(−b₁ :−1)

)

. Then rsf(α) = α^L,−k, . . . ,−(t−1),−t, α^R

, where −t /∈α^R and −k,−(k−1)∈/ α^L. Thus (−k, . . . ,−(t −1),−t) is a subtuple of α and (−(k − 1),−k, . . . ,−(t −1),−t) is not a subtuple of α. So t − k = i−t(α), i.e., t − k = q. Thus k = t − q and

−(t−q),−(t−q−1)∈/α^L. This proves (b) as α∼rsf(α).

We need the following result to derive the recovery of eigenvectors of P(λ) corre- sponding to an eigenvalue at ∞ from those of the GFPRs of P(λ).

Lemma 2.1.22. Let L(λ) := Mτ1(Y1)Mσ1(X1) (λM_τ^P − M_σ^P)Mσ2(X2)Mτ2(Y2) be a GFPR of P(λ). Then we have the following.

(a) (e^T_c

−m(τ)+1⊗I_n)M_σ2(X₂)M_τ2(Y₂) =e^T_c

−m(τ,τ2)+1⊗I_n. (b) M_τ1(Y₁)M_σ1(X₁) (e_i−m(τ)+1⊗I_n) = e_{i−m(τ1,τ)+1}⊗I_n.

Proof. LetZ be an n×n arbitrary matrix. For j = 1 :m−1, we have

(e^T_m−j⊗I_n)M_−j(Z) = (e^T_m−j ⊗I_n)

















I(m−j−1)n

0 I_n I_n Z

I_(j−1)n

















=e^T_m−(j−1)⊗I_n

and

M−j(Z)(em−j⊗I_n) =

















I(m−j−1)n

0 In

I_n Z

I(j−1)n

















(em−j ⊗I_n) = em−(j−1)⊗I_n.

This shows that (e^T_m−j ⊗In)M−k(Z) =







e^T_m−(j−1)⊗I_n for k =j and j = 1 :m−1, e^T_m−j ⊗I_n for k /∈ {j, j+ 1}, j = 0 : m−1,

(2.12) and

M−k(Z)(em−j⊗I_n) =







em−(j−1)⊗In for k =j and j = 1 :m−1, em−j ⊗I_n for k /∈ {j, j+ 1}, j = 0 : m−1.

(2.13) (a) Now we show that

(e^T_c−m(τ)+1⊗In)Mσ2(X2)Mτ2(Y2) =e^T_{c−m(τ,τ2)+1}⊗In. (2.14) Since τ is a permutation of {−m : −(h+ 1)}, we have c_−m(τ) ≤ m − h − 1 and c−m(τ)+1≤m−h. Further, sinceσ₂is an index tuple containing indices from{0 :h−1}, we haveM_σ2(X₂) = diag(I(m−h)n, ∗). Consequently, we have (e^T_c

−m(τ)+1⊗I_n)M_σ2(X₂) = e^T_c

−m(τ)+1⊗I_n. Hence to prove (2.14) we only need to show that (e^T_c

−m(τ)+1⊗I_n)M_τ2(Y₂) =

e^T_c−m(τ,τ

2)+1⊗In. Observe that e^T_c−m(τ)+1⊗In =e^T_{m−(m−c−m(τ)−1)} ⊗In. Set s :=c−m(τ).

We show that

(e^T_{m−(m−s−1)}⊗I_n)M_τ2(Y₂) = e^T_c−m(τ,τ

2)+1⊗I_n (2.15)

which gives (2.14). Since τ has s consecutions at −m, we have τ ∼

(

τ^L,−m,−(m− 1), . . . ,−(m−s)

)

. Note that

(τ, τ₂)∼

(

τ^L,−m,−(m−1), . . . ,−(m−s), τ₂

)

satisfies the SIP. (2.16) Case-I: Suppose that −(m−s−1) ∈/ τ2. Then, since (τ, τ2) satisfies the SIP, it is clear from (2.16) that−(m−s)∈/ τ₂. Now, as−(m−s),−(m−s−1)∈/ τ₂, by (2.12), we have (e^T_{m−(m−s−1)}⊗I_n)M_τ2(Y₂) =e^T_{m−(m−s−1)}⊗I_n =e^T_s+1⊗I_n. Since −(m−s−1)∈/ τ₂, it is clear from (2.16) that c−m(τ, τ₂) = s. This proves (2.15).

Case-II: Suppose that −(m − s− 1) ∈ τ₂. Set p := c−(m−s−1)(τ₂), i.e., τ₂ has p consecutions at −(m−s−1). Then by Lemma 2.1.21,

τ₂ ∼

(

τ₂^L,−(m−s−1),−(m−s−2), . . . ,−(m−s−p−1), τ₂^R

)

,

where −(m−s−1) ∈/ τ₂^L and −(m−s−p−1),−(m−s−p−2) ∈/ τ₂^R. By setting t :=m−s−p−1, we have −t,−(t−1)∈/ τ₂^R. Note that

(τ, τ₂)∼

(

^{τL,−m:−(m−s), τ}₂^{L,−(m−s−1) : −t, τ}₂^R

)

satisfies the SIP, (2.17) where −(m−s −1) ∈/ τ₂^L and −t,−(t−1) ∈/ τ₂^R. As −(m −s −1) ∈/ τ₂^L, we have

−(m −s) ∈/ τ₂^L since otherwise (τ, τ₂) would not satisfy the SIP which is clear from (2.17). We denote by (∗) any arbitrary matrix assignment. Then

(e^T_{m−(m−s−1)}⊗I_n)M_τ^L

2 (∗)M(−(m−s−1):−t)(∗)M_τ^R

2 (∗) = (e^T_{m−(m−s−1)}⊗I_n)M(−(m−s−1):−t)(∗)M_τ^R

2 (∗)







by (2.12) since

−(m−s),−(m−s−1)∈/ τ₂^L

= (e^T_m−(t−1)⊗I_n)M_τR

2 (∗) by applying (2.12) repeatedly

= e^T_m−(t−1)⊗I_n by (2.12) as −t,−(t−1)∈/ τ₂^R.

Hence (e^T_{m−(m−s−1)}⊗In)Mτ2(Y2) = e^T_m−(t−1)⊗In asτ2 ∼ τ₂^L,−(m−s−1) : −t, τ₂^R . It is clear from (2.17) that c_−m(τ, τ₂) =m−t, that is, m−(t−1) = c_−m(τ, τ₂) + 1. This proves (2.15) and hence (2.14) holds. This completes the proof of (a).

(b) Next, we show that

M_τ1,σ1(Y₁, X₁) (e_i−m(τ)+1⊗I_n) = e_{i−m(τ1,τ)+1}⊗I_n. (2.18)

Since τ is a permutation of {−m : −(h + 1)}, we have i−m(τ) ≤ m − h − 1 and i−m(τ)+1 ≤m−h. Further, sinceσ1 is an index tuple containing indices from{0 :h−1}, we haveM_σ1(X₁) = diag(I_(m−h)n, ∗). Consequently, we haveM_σ1(X₁) (e_i−m(τ)+1⊗I_n) = e_i−m(τ)+1⊗I_n.Thus, to prove (2.18) we only need to show thatM_τ1(Y₁) (e_i−m(τ)+1⊗I_n) = e_{i−m(τ1,τ)+1}⊗I_n.

Observe that e_i−m(τ)+1⊗I_n=em−(m−i−m(τ)−1)⊗I_n. Set s:=i−m(τ).Next, we show that

M_τ1(Y₁) (em−(m−s−1)⊗I_n) =e_{i−m(τ1,τ)+1}⊗I_n (2.19) which gives (2.18). Sinceτ hass inversions at−m, we haveτ ∼

(

−(m−s), . . . ,−(m− 1),−m, τ^R

)

. Note that

(τ₁, τ)∼

(

τ₁,−(m−s), . . . ,−(m−1),−m, τ^R

)

satisfies the SIP. (2.20) Case-I: Suppose that −(m−s−1) ∈/ τ₁. Then, since (τ₁, τ) satisfies the SIP, it is clear from (2.20) that−(m−s)∈/ τ₁. Now, as−(m−s),−(m−s−1)∈/ τ₁, by (2.13), we haveMτ1(Y1) (em−(m−s−1)⊗In) =em−(m−s−1)⊗In =es+1⊗In. Since−(m−s−1)∈/ τ1, it is clear from (2.20) thati−m(τ1, τ) = s. This proves (2.19).

Case-II: Suppose that −(m − s− 1) ∈ τ₁. Set p := i_{−(m−s−1)}(τ₁), i.e., τ₁ has p inversions at −(m−s−1). Then by Lemma 2.1.21,

τ₁ ∼

(

τ₁^L,−(m−s−p−1), . . . ,−(m−s−2),−(m−s−1), τ₁^R

)

,

where −(m −s −1) ∈/ τ₁^R and −(m − s− p− 1),−(m −s −p −2) ∈/ τ₁^L. Setting t:=m−s−p−1, we have −t,−(t−1)∈/ τ₁^L.Note that

(τ₁, τ)∼

(

τ₁^L,−t, . . . ,−(m−s−1), τ₁^R,−(m−s), . . . ,−m, τ^R

)

(2.21) satisfies the SIP, where−(m−s−1)∈/ τ₁^Rand −t,−(t−1)∈/ τ₁^L.As−(m−s−1)∈/ τ₁^R, we have −(m−s)∈/ τ₁^R since otherwise (τ₁, τ) would not satisfy the SIP which is clear from (2.21). Now we have

M_τ^L

1 (∗)M(−t,...,−(m−s−2),−(m−s−1))(∗)M_τ^R

1 (∗) (em−(m−s−1)⊗I_n)

= M_τ^L

1 (∗)M(−t,...,−(m−s−2),−(m−s−1))(∗) (em−(m−s−1)⊗I_n)







by (2.13) as −(m−s),

−(m−s−1)∈/ τ₁^R

= M_τ^L

1 (∗) (em−(t−1)⊗I_n) by applying (2.13) repeatedly

= em−(t−1)⊗In by (2.13) as −t,−(t−1)∈/ τ₁^L.

Hence M_τ1(Y₁) (em−(m−s−1)⊗I_n) =em−(t−1)⊗I_n asτ₁ ∼ τ₁^L,−t, . . . ,−(m−s−1), τ₁^R . It is clear from (2.21) thati−m(τ₁, τ) =m−t, that is, m−(t−1) =i−m(τ₁, τ) + 1. This proves (2.19) and hence (2.18) holds. This completes the proof of (b).

We are now ready to describe the recovery of eigenvectors ofP(λ) corresponding to an eigenvalue at ∞ from those of the GFPRs ofP(λ).

Theorem 2.1.23 (Eigenvector recovery at ∞). Let P(λ) be regular and let L(λ) :=

M_τ1(Y₁)M_σ1(X₁)(λM_τ^P −M_σ^P)M_σ2(X₂)M_τ2(Y₂) be a GFPR of P(λ). Suppose that ∞ is an eigenvalue of P(λ). Then we have the following.

Right eigenvectors. If x₁, . . . , x_k

is a basis of the right eigenspace ofL(λ)at ∞, then

(

(e^T_c

−m(τ,τ2)+1⊗I_n)x₁, . . . ,(e^T_c

−m(τ,τ2)+1⊗I_n)x_k

)

is a basis of the right eigenspace of P(λ) at ∞.

Left eigenvectors. If y₁, . . . , y_k

is a basis of the left eigenspace of L(λ) at ∞, then

(

(e^T_i−m(τ

1,τ)+1⊗I_n)y₁, . . . ,(e^T_i−m(τ

1,τ)+1⊗I_n)y_k

)

is a basis of the left eigenspace of P(λ) at ∞.

Proof. We have L(λ) = λL₁−L₀, where L₁ :=M_(τ1,σ1)(Y₁, X₁)M_τ^PM_(σ2,τ2)(X₂, Y₂) and L₀ :=M_(τ1,σ1)(Y₁, X₁)M_σ^PM_(σ2,τ2)(X₂, Y₂) . Note that ∞ is an eigenvalue of L(λ)⇐⇒0 is an eigenvalue of rev(L(λ)) ⇐⇒ 0 is an eigenvalue of L₁. Since M_(τ1,σ1)(Y₁, X₁) is invertible, we have N_r(L₁) = N_r

(

M_τ^PM_(σ2,τ2)(X₂, Y₂)

)

. Further, sinceM_(σ2,τ2)(X₂, Y₂) is nonsingular, it is easily seen that the map N_r

(

M_τ^PM_(σ2,τ2)(X₂, Y₂)

)

−→ N_r(M_τ^P), z 7→

(

M_(σ2,τ2)(X₂, Y₂)

)

z is an isomorphism. Define T(λ) := λM_τ^P −M_σ^P. Then N_r(M_τ^P) = N_r

(

rev(T(0))

)

. Now, by Theorem 1.2.28, the map

Nr

(

rev(T(0))

)

−→ Nr

(

rev(P(0))

)

, u7→(e^T_c−m(τ)+1⊗In)u is an isomorphism. Hence the map

N_r

(

rev(L(0))

)

−→ N_r

(

rev(P(0))

)

, x7→

(

(e^T_c−m(τ)+1⊗I_n)M_(σ2,τ2)(X₂, Y₂)

)

x (2.22) is an isomorphism. Now, by Lemma 2.1.22, we have

(e^T_c−m(τ)+1⊗I_n)M_σ2(X₂)M_τ2(Y₂) =e^T_{c−m(τ,τ2)+1}⊗I_n.

Hence the desired result for the right eigenspace of P(λ) at ∞follows from (2.22).

Next, we prove the result for left eigenspace ofP(λ) at ∞. Since M_(σ2,τ2)(X₂, Y₂) is invertible, we have Nl

(

rev(L(0))

)

=Nl(L1) =Nl

(

M_(τ1,σ1)(Y1, X1)M_τ^P

)

. Further, since M(τ1,σ1)(Y1, X1) is nonsingular, the map Nl

(

M(τ1,σ1)(Y1, X1)M_τ^P

)

−→ Nl(M_τ^P), z 7→

(

M_(τ1,σ1)(Y1, X1)

)

^Tz is an isomorphism. Recall that T(λ) := λM_τ^P −M_σ^P and hence N_l(M_τ^P) =N_l

(

rev(T(0))

)

. Now, by Theorem 1.2.28, the map

N_l

(

rev(T(0))

)

−→ N_l

(

rev(P(0))

)

, v 7→(e^T_i−m(τ)+1⊗I_n)v

is an isomorphism. Hence the map

N_l

(

rev(L(0))

)

−→ N_l

(

rev(P(0))

)

, y 7→

(

(e^T_i−m(τ)+1⊗I_n)(M_(τ1,σ1)(Y₁, X₁))^T

)

y (2.23) is an isomorphism. Now, by Lemma 2.1.22, we have

(e^T_i−m(τ)+1⊗I_n)

(

M_(τ1,σ1)(Y₁, X₁)

)

^T =

(

M_(τ1,σ1)(Y₁, X₁) (e_i−m(τ)+1⊗I_n)

)

^T =e^T_{i−m(τ1,τ)+1}⊗I_n. Hence the result for the left eigenspace of P(λ) at ∞ follows from (2.23).

Remark 2.1.24. Notice that the proof of Lemma 2.1.22 and Theorem 2.1.23 does not use the fact that the index tuple (σ₁, σ, σ₂) satisfies the SIP and hence the results in Theorem 2.1.23 remain valid for GFPR-like pencils of the form

L(λ) := M_τ1(Y₁)M_σ1(X₁)(λM_τ^P −M_σ^P)M_σ2(X₂)M_τ2(Y₂)

in which only the index tuple (τ₁, τ, τ₂) satisfies the SIP but not the tuple (σ₁, σ, σ₂).

This shows that the SIP of (τ₁, τ, τ₂) is enough for operation-free recovery of eigenvec- tors of P(λ) corresponding to the eigenvalue ∞ from those of L(λ). Compare this with Remark 2.1.14.

We illustrate the recovery of eigenvectors of P(λ) corresponding to an eigenvalue at

∞ from those of the GFPRs ofP(λ) by considering an example.

Example 2.1.25. Let P(λ) := P5

i=0λⁱA_i. Suppose that P(λ) is regular and ∞ is an eigenvalue of P(λ). Let σ := (0,1), σ₁ := ∅, σ₂ = (0), τ := (−4,−5,−3,−2), τ₂ = (−4,−3)andτ₁ =∅. LetX and(Y, Z)be any nonsingular matrix assignments forσ₂ and τ₂, respectively. Then the GFPRL(λ) = λM(−4,−5,−3,−2)^P −M_(0,1)^P

M₀(X)M(−4,−3)(Y, Z) =:

λL₁−L₀ of P(λ) is given by

L(λ) = λ























0 0 0 I_n 0

0 0 A₅ A₄ 0 I_n 0 Y A₃ 0 0 I_n Z A₂ 0

0 0 0 0 X























−























0 0 I_n 0 0

I_n 0 Y 0 0

0 I_n Z 0 0

0 0 0 −A₁ X

0 0 0 −A₀ 0





















 .

Let x and y, respectively, be right and left eigenvectors of L(λ) corresponding to the eigenvalue ∞. Define x_i := (e^T_i ⊗I_n)x and y_i := (e^T_i ⊗I_n)y, i = 1 : 5. We have c−5(τ, τ₂) =c−5(−4,−5,−3,−2,−4,−3) = 2. Hence by Theorem 2.1.23, (e^T_c

−m(τ,τ2)+1⊗ I_n)x= (e^T₃⊗I_n)x=x₃ is a right eigenvector ofP(λ)corresponding to the eigenvalue∞.

Similarly, i−5(τ1, τ) =i−5(−4,−5,−3,−2) = 1.Hence by Theorem 2.1.23,(e^T_i−m(τ

1,τ)+1⊗ In)y = (e^T₂ ⊗In)y =y2 is a left eigenvector of P(λ) corresponding to the eigenvalue ∞.

To verify the recovery rule, consider L₁x= 0. This gives x₄ = 0 =x₅ and A₅x₃ = 0.

Further, if x₃ = 0 then x₁ = 0 = x₂. Thus x₃ = 0 ⇒ x = 0. Hence x₃ 6= 0 and is a right eigenvector of P(λ) corresponding to the eigenvalue ∞.

Similarly, y^TL₁ = 0 implies that y_i^T = 0 for i = 3,4,5, and y^T₂A₅ = 0. Further, if y₂^T = 0 then y₁^T = 0. Thus y^T₂ ⇒y= 0. Hence y^T₂ 6= 0 and is a left eigenvector of P(λ) corresponding to the eigenvalue ∞.

2.2 Recovery of eigenvectors and minimal bases from struc-