A convergence analysis of SOR iterative methods for linear systems with weak H-matrices

Assume that either A or B is the coefficient matrix of linear system (1), where

It is verified that both A and B are weak H−matrices and not strong H−matrices, since their comparison matrices are both singular M−matrices. How can we get the convergence on FSOR-, BSOR- and SSOR iterative methods for linear systems with this class of matrices without direct computations?

A matrix A ∈ ℂ^n×nis called diagonally dominant by row if

A matrix A ∈ ℂ^n×nis called generalized diagonally dominant if there exist positive constants α_i, i ∈ 〈n〉, such that

(See [30–32]). Let A ∈ D_n(GD_n). ThenA∈HnIif and only if A has no (generalized) diagonally equipotent principal submatrices. Furthermore, if A ∈ D_n) ∩ Z_n (GD_n ∩ Z_n), thenA∈Mn∙if and only if A has no (generalized) diagonally equipotent principal submatrices.

(See [31, 32]). DEn⊂GDEn⊂HnW and SDn∪IDn⊂HnS=GSDn.

(See [6]). GD_n ⊂ H_n

(See [6]). Let A ∈ ℂ^n×nbe irreducible. Then A ∈ H_n if and only if A ∈ GD_n

More importantly, under the condition of "reducibility", we have the following conclusion.

(See [6, 29]). Let A ∈ ℂ^{n × n}be reducible. Then A ∈ H_n if and only if in the Frobenius normal form of A

(See [6, 29]). A matrixA∈HnWif and only if in the Frobenius normal from (15) of A, each irreducible diagonal square block R_ii is generalized diagonally dominant and has at least one generalized diagonally equipotent principal submatrix.

Let E^iθ = (e^iθrs) ∈ ℂ^n×n, wheree^iθrs = cos⊆_rs + i sin θ_rs, i=−1and θ_rs ∈ ℝ for all r, s ∈ 〈n〉. The matrix E^iθ = (e^iθrs) ∈ ℂ^{n × n}is called a π−ray pattern matrix if

1. θ_rs+ θ_sr = 2kβ holds for all r, s ∈ 〈n〉, r ≠ s, where k ∈ ℤ;

2. θ_rs−θ_rt = θ_ts + (2k + 1)π holds for all r, s, t ∈ 〈n〉 and r ≠ s, r ≠ t, t ≠ s, where k ∈ ℤ;

3. θ_rr = 0 for all r ∈ 〈n〉.

Any complex matrix A = (a_rs) ∈ ℂ^{n × n} has the following form:

Let a matrix A = D_A − L_A − U_A = (a_rs) ∈ ℂ^{n ×n} with D_A = diag(a₁₁, a₂₂, ... , a_nn). ThenA∈Rnπif and only if there exists an n × n unitary diagonal matrix D such that D⁻¹AD = e^iη.(|D_A|−|L_A|−|U_A|) for η ∈ ℝ.

A matrix A∈HnW is singular if and only if the matrix A has at least either one zero principal submatrix or one irreducible principal submatrix A _k = A(i₁, i₂, ... ,i_k), 1 < k ≤ n, such thatDAK−1AK∈GDEK∩RKπ, where D_Ak = diag(a_i1i1,... ,a_ikik).

(See [13–15, 23]). Let A ∈ SD_n ∪ ID_n. Then ρ(H_FSOR) < 1, ρ(H_BSOR) < 1 and ρ(H_SSOR) < 1, where H_FSOR, H_BSOR and H_SSOR are defined in (5) , (6) and (7) , respectively, and therefore the sequence {x⁽ⁱ⁾} generated by the FSOR-, BSOR- and SSOR-scheme (3) , respectively, converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾.

(See [37]). LetA∈HnSThen the sequence {x⁽ⁱ⁾} generated by the FSOR-, BSOR- and SSOR-scheme (3) , respectively, converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾.

(See [4, 5, 38]). Let A ∈ ℂ^{n × n}be a Hermitian positive definite matrix. Then the sequence {x⁽ⁱ⁾} generated by the FSOR-, BSOR- and SSOR-scheme (3) , respectively, converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾.

Let A = I − L− U ∈ DE _nbe irreducible. Then for ω ∈ (0, 1), ρ(H_FSOR(ω)) < 1 and ρ(H_BSOR(ω)) < 1, i.e. the sequence {x⁽ⁱ⁾)} generated by the FSOR and BSOR iterative schemes (5) and (6) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾if and only if AA∉Rnπ.

The sufficiency can be proved by contradiction. We assume that there exists an eigenvalue λ of H_FSOR such that |λ| ≥ 1. According to Equality (5),

Since |λ ≥ 1, λ − 1 + ω ≠ 0. Hence, equality (17) yields

i.e.

is singular. Set λ = μe^iθ with μ ≥ 1 and θ ∈ R. Then 1 − cosθ ≥ 0, μ − 1 ≥ 0 and μ² − 1 ≥ 0. Again, ω ∈ (0, 1) shows 1 − ω > 0. Therefore, we have

and

This shows that

Since A = I − L− U ∈ DE_n is irreducible, both L ≠ 0 and U ≠ 0. As a result, (19) and (22) indicate that A(λ, ω) ∈ D_n and irreducible. Again, since A(λ, ω) is singular and hence A(λ,ω)∈HnW, it follows from Lemma 2.12 that A(λ,ω)∈An0∩DEn, i.e. there exists an unitary diagonal matrix D such that

Since A = I − L− U ∈ DE _n,

(23) and (24) shows

Because of |λ ≥ 1 and ω ∈ (0,1), the latter equality of (25) implies |λ| = 1. As a result, (25) and (19) show A(λ, ω) = I − L− U = A. From (23), it is easy to see A(λ,ω)=A∈Rnπ. This contradicts A∉Rnπ. Thus, ρ(H_FSOR(ω)) < 1, i.e. FSOR-method converges.

The following will prove the necessity by contradiction. Assume that A∈Rnπ. Then it follows from Lemma 2.11 that there exists an n × n unitary diagonal matrix D such that A = I − L− U = A = I − D |L| D⁻¹ − D |U| D⁻¹ and

Hence,

Sice A = I − L− U ∈ DE _n and is irreducible, Lemma 2.4 shows that I−|L|−|U|∈HnW∩Rnπ and is irreducible. Lemma 2.12 shows that I − |L| − |U| is singular and hence det(I − |L| − |U|) = 0. Therefore, (27) yields det(I − H_FSOR(ω)) = 0, which shows that 1 is an eigenvalue of H_FSOR(ω). Then, we have that ρ(H_FSOR(ω)) ≥ 1, i.e. FSOR-method doesn’t converge. This is a contradiction. Thus, the assumption is incorrect and hence, A∉Rnπ. This completes the necessity.

In the same way, we can prove that for ω ∈ (0,1), BSOR-method converges, i.e. ρ(_HBSOR(ω)) < 1 if and only if A∉An0. Here, we finish the proof. □

Let A = I − L − U = (a_ij) ∈ D_n with a_ii ≠ 0 for all i ∈ 〈n〉. Then for ω ∈ (0, 1), ρ(H_FSOR(ω)) < 1 and ρ(H_BSOR(ω) < 1, i.e. the sequence {x⁽ⁱ⁾)} generated by the FSOR and BSOR iterative schemes (5) and (6) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾if and only if A is nonsingular.

The conclusion of this theorem is not difficult to be obtained form Lemma 2.1 2, Theorem 3.1 and Theorem 3.4. □

In what follows we will propose the convergence result for SSOR iterative method for linear system with weak H-matrices including nonstrictly diagonally dominant matrices. Firstly, the following lemma will be given for the convenience of the proof.

([32]). LetA=[EULF]∈C2n×2n, where E, F, L, U ∈ C^{n × n}and E is nonsingular. Then A/E is nonsingular if and only if A is nonsingular, where A/E = F − LE⁻¹U is the Schur complement of A with respect to E.

Let A = I − L − U ∈ DE_n be irreducible. Then for ω ∈ (0, 1), ρ(H_SOR(ω)) < 1, i.e. the sequence {x⁽ⁱ⁾)} generated by the SSOR iterative schemes (7) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾if and only if AA∉Rnπ.

The sufficiency can be proved by contradiction. We assume that there exists an eigenvalue λ of H_SSOR such that |λ| ≥ 1. According to equalities (5), (6) and (7),

Equality (28) gives

i.e.

is singular. Let R = I − ωL, S = λ.I − ωU) , T = λ⁻¹[(1 − ω)I + ωL], V = (1 − ω)I + ωU and

Then, B(λ, ω) = ℬ/R is the Schur complement of ℬ with respect to the principal submatrix R. Since B(λ, ω) = ℬ/R is singular, Lemma 3.6 shows that ℬ is also singular. Again since A is irreducible, both L ≠ 0 and U ≠ 0. As a result, ℬ is also irreducible. Since A = I − L − U = (a_ij) ∈ DE _n with unit diagonal entries,

Thus, for all ω ∈ (0, 1) and |λ ≥ 1, we have that both

and

hold for all i ∈ N = {1, 2,..., n}. Immediately, we obtain

(33) shows that ℬ ∈ D_2n. Again, ℬ is irreducible and singular, and hence Lemma 2.4 shows that A(λ,ω)∈H2nW. Then, it follows from Lemma 2.12 that B∈DE2nB2nπ. Consequently |λ| = 1. Let λ = e^iθ with θ ∈ R. Since B∈R2nπ, Lemma 2.12 shows that there exists an n × n unitary diagonal matrix D such that D̃ = diag(D, D) and

The latter two equalities of (34) indicate that θ = 2kβ; where k is an integer and thus λ = eⁱ2kβ = 1, and there exists an n × n unitary diagonal matrix D such that D⁻¹AD = I − |L| − |U|, i.e. A∈An0. However, this contradicts A∉An0. According to the proof above, we have that ρ(H_SSOR(ω)) < 1, i.e. SSOR-method converges.

The following will prove the necessity by contradiction. Assume that A∈An0. Then there exists an n × n unitary diagonal matrix D such that A = I − L− U = I − D |L| D⁻¹ − D |U| D⁻¹ and hence

Thus,

Set C(ω) = (I − ω |U|)− [(1−ω)I + ω|L|](I− ω |L|)⁻¹ [(1−ω)I + ω |U|] and let R̂ = I − ω |L|, Ŝ = I − ω |U|, T̂ = (1 − ω)I + ω |L|, V̂ = (1 − ω)I + ω |U| and

Then, C(ω) = ℬ̂/R̂ is the Schur complement of ℬ̂ with respect to the principal submatrix R̂. (33) and (37) show B^∈H2nW∩R2nπ. It comes from Lemma 2.12 that ℬ̂ is singular. Therefore, Lemma 3.6 yields that C(ω) is singular, i.e.

(36) gives det(I − H_SSOR) = 0, which shows that 1 is an eigenvalue of H_SSOR(ω). Then, we have that ρ(H_SSOR(ω)) ≥ 1, i.e. SSOR-method doesn’t converge. This is a contradiction. Thus, the assumption is incorrect and A∉An0. This completes the proof. □

Let A = I − L − U = (a_ij) ∈ D_n with a_ii ≠ 0 for all i ∈ 〈n〉. Then for ω ∈ (0, 1), ρ(H_SSOR(ω)) < 1, i.e. the sequence {x⁽ⁱ⁾} generated by the SSOR iterative schemes (7) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾if and only if A is nonsingular.

The conclusion of this theorem is easy to be obtained form Lemma 2.1 2, Theorem 3.1 and Theorem 3. 7. □

LetA=I−L−U∈HnWbe irreducible. Then for ω ∈ (0, 1), ρ(H_FSOR(ω)) < 1 and ρ(H_BSOR(ω)) < 1, i.e. the sequence {x⁽ⁱ⁾} generated by the FSOR and BSOR iterative schemes (5) and (6) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾if and only if AA∉Rnπ.

Since A=I−L−U∈HnW is irreducible, it follows form Lemma 2.3 and Lemma 2.6 that A = I − L − U ∈ GDE_n. Then, from Definition 2. 2, there exists a positive diagonal matrix D such that  = D⁻¹AD = I − D⁻¹LD− D⁻¹UD = I − L̂ − Û ∈ DE_n and is irreducible, where L̂ = D⁻¹LD and Û = D⁻¹UD. Theorem 3.4 shows that for ω ∈ (0, 1), ρ(Ĥ_FSOR(ω)) < 1 and ρ(Ĥ_BSOR(ω)) < 1, i.e. the sequence {x⁽ⁱ⁾} generated by the FSOR and BSOR iterative schemes (5) and (6) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾ if and only if A^∉Rnπ. Again, from (5), we have

In the same way, we can get

from (6). Since A^=D−1AD∉Rnπ and D is a positive diagonal matrix, it follows from Definition 2.9 and Definition 2.10 that A∉Rnπ. Therefore, for ω ∈ (0, 1), ρ(H_FSOR(ω)) = ρ(Ĥ_FSOR(ω)) < 1 and ρ(H_BSOR(ω)) = ρ(Ĥ_BSOR(ω)) < 1, i.e. the sequence {x⁽ⁱ⁾} generated by the FSOR and BSOR iterative schemes (5) and (6) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾ if and only if A∉Rnπ. This completes the proof.□

LetA=I−L−U∈HnWbe irreducible. Then for ω ∈ (0, 1), ρ(H_SSOR(ω)) < 1, i.e. the sequence {x⁽ⁱ⁾} generated by the SSOR iterative schemes (7) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾ if and only ifA∉Rnπ.

From (5), (6), (7), (38) and (39),

Therefore, similarly as in the proof of Theorem 3. 9, we have with Theorem 3.7 that for ω ∈ (0, 1), ρ(H_SSOR(ω)) = ρ(Ĥ_SSOR(ω)) < 1 i.e. the sequence {x⁽ⁱ⁾} generated by the SSOR iterative schemes (7) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾ if and only if A∉Rnπ. □

LetA=I−L−U=(aij)∈HnWwith a_ii ≠ 0 for all i ∈ 〈n〉. Then for ω ∈ (0, 1), ρ(H_FSOR(ω)) < 1 and ρ(H_BSOR(ω)) < 1, i.e. the sequence {x⁽ⁱ⁾} generated by the FSOR and BSOR iterative schemes (5) and (6) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾if and only if A is nonsingular.

If A∈HnW is irreducible, it follows from Theorem 3.9 that the conclusion of this theorem is true. If A∈HnW is reducible, since A∈HnW with a_ii ≠ 0 for all i ∈ 〈n〉, H_FSOR(ω) and H_BSOR(ω) exist and Theorem 2.7 shows that there exists some i ∈ 〈n〉 such that diagonal square block R_ii in the Frobenius normal from (15) of A is irreducible and generalized diagonally equipotent. Let HFSORRii and HBSORRii denote the FSOR- and BSOR-iteration matrices associated with diagonal square block R_ii. Direct computations give

Since R_ii ∈ GDE_n is irreducible, Theorem 3.9 shows that for ω ∈ (0, 1), if ρ(HFSOR(ω))=maxi≤i≤sρ(HFSORRii)<1 and ρ(HBSOR(ω))=maxi≤i≤sρ(HBSORRii)<1, i.e. the sequence {x⁽ⁱ⁾} generated by the FSOR and BSOR iterative schemes (5) and (6) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾, then Rii=A(αi)∉R|ai|π. Again, R_ii ∈ GDE_n but Rii∉R|ai|π. Lemma 2.12 shows R_ii = A(α_i) is nonsingular. However, it is easy to obtain that R_jj = A(α_j) is nonsingular that satisfies ρ(HFSOR(ω)Rjj)=maxi≤i≤sρ(HFSORRii)<1 and ρ(HBSOR(ω)Rjj)=maxi≤i≤sρ(HBSORRii)<1. Thus, (15) shows that A is nonsingular. This completes the proof of the necessity.

Let us prove the sufficiency. Assume that A is nonsingular, then each diagonal square block R_ii in the Frobenius normal from (15) of A is nonsingular for all i ∈ 〈n〉. Since A∈HnW, Theorem 2.7 shows that some diagonal square block R_ii in the Frobenius normal from (15) of A is irreducible and generalized diagonally equipotent and the other diagonal square block R_jj is generalized strictly diagonally dominant or generalized irreducibly diagonally dominant. Again, each irreducible and generalized diagonally equipotent diagonal square block R_ii is nonsingular. Lemma 2.12 yields that Rii=A(αi)∉R|αi|π. Then it follows from Theorem 3. 1, Theorem 3.2 and Theorem 3.9 that ρ(HFSOR(ω))=max1<i<sρ(HFSORRii)<1 and ρ(HBSOR(ω))=max1<i<sρ(HBSORRii)<1, i.e. the sequence x⁽ⁱ⁾)} generated by the FSOR and BSOR iterative schemes (5) and (6) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾. This completes the proof. □

A=I−L−U=(aij)∈HnWwitha_ii ≠ 0 for all i ∈ 〈n〉. Then for ω ∈ (0, 1), ρ(H_SSOR(ω)) < 1, i.e. the sequence {x⁽ⁱ⁾} generated by the SSOR iterative schemes (7) converges to the unique solution of (1) for any choice of the initial guess x⁽⁰⁾if and only if A is nonsingular.

Similar to the proof of Theorem 3.1 1, the conclusion of this theorem is easy to be obtained form Theorem 3.1, Theorem 3.2 and Theorem 3.1 0. □

Let the coefficient matrix A of linear system (1) be given by the following n × n matrix

It is easy to see that A_n ∈ DE_n is irreducible. Since

where