2.3 Linearizations arising from g-linearizations
3.1.2 Global backward error analysis: A detailed approach
t
As Lt(λ) + ∆Lt(λ) = ˜D( ˆLt(λ) + ˜D−1∆Lt(λ)), the pencils Lt(λ) + ∆Lt(λ) and Lˆt(λ) + ˜D−1∆Lt(λ) are strictly equivalent. Hence Lt(λ) + ∆Lt(λ) is a strong lin- earization of the polynomialP(λ) + ∆P(λ) and the rules for recovering the minimal indices of P(λ) + ∆P(λ) from those of Lt(λ) + ∆Lt(λ) are the same as the ones applied to ˆLt(λ) + ˜D−1∆Lt(λ). Therefore it follows from Corollary 3.1.2, that the right minimal indices of Lt(λ) + ∆Lt(λ) are those ofP(λ) + ∆P(λ) shifted byk−1 and left minimal indices ofLt(λ) + ∆Lt(λ) are same as those ofP(λ) + ∆P(λ).
Clearly, the solution of the complete eigenvalue problem for P(λ) via lineariza- tionsLt(λ) is a globally backward stable process if the constantCLbt,P given by (3.1.5) is moderate. For instance, one such choice is any Lt(λ) for which κ2( ˜D) u 1,
|||A|||F u|α||||P|||F u1 and κ2( ˜R)uσmin( ˜R)u1.
Recall that for any matrix M, σmin(M) denotes the smallest singular value of M. Observing that B(λ) = −R(H˜ k−1(λ)⊗In) where Hk−1(λ) is given by (2.3.6), the next lemma will be useful in establishing a bound on |||∆B|||F that achieves the desired objectives.
Lemma 3.1.5. For k >2, letτ(λ) = Hk−1(λ)⊗In, where Hk−1(λ) is as in (2.3.6).
Then,
σmin(Cbk−2(τ)) = σmin(Cbk−1(τ)) = 2 sin π
4k−2
> 3 2k.
Proof. SinceCbj(τ) = Cbj(Hk−1)⊗Inforj =k−1, k−2,the first two equalities follow from Lemma 1.3.2 and the last inequality follows from the fact that sin(x)> 3xπ for 06x6 π6.
The following result bounds |||∆B|||F such that B(λ) + ∆B(λ) is a minimal basis with all row degrees equal to 1.
Theorem 3.1.6. Let B(λ) be as in (3.1.2), and ∆B(λ) ∈ C[λ](k−1)n×kn be any pencil such that
|||∆B|||F < 3σmin( ˜R)
2k3/2 . (3.1.6)
Then B(λ) + ∆B(λ) is a minimal basis with all its row degrees equal to 1 and all row degrees of any minimal basis dual to it equal to k−1.
Proof. In view of Theorem 3.1.4, the proof follows by establishing thatCbk−2(B+∆B) is nonsingular and Cbk−1(B+ ∆B) has full row rank. For j =k−1 or k−2,
σmin(Cbj(B)) = σmin
−R˜ . ..
−R˜
Cbj(τ)
>σmin( ˜R)σmin(Cbj(τ)),
where τ(λ) is as in Lemma 3.1.5. Now by Lemma 3.1.5, σmin(Cbj(B))>2σmin( ˜R) sin
π 4k−2
>σmin( ˜R) 3
2k. (3.1.7) Since ˜R is nonsingular, Cbk−2(B) is nonsingular and Cbk−1(B) has full row rank. By Lemma 1.3.1(a),
Cbj(B+ ∆B) =Cbj(B) +Cbj(∆B) for j =k−1 and k−2.Again for j =k−1 and k−2,
kCbj(∆B)kF =p
j + 1|||∆B|||F <p
j+ 13σmin( ˜R) 2k3/2 <
√
k3σmin( ˜R)
2k3/2 6σmin(Cbj(B)),
t
where the equality follows from Lemma 1.3.1(b), the first and last inequality is due to equations (3.1.6) and (3.1.7) respectively. As kCbj(∆B)kF < σmin(Cbj(B)), for j =k−1 and k−2,Cbk−2(B+ ∆B) is nonsingular andCbk−1(B+ ∆B) has full row rank.
To establish the required upper bound on |||∆B|||F such that both Condition (A) and Condition (B) are fulfilled, we use [22, Corollary 5.16] which is stated below as a lemma.
Lemma 3.1.7. Let ∆D(λ) ∈ Ckn×n be a matrix polynomial of grade k −1 and
|||∆D|||F < √1
2. Then Λk,n(λ)T + ∆D(λ)T is a minimal basis with all the row degrees equal to k−1 and with all row degrees of any minimal basis dual to it equal to 1.
The following result gives the desired upper bound on |||∆B|||F.
Theorem 3.1.8. Let B(λ) as in (3.1.2) and ∆B(λ) ∈ C[λ](k−1)n×kn be any pencil such that
|||∆B|||F < σmin( ˜R)
2k3/2 . (3.1.8)
Then there exists a matrix polynomial ∆D(λ)∈C[λ]kn×n of grade k−1 such that (a) B(λ) + ∆B(λ) and Λk,n(λ)T + ∆D(λ)T are dual minimal bases, with all the
row degrees equal to 1 and k−1 respectively, and (b) |||∆D|||F 6 k
√ 2
σmin( ˜R)|||∆B|||F < √1
2k. Proof. As
|||∆B|||F < σmin( ˜R)
2k3/2 < 3σmin( ˜R) 2k3/2 ,
therefore, by Theorem 3.1.6,B(λ) + ∆B(λ) is minimal basis with all its row degrees equal to 1 and all row degrees of any minimal basis dual to it equal to k −1. If possible, suppose ∆D(λ) is a matrix polynomial of gradek−1 satisfying
(B(λ) + ∆B(λ))(Λk,n(λ) + ∆D(λ)) = 0 (3.1.9) Since B(λ)Λk,n(λ) = 0, so (3.1.9) becomes
(B(λ) + ∆B(λ))∆D(λ) = −∆B(λ)Λk,n(λ)
⇔Cb0((B+ ∆B)∆D) =−Cb0(∆BΛk,n)
⇔Cbk−1(B+ ∆B)C0(∆D) = −Cb0(∆BΛk,n). (3.1.10)
where (3.1.10) follows from Lemma 1.3.1(c). As Cbk−1(B + ∆B) has full row rank, we have
Cb0(∆D) = −(Cbk−1(B+ ∆B))†Cb0(∆BΛk,n),
which solves (3.1.10). This gives the matrix polynomial ∆D(λ) of grade k − 1 satisfying (3.1.9). To prove part (b) note that,
|||∆D|||F =kCb0(∆D)kF 6k(Cbk−1(B+ ∆B))†k2kCb0(∆BΛk,n)kF. (3.1.11) As k(Cbk−1(B+ ∆B))†k2 = 1
σmin(Cbk−1(B+∆B)), using (3.1.7), (3.1.8) and standard re- sults for perturbation of singular values, it follows that
k(Cbk−1(B+ ∆B))†k2 6 1
3σmin( ˜R)
2k −√
k|||∆B|||F
6 k
σmin( ˜R), and kCb0(∆BΛk,n)kF =|||∆BΛk,n|||F 6√
2|||∆B|||F < σmin( ˜R)
√2k3/2 . Hence from (3.1.11),|||∆D|||F 6
√2k
σmin( ˜R)k∆BkF < √1
2k. Now as the matrix polynomial
∆D(λ) of grade k−1 satisfies (3.1.9) with |||∆D|||F < √1
2k < √1
2, part (a) follows from Lemma 3.1.7.
Next we have the main result which completes the global backward error anal- ysis for solutions of the complete eigenvalue problem for P(λ) obtained from the linearizations ˆLt(λ) given by (2.3.4).
Theorem 3.1.9. Let Lˆt(λ) be any linearization of P(λ) = Pk
i=0λiAi ∈ C[λ]m×n of grade k with m > n, of the form (2.3.4). Let A(λ) and B(λ) be the blocks of Lˆt(λ) as specified by (3.1.1) and (3.1.2) respectively and R˜ ∈ C(k−1)n×(k−1)n be the nonsingular upper triangular matrix appearing in the block B(λ). If ∆ ˆLt(λ) is any pencil of the same size as Lˆt(λ) such that
|||∆ ˆLt|||F < σmin( ˜R) 2k3/2 ,
then Lˆt(λ) + ∆ ˆLt(λ) is a strong linearization of a matrix polynomial P(λ) + ∆P(λ) of grade k and
|||∆P|||F
|||P|||F 6CLdˆ
t,P
|||∆ ˆLt|||F
|||Lˆt|||F
where Cdˆ
Lt,P = |α|1 ||||||PLˆt||||||F
F
3 + 2k |||A|||F
σmin( ˜R)
.
The right minimal indices of Lˆt(λ) + ∆ ˆLt(λ) are those of P(λ) + ∆P(λ) shifted byk−1 and left minimal indices ofLˆt(λ) + ∆ ˆLt(λ) are the same as those ofP(λ) +
t
∆P(λ). This coincides with the corresponding relationship between the minimal indices of Lˆt(λ) and P(λ).
Proof. Clearly, |||∆ ˆLt|||F < σmin2k3/2( ˜R) ⇒ |||∆B|||F < σmin2k3/2( ˜R). By Theorem 3.1.8, there exists ∆D(λ) of grade k−1 such that B(λ) + ∆B(λ) and Λk,n(λ)T + ∆D(λ)T are dual minimal bases with all the row degrees 1 and k −1 respectively. Therefore Lˆt(λ) + ∆ ˆLt(λ) is a strong block minimal bases pencil and Theorem 2.3.5 implies that ˆLt(λ) + ∆ ˆLt(λ) is a strong block minimal bases linearization of
1
α(A(λ) + ∆A(λ))(Λk,n(λ) + ∆D(λ)) =:P(λ) + ∆P(λ) of grade k.As P(λ) = α1A(λ)(Λk,n(λ)), we have,
∆P(λ) = 1
α{(A(λ) + ∆A(λ))∆D(λ) + ∆A(λ)Λk,n(λ)}.
Therefore, by applying Lemma 3.0.1 and Theorem 3.1.8 (b),
|||∆P|||F 6 1
|α|{|||A(∆D)|||F +|||(∆A)(∆D)|||F +|||(∆A)Λk,n|||F}
6 1
|α|{√
2|||A|||F|||∆D|||F +√
2|||∆A|||F|||(∆D)|||F +√
2|||∆A|||F}
6 1
|α|
(√
2|||A|||F k√ 2
σmin( ˜R)|||∆B|||F + 3|||∆A|||F )
6 1
|α|
3 + 2k |||A|||F σmin( ˜R)
|||∆ ˆLt|||F.
This implies that, |||∆P|||P||||||F
F 6 |α|1 ||||||PLˆt||||||FF
3 + 2k |||A|||F
σmin( ˜R)
|||∆ ˆL
t|||F
|||Lˆt|||F . Also as the block ˆB(λ) is absent in the linearization ˆLt(λ) + ∆ ˆLt(λ), we have ˆC(λ) = Im in (2.3.8) and consequently by Theorem 2.3.5, the right minimal indices of ˆLt(λ) + ∆ ˆLt(λ) are those of P(λ) + ∆P(λ) shifted by k−1 and left minimal indices of ˆLt(λ) + ∆ ˆLt(λ) are the same as those of P(λ) + ∆P(λ). By Theorem 2.3.6, the shifting relations between the left and right minimal indices ofP(λ) + ∆P(λ) and ˆLt(λ) + ∆ ˆLt(λ) are exactly the same as those between P(λ) and ˆLt(λ).
Now we extend the above analysis to solutions of the complete eigenvalue prob- lem for P(λ) obtained via any linearization Lt(λ) arising from a g-linearization in L1(P).Using this fact that any such linearizationLt(λ) is strictly equivalent to a lin- earization ˆLt(λ) of the form (2.3.4), and the results for ˆLt(λ),we have the following theorem.
Theorem 3.1.10. LetLt(λ)be any linearization of P(λ) =Pk
i=0λiAi ∈C[λ]m×n of grade k withm >n, arising from a g-linearization in L1(P). Let Lˆt(λ) = ˜D−1Lt(λ) where D˜ is as given in (2.3.5). Then Lˆt(λ) is of the form (2.3.4). Let A(λ) and B(λ) be the blocks of Lˆt(λ) as specified by (3.1.1) and (3.1.2) respectively and R˜ be the nonsingular upper triangular matrix appearing in the block B(λ). If ∆Lt(λ) is any pencil of the same size as Lt(λ) such that
|||∆Lt|||F < σmin( ˜R)σmin( ˜D)
2k3/2 , (3.1.12)
then Lt(λ) + ∆Lt(λ) is a strong linearization of a matrix polynomial P(λ) + ∆P(λ) of grade k such that
|||∆P|||F
|||P|||F 6CLdt,P|||∆Lt|||F
|||Lt|||F , (3.1.13) whereCLdt,P = κ2|α|( ˜D)||||||PLˆt||||||F
F
3 + 2k |||A|||F
σmin( ˜R)
, κ2( ˜D)being the 2-norm condition number of D.˜ The right minimal indices of Lt(λ) + ∆Lt(λ) are those of P(λ) + ∆P(λ) shifted by k − 1 and left minimal indices of Lt(λ) + ∆Lt(λ) are same as those of P(λ) + ∆P(λ). This coincides with the corresponding relationship between the minimal indices of Lt(λ) and P(λ).
Proof. Evidently, ˆLt(λ) = ˜D−1Lt(λ) is of the form (2.3.4). Using (3.1.12) and the fact that
Lt(λ) + ∆Lt(λ) = ˜D( ˆLt(λ) + ˜D−1∆Lt(λ)), we have,
|||D˜−1∆Lt|||F < σmin( ˜R) 2k3/2 .
Therefore by Theorem 3.1.9, ˆLt(λ) + ˜D−1∆Lt(λ) is a strong block minimal bases linearization of some polynomial P(λ) + ∆P(λ) of gradek such that
|||∆P|||F
|||P|||F 6 1
|α|
|||Lˆt|||F
|||P|||F
3 + 2k |||A|||F σmin( ˜R)
|||D˜−1∆Lt|||F
|||Lˆt|||F .
Using the relations, |||D˜−1∆Lt|||F 6 kD˜−1k2|||∆Lt|||F and |||Lt|||F 6 kDk˜ 2|||Lˆt|||F, we get,
|||∆P|||F
|||P|||F 6 1
|α|
|||Lˆt|||F
|||P|||F
3 + 2k |||A|||F σmin( ˜R)
kD˜−1k2|||∆Lt|||FkDk˜ 2
|||Lt|||F
= κ2( ˜D)
|α|
|||Lˆt|||F
|||P|||F
3 + 2k |||A|||F σmin( ˜R)
|||∆Lt|||F
|||Lt|||F .
t
As Lt(λ) + ∆Lt(λ) = ˜D( ˆLt(λ) + ˜D−1∆Lt(λ)), the pencils Lt(λ) + ∆Lt(λ) and Lˆt(λ) + ˜D−1∆Lt(λ) are strictly equivalent. Hence Lt(λ) + ∆Lt(λ) is a strong lin- earization of the polynomialP(λ) + ∆P(λ) and the rules for recovering the minimal indices of P(λ) + ∆P(λ) from those of Lt(λ) + ∆Lt(λ) are the same as the ones applied to ˆLt(λ) + ˜D−1∆Lt(λ). Therefore it follows from Theorem 3.1.9, that the right minimal indices of Lt(λ) + ∆Lt(λ) are those ofP(λ) + ∆P(λ) shifted byk−1 and left minimal indices ofLt(λ) + ∆Lt(λ) are same as those ofP(λ) + ∆P(λ).
If the complete eigenvalue problem forLt(λ) is solved by using a backward stable algorithm, then |||∆L|||Lt|||F
t|||F =O(u).In such a situation (3.1.13) shows that the process of solving the complete eigenvalue problem forP(λ) via linearizationsLt(λ),is globally backward stable ifCLdt,P is not very large. AsCLdt,P =κ2( ˜D)CLdˆ
t,P, so a good choice of Lt(λ) would be one for whichκ2( ˜D)u1 andCdˆ
Lt,P is not large for the corresponding pencil ˆLt(λ) = ˜D−1Lt(λ). To identify such linearizations, we first note that for the block A(λ) of ˆLt(λ),
αP(λ) = A(λ)Λk,n(λ)⇒ |α| |||P|||F =|||AΛk,n|||F 6√
2|||A|||F. This implies that
|||Lˆt|||F
|||P|||F > |||A|||F
|||P|||F > |α|
√2. Now if |α| |||P|||F σmin( ˜R) then
√2|||A|||F
σmin( ˜R) 1 and since |α||||P|||Lˆt||||||F
F > √12, so Cdˆ
Lt,P is big. Again if |α| |||P|||F σmin( ˜R) then |α||||P|||Lˆt||||||F
F > |α||||P|||kRk˜ F
F 1 and once again Cdˆ
Lt,P
is big. So, a good choice of Lt(λ) would be one for which |α| |||P|||F uσmin( ˜R).
Besides, if |||A|||F u|α| |||P|||F uσmin( ˜R), and κ2( ˜R)u1 then CLdt,P u(3 + 2k)p
1 + 2(k−1)n and then
|||∆P|||F
|||P|||F /(3 + 2k)p
1 + 2(k−1)n|||∆Lt|||F
|||Lt|||F . In summary, by using linearizations Lt(λ) satisfying
(i) κ2( ˜D)u1 and κ2( ˜R)u1 and (ii) |||A|||F u|α| |||P|||F uσmin( ˜R), we will have |||∆P|||P||||||F
F = O(u) if |||∆L|||Lt|||F
t|||F =O(u). So the complete eigenvalue problem for P(λ) can be solved in a globally backward stable manner by using backward stable algorithms to solve the complete eigenvalue problem for such choices ofLt(λ).
The optimal block Kronecker linearizations of the form
A(λ) B(λ)
ensuring global backward stability that were identified in [22] are included in the above choices.
In fact they are the ones for which ˜D =Im+(k−1)n, |α| = 1/|||P|||F, R˜ =I(k−1)n and kX12k2F+kY11k2F u |||P1|||2F
Pk−1
i=1 kAik2F in (2.3.9) and include the Frobenius companion formC1(λ). Our analysis shows that there exist many more choices of linearizations among the pencils Lt(λ) with which the complete eigenvalue problem for P(λ) can be solved in a globally backward stable manner.