In numerical linear algebra, the successive overrelaxation iterative method, simultaneously introduced by Frankel (1950) [33] and Young (1950) [34], is a famous iterative method used to solve a linear system of equations. This iterative method is also called the *accelerate Liebmann method* by Frankel (1950) [33] and the other many subsequent researchers. Kahan (1958) [35] calls it the *extrapolated Gauss-Seidel method*. It is often called the method of *systematic overrelaxation*. Frankel showed that for the numerical solutation of the Dirichlet problem for a rectangle, successive overrelaxation iterative method gave substantially larger (by an order of magnitude) asymptotic rates of convergence than those for the point Jacobi and point Gauss-Seidel iterative methods with suitable chosen relaxation factor. Young (1950) [34] and Young (1954) [36] showed that these conclusions held more generally for matrices satisfying his definition of propertly A, and that these results could be rigorously applied to the iterative solution of matrix equations arising from discrete approximations to a large class of elliptic partial differential equations for general regions.

Later, this iterative method was developed as three iterative methods, i.e., the forward, backward and symmetric successive overrelaxation (FSOR-, BSOR- and SSOR-) iterative methods. Though these iterative methods can be applied to any matrix with non-zero elements on the diagonals, convergence is only guaranteed if the matrix is strictly or irreducibly diagonally dominant matrix, Hermitian positive definite matrix, strong *H*−matrix and consistently ordered *p*-cyclic matrix. Some classic results on convergence of SOR iterative methods are as follows:

(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]). *Let* $A\in {H}_{n}^{S}$ *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)}.

(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)}.

In this section, we mainly study convergence of SOR iterative methods for the linear systems with weak *H*-matrices.

*Let A* = *I* − *L*− *U* ∈ DE _{n} *be 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 A* $A\notin {\mathcal{R}}_{n}^{\pi}$.

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),
$$\begin{array}{ccc}det\left(\lambda I-{H}_{FSOR(\omega )}\right)\hfill & =\hfill & det\left\{\lambda I-{\left(I-\omega L\right)}^{-1}\left[\left(1-\omega \right)I+\omega U\right]\right\}\hfill \\ \hfill & =\hfill & det\left[{\left(I-\omega L\right)}^{-1}\right]det\left\{\lambda \left(I-\omega L\right)-\left[\left(1-\omega \right)I+\omega U\right]\right\}\hfill \\ \hfill & =\hfill & \frac{det\left[\left(\lambda -1+\omega \right)I-\lambda \omega L-\omega U\right]}{det\left(I-\omega L\right)}\hfill \\ \hfill & =\hfill & \left(\lambda -1+\omega \right)\frac{det\phantom{\rule{thinmathspace}{0ex}}[I-\frac{\lambda \omega}{\lambda -1+\omega}L-\frac{\omega}{\lambda -1+\omega}U]}{det\left(I-\omega L\right)}\hfill \\ \hfill & =\hfill & 0.\hfill \end{array}$$(17)

Since |*λ* ≥ 1, *λ* − 1 + *ω* ≠ 0. Hence, equality (17) yields
$$det\phantom{\rule{thinmathspace}{0ex}}A\left(\lambda ,\omega \right)=det\phantom{\rule{thinmathspace}{0ex}}[I-\frac{\lambda \omega}{\lambda -1+\omega}L-\frac{\omega}{\lambda -1+\omega}U]=0$$(18)
i.e.
$$A\left(\lambda ,\omega \right)=I-\frac{\lambda \omega}{\lambda -1+\omega}L-\frac{\omega}{\lambda -1+\omega}U$$(19)
is singular. Set *λ* = *μe*^{iθ} with *μ* ≥ 1 and *θ* ∈ *R*. Then 1 − *cosθ* ≥ 0, *μ* − 1 ≥ 0 and *μ*^{2} − 1 ≥ 0. Again, *ω* ∈ (0, 1) shows 1 − *ω > 0*. Therefore, we have
$$\begin{array}{ccc}{\left|\lambda -1+\omega \right|}^{2}-{\left|\lambda \omega \right|}^{2}\hfill & =\hfill & \left|\mu {e}^{i\theta}-1+{\omega}^{2}\right|-{\left|\mu {e}^{i\theta}\omega \right|}^{2}\hfill \\ \hfill & =\hfill & {\mu}^{2}-2\left(1-\omega \right)\mu cos\theta +{\left(1-\omega \right)}^{2}-{\mu}^{2}{\omega}^{2}\hfill \\ \hfill & =\hfill & \left[{\mu}^{2}\left(1+\omega \right)-2\mu \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\phantom{\rule{thinmathspace}{0ex}}\theta +1-\omega \right]\left(1-\omega \right)\hfill \\ \hfill & =\hfill & \left[{\mu}^{2}+1-2\mu \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\theta +{\mu}^{2}\omega -\omega \right]\left(1-\omega \right)\hfill \\ \hfill & \ge \hfill & \left[2\mu -2\mu \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\phantom{\rule{thinmathspace}{0ex}}\theta +\left({\mu}^{2}-1\right)\omega \right]\left(1-\omega \right)\hfill \\ \hfill & =\hfill & \left[2\mu \left(1-\mathrm{cos}\theta \right)+\left({\mu}^{2}-1\right)\omega \right]\left(1-\omega \right)\hfill \\ \hfill & \ge \hfill & 0.\hfill \end{array}$$(20)
and
$$\begin{array}{ccc}{\left|\lambda -1+\omega \right|}^{2}-{\omega}^{2}\hfill & =\hfill & {\left|\mu {e}^{i\theta}-1+\omega \right|}^{2}-{\omega}^{2}\hfill \\ \hfill & =\hfill & {\mu}^{2}-2\left(1-\omega \right)\mu cos\theta +{\left(1-\omega \right)}^{2}-{\omega}^{2}\hfill \\ \hfill & =\hfill & {\left[\mu -\left(1-\omega \right)\right]}^{2}-{\omega}^{2}+2\mu \left(1-\omega \right)-2\mu \left(1-\omega \right)\mathrm{cos}\phantom{\rule{thinmathspace}{0ex}}\theta \hfill \\ \hfill & =\hfill & {\left[\mu -1+\omega \right]}^{2}-{\omega}^{2}+2\mu \left(1-\omega \right)\left(1-\mathrm{cos}\phantom{\rule{thinmathspace}{0ex}}\theta \right)\hfill \\ \hfill & \ge \hfill & {\omega}^{2}-{\omega}^{2}\hfill \\ \hfill & \ge \hfill & 0.\hfill \end{array}$$(21)

This shows that
$$\left|\frac{\lambda \omega}{\lambda -1+\omega}\right|\le 1,\phantom{\rule{thinmathspace}{0ex}}\left|\frac{\omega}{\lambda -1+\omega}\right|\le 1.$$(22)

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 $A(\lambda ,\omega )\in {H}_{n}^{W}$, it follows from Lemma 2.12 that $A\left(\lambda ,\omega \right)\in {\mathcal{A}}_{n}^{0}\cap D{E}_{n}$, i.e. there exists an unitary diagonal matrix *D* such that
$$\begin{array}{ccc}{D}^{-1}A\left(\lambda ,\omega \right)D\hfill & =\hfill & I-\frac{\lambda \omega}{\lambda -1+\omega}{D}^{-1}LD-\frac{\omega}{\lambda -1+\omega}{D}^{-1}UD\hfill \\ \hfill & =\hfill & I-\left|\frac{\lambda \omega}{\lambda -1+\omega}\right|\left|L\right|-\left|\frac{\omega}{\lambda -1+\omega}\right|\left|U\right|\hfill \\ \hfill & \in \hfill & D{E}_{n}.\hfill \end{array}$$(23)

Since *A* = *I* − *L*− *U* ∈ DE _{n},
$$\mu \left(A\right)=I-\left|L\right|-\pi \left|U\right|\in D{E}_{n}$$(24)
(23) and (24) shows
$$\left|\frac{\lambda \omega}{\lambda -1+\omega}\right|=1,\phantom{\rule{thinmathspace}{0ex}}\left|\frac{\omega}{\lambda -1+\omega}\right|=1.$$(25)

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 $A\left(\lambda ,\omega \right)=A\in {\mathcal{R}}_{n}^{\pi}$. This contradicts $A\notin {\mathcal{R}}_{n}^{\pi}$. Thus, *ρ*(*H*_{FSOR(ω)}) < 1, i.e. FSOR-method converges.

The following will prove the necessity by contradiction. Assume that $A\in {\mathcal{R}}_{n}^{\pi}$. 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
$$\begin{array}{ccc}{H}_{FSOR(\omega )}\hfill & =\hfill & {\left(I-\omega L\right)}^{-1}\left[\left(1-\omega \right)I+\omega U\right]\hfill \\ \hfill & =\hfill & D{\left(I-\omega \left|L\right|\right)}^{-1}\left[\left(1-\omega \right)I+\omega \left|U\right|\right]{D}^{-1}.\hfill \end{array}$$(26)

Hence,
$$\begin{array}{ccc}det\left(I-{H}_{FSOR\left(\omega \right)}\right)\hfill & =\hfill & det\left\{I-{\left(I-\omega L\right)}^{-1}\left[\left(1-\omega \right)I+\omega U\right]\right\}\hfill \\ \hfill & =\hfill & det\left\{I-{\left(I-\omega \left|L\right|\right)}^{-1}\left[\left(1-\omega \right)I+\omega \left|U\right|\right]\right\}\hfill \\ \hfill & =\hfill & \frac{det\left[I-\omega \left|L\right|-\left(1-\omega \right)I-\omega \left|U\right|\right]}{det\left(I-\omega \left|L\right|\right)}\hfill \\ \hfill & =\hfill & \frac{\omega det\left[I-\left|U\right|-\left|L\right|\right]}{det\left(I-\omega \left|L\right|\right)}.\hfill \end{array}$$(27)

Sice *A* = *I* − *L*− *U* ∈ DE _{n} and is irreducible, Lemma 2.4 shows that $I-\left|L\right|-\left|U\right|\in {H}_{n}^{W}\cap {\mathcal{R}}_{n}^{\pi}$ 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, $A\notin {\mathcal{R}}_{n}^{\pi}$. 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 $A\notin {\mathcal{A}}_{n}^{0}$. 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]). *Let* $A=\left[\begin{array}{cc}E\hfill & U\hfill \\ L\hfill & F\hfill \end{array}\right]\in {C}^{2n\times 2n}$, *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*^{−1} *U 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 A* $A\notin {\mathcal{R}}_{n}^{\pi}$.

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),
$$\begin{array}{rl}& det\left(\lambda I-{H}_{SSOR\left(\omega \right)}\right)\\ & =det\left\{\lambda I-{\left(I-\omega U\right)}^{-1}\left[\left(1-\omega \right)I+\omega L\right]{\left(I-\omega L\right)}^{-1}\left[\left(1-\omega \right)I+\omega U\right]\right\}\\ & =\frac{det\left\{\lambda \left(I-\omega U\right)-\left[\left(1-\omega \right)I+\omega L\right]{\left(I-\omega L\right)}^{-1}\left[\left(1-\omega \right)I+\omega U\right]\right\}}{det\left(I-\omega U\right)}\\ & =\frac{\lambda \phantom{\rule{thinmathspace}{0ex}}det\left\{\left(I-\omega U\right)-{\lambda}^{-1}\left[\left(1-\omega \right)I+\omega L\right]{\left(I-\omega L\right)}^{-1}\left[\left(1-\omega \right)I+\omega U\right]\right\}}{det\left(I-\omega U\right)}\\ & =0.\end{array}$$(28)

Equality (28) gives
$$\begin{array}{lll}detB(\lambda ,\omega )& =& det\{(I-\omega U)-{\lambda}^{-1}[(1-\omega )I+\omega L]{(I-\omega L)}^{-1}[(1-\omega )I+\omega U]\}\\ & =& 0,\end{array}$$(29)
i.e.
$$B(\lambda ,\omega )\phantom{\rule{thinmathspace}{0ex}}=(I-\omega U)\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}{\lambda}^{-1\phantom{\rule{thinmathspace}{0ex}}}[(1\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\omega )I+\omega L](I\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\omega L{)}^{-1\phantom{\rule{thinmathspace}{0ex}}}[(1\phantom{\rule{thinmathspace}{0ex}}-\omega )I+\omega U]$$(30)
is singular. Let *R* = *I* − *ωL*, *S* = *λ.I* − *ωU*) , *T* = *λ*^{−1}[(1 − *ω*)*I* + *ωL*], *V* = (1 − *ω*)*I* + *ωU* and
$$\mathcal{B}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\left[\begin{array}{cc}R\hfill & -V\hfill \\ -T\hfill & S\hfill \end{array}\right]\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\left[\begin{array}{cc}I-\omega L& -[(1-\omega )I+\omega U]\\ -{\lambda}^{-1}[(1-\omega )I+\omega L]& I-\omega U\end{array}\right].$$(31)

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,
$$1\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\sum _{j\phantom{\rule{thinmathspace}{0ex}}=1}^{i-1}\left|aij\right|+\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sum _{j\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}i+1}^{n}\left|aij\right|,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}i=1,2,...,\phantom{\rule{thinmathspace}{0ex}}n.$$(32)

Thus, for all *ω* ∈ (0, 1) and |*λ* ≥ 1, we have that both
$$1-\omega \phantom{\rule{thinmathspace}{0ex}}\sum _{j=1}^{i-1}\left|aij\right|-(1-\omega )-\omega \phantom{\rule{thinmathspace}{0ex}}\sum _{j=i+1}^{n}\left|aij\right|\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\omega (1-\sum _{j=1}^{i-1}\left|aij\right|-\sum _{j\phantom{\rule{thinmathspace}{0ex}}=i+1}^{n}\left|aij\right|)\phantom{\rule{thinmathspace}{0ex}}=0$$
and
$$1-\omega \phantom{\rule{thinmathspace}{0ex}}\sum _{j=1+1}^{n}\left|aij\right|-|\lambda {|}^{-1}[(1-\omega )+\omega \phantom{\rule{thinmathspace}{0ex}}\sum _{j=1}^{i-1}\left|aij\right|\phantom{\rule{thinmathspace}{0ex}}]\ge \phantom{\rule{thinmathspace}{0ex}}\omega (1-\sum _{j=1}^{i-1}\left|aij\right|-\sum _{j\phantom{\rule{thinmathspace}{0ex}}=i+1}^{n}\left|aij\right|)\phantom{\rule{thinmathspace}{0ex}}=0$$
hold for all *i* ∈ *N* = {1, 2,..., *n*}. Immediately, we obtain
$$\begin{array}{rl}& 1\phantom{\rule{thinmathspace}{0ex}}=\omega \phantom{\rule{thinmathspace}{0ex}}\sum _{j\phantom{\rule{thinmathspace}{0ex}}=1}^{i-1}\left|aij\right|+\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(1-\omega )+\omega \sum _{j\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}i+1}^{n}\left|aij\right|,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\\ & 1\ge \phantom{\rule{thinmathspace}{0ex}}\omega \phantom{\rule{thinmathspace}{0ex}}\sum _{j\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}i+1}^{n}\left|aij\right|\phantom{\rule{thinmathspace}{0ex}}+|\lambda {|}^{-1}[(1-\omega )+\omega \sum _{j\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}1}^{i-1}\left|aij\right|],\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}i\phantom{\rule{thinmathspace}{0ex}}=1,2,...,n.\end{array}$$(33)
(33) shows that *ℬ* ∈ *D*_{2n}. Again, *ℬ* is irreducible and singular, and hence Lemma 2.4 shows that $A(\lambda ,\omega )\in {H}_{2n}^{W}$. Then, it follows from Lemma 2.12 that $\mathcal{B}\in \phantom{\rule{thinmathspace}{0ex}}D{E}_{2n}{\mathcal{B}}_{2n}^{\pi}$. Consequently |*λ*| = 1. Let *λ* = *e*^{iθ} with *θ* ∈ *R*. Since $\mathcal{B}\in \phantom{\rule{thinmathspace}{0ex}}{\mathcal{R}}_{2n}^{\pi}$, Lemma 2.12 shows that there exists an *n* × *n* unitary diagonal matrix *D* such that *D̃* = *diag*(*D, D*) and
$$\begin{array}{ccc}{\stackrel{~}{D}}^{-1}\mathcal{B}\stackrel{~}{D}\hfill & =\hfill & {\stackrel{~}{D}}^{-1}\left[\begin{array}{cc}I-\omega L& (1-\omega )I+\omega U\\ {\lambda}^{-1}[(1-\omega )I+\omega L]& I-\omega U\end{array}\right]\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{D}}^{-1}\hfill \\ \hfill & =\hfill & {\stackrel{~}{D}}^{-1}\left[\begin{array}{cc}I-\omega L& -[(1-\omega )I+\omega U]\\ -{e}^{-i\theta}[(1-\omega )I+\omega L]& I-\omega U\end{array}\right]\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{D}}^{-1}\hfill \\ \hfill & =\hfill & \left[\begin{array}{cc}I-\omega {D}^{-1}LD& -[(1-\omega )I+\omega {D}^{-1}UD]\\ -{e}^{-i\theta}[(1-\omega )I+\omega {D}^{-1}LD]& I-\omega {D}^{-1}UD\end{array}\right]\hfill \\ \hfill & =\hfill & \left[\begin{array}{cc}I-\omega |L|& -[(1-\omega )I+\omega |U|]\\ -{e}^{-i\theta}[(1-\omega )I+\omega |L|]& I-\omega |U|\end{array}\right]\hfill \\ \hfill & =\hfill & \left[\begin{array}{cc}I-\omega |L|& -\phantom{\rule{thinmathspace}{0ex}}[(1-\omega )I+\omega |U|]\\ -\phantom{\rule{thinmathspace}{0ex}}[(1-\omega )I+\omega |L|]& I-\omega |U|\end{array}\right].\hfill \end{array}\phantom{\rule{thinmathspace}{0ex}}$$(34)

The latter two equalities of (34) indicate that *θ* = *2kβ;* where *k* is an integer and thus *λ* = *e*^{i}2kβ = 1, and there exists an *n* × *n* unitary diagonal matrix *D* such that *D*^{−1}*AD* = *I* − |*L*| − |*U*|, i.e. $A\in \phantom{\rule{thinmathspace}{0ex}}{\mathcal{A}}_{n}^{0}$. However, this contradicts $A\notin \phantom{\rule{thinmathspace}{0ex}}{\mathcal{A}}_{n}^{0}$. 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 $A\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}{\mathcal{A}}_{n}^{0}$. 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
$$\begin{array}{ll}H{\phantom{\rule{thinmathspace}{0ex}}}_{S\phantom{\rule{thinmathspace}{0ex}}S\phantom{\rule{thinmathspace}{0ex}}O\phantom{\rule{thinmathspace}{0ex}}R(\omega )\phantom{\rule{thinmathspace}{0ex}}}& ={(I-\omega U)}^{-1}[(1-\omega )I+\omega L]{(I-\omega L)}^{-1}[(1-\omega )I+\omega U]\\ & =D\{{(I-\omega |U|)}^{-1}[(1-\omega )I+\omega |L|]{(1-\omega |L|)}^{-1}[(1-\omega )I+\omega |U|]\}{D}^{-1}.\end{array}$$(35)

Thus,
$$\begin{array}{l}det(I-H{S\text{\hspace{0.05em}}S\text{\hspace{0.05em}}O\text{\hspace{0.05em}}R(\omega )\text{\hspace{0.05em}}}_{})\hfill \\ =det\{I-{(I-\omega |U|)}^{-1}[(1-\omega )I+\omega \left|L\right|]{(1-\omega |L|)}^{-1}[(1-\omega )I+\omega \left|U\right|]\}\hfill \\ =\frac{det\{(I-\omega \left|U\right|)-[(1-\omega )I+\omega \left|L\right|]{(1-\omega |L|)}^{-1}[(1-\omega )I+\omega \left|U\right|]\}}{det(I-\omega |U|)}.\hfill \end{array}$$(36)

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
$$\hat{\mathcal{B}}\phantom{\rule{thinmathspace}{0ex}}=\left[\begin{array}{cc}\hat{R}& -\hat{V}\\ -\hat{T}& \hat{S}\end{array}\right]\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\left[\begin{array}{cc}I-\omega \left|L\right|& -[(1-\omega )I+\omega \left|U\right|]\\ -[(1-\omega )I+\omega \left|L\right|& I-\omega \left|U\right|\end{array}\right].$$(37)

Then, *C*(*ω*) = *ℬ̂*/*R̂* is the Schur complement of *ℬ̂* with respect to the principal submatrix *R̂*. (33) and (37) show $\hat{\mathcal{B}}\in \phantom{\rule{thinmathspace}{0ex}}{H}_{2n}^{W}\phantom{\rule{thinmathspace}{0ex}}\cap {\mathcal{R}}_{2n}^{\pi}$. It comes from Lemma 2.12 that *ℬ̂* is singular. Therefore, Lemma 3.6 yields that *C*(*ω*) is singular, i.e.
$$\mathit{\text{det}}\phantom{\rule{thinmathspace}{0ex}}C(\omega )\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\mathit{\text{det}}\phantom{\rule{thinmathspace}{0ex}}\{(I-\omega |U|)-[(1-\omega )I+\omega |L|]{(I-\omega |L|)}^{-1}[(1-\omega )I+\omega |U|]\}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}0.$$
(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 $A\notin {\mathcal{A}}_{n}^{0}$. 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. □

*Let* $A=I-L-U\in {H}_{n}^{W}$ *be 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 A* $A\notin {\mathcal{R}}_{n}^{\pi}$.

Since $A=I-L-U\in {H}_{n}^{W}$ 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 *Â* = *D*^{−1} *AD* = *I* − *D*^{−1} *LD*− *D*^{−1} *UD* = *I* − *L̂* − *Û* ∈ *DE*_{n} and is irreducible, where *L̂* = *D*^{−1} *LD* and *Û* = *D*^{−1} *UD*. 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 $\hat{A}\phantom{\rule{thinmathspace}{0ex}}\notin {\mathcal{R}}_{n}^{\pi}$. Again, from (5), we have
$$\begin{array}{ll}\hat{H}{\phantom{\rule{thinmathspace}{0ex}}}_{F\phantom{\rule{thinmathspace}{0ex}}S\phantom{\rule{thinmathspace}{0ex}}O\phantom{\rule{thinmathspace}{0ex}}R(\omega )\phantom{\rule{thinmathspace}{0ex}}}& =\phantom{\rule{thinmathspace}{0ex}}{(I-\omega \hat{L})}^{-1}[(1-\omega )I+\omega \hat{U}]\\ & ={(I-\omega {D}^{-1}LD)}^{-1}[(1-\omega )I+\omega {D}^{-1}UD]\\ & =\phantom{\rule{thinmathspace}{0ex}}{D}^{-1}\{{(I-\omega L)}^{-1}[(1-\omega )I+\omega U]\}D\\ & =\phantom{\rule{thinmathspace}{0ex}}{D}^{-1}H{\phantom{\rule{thinmathspace}{0ex}}}_{F\phantom{\rule{thinmathspace}{0ex}}S\phantom{\rule{thinmathspace}{0ex}}O\phantom{\rule{thinmathspace}{0ex}}R(\omega )\phantom{\rule{thinmathspace}{0ex}}}D.\end{array}$$(38)

In the same way, we can get
$$\hat{H}{\phantom{\rule{thinmathspace}{0ex}}}_{B\phantom{\rule{thinmathspace}{0ex}}S\phantom{\rule{thinmathspace}{0ex}}O\phantom{\rule{thinmathspace}{0ex}}R(\omega )\phantom{\rule{thinmathspace}{0ex}}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{D}^{-1}H{\phantom{\rule{thinmathspace}{0ex}}}_{B\phantom{\rule{thinmathspace}{0ex}}S\phantom{\rule{thinmathspace}{0ex}}O\phantom{\rule{thinmathspace}{0ex}}R(\omega )\phantom{\rule{thinmathspace}{0ex}}}D.$$(39)
from (6). Since $\hat{A}={D}^{-1}AD\notin {\mathcal{R}}_{n}^{\pi}$ and *D* is a positive diagonal matrix, it follows from Definition 2.9 and Definition 2.10 that $A\notin {\mathcal{R}}_{n}^{\pi}$. 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 $A\notin {\mathcal{R}}_{n}^{\pi}$. This completes the proof.□

*Let* $A=I-L-U\in {H}_{n}^{W}$ *be 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 if $A\notin {\mathcal{R}}_{n}^{\pi}$.

From (5), (6), (7), (38) and (39),
$${\hat{H}}_{SSOR\left(\omega \right)}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{D}^{-1}\phantom{\rule{thinmathspace}{0ex}}\left[{H}_{BSOR\left(\omega \right)}{H}_{FSOR\left(\omega \right)}\right]D\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{D}^{-1}\phantom{\rule{thinmathspace}{0ex}}{H}_{SSOR\left(\omega \right)}\phantom{\rule{thinmathspace}{0ex}}D.$$(40)

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 $A\notin {\mathcal{R}}_{n}^{\pi}$. □

*Let* $A=I-L-U=({a}_{ij})\in {H}_{n}^{W}$ *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*.

If $A\in {H}_{n}^{W}$ is irreducible, it follows from Theorem 3.9 that the conclusion of this theorem is true. If $A\in {H}_{n}^{W}$ is reducible, since $A\in {H}_{n}^{W}$ 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 ${H}_{FSOR}^{{R}_{ii}}$ and ${H}_{BSOR}^{{R}_{ii}}$ denote the FSOR- and BSOR-iteration matrices associated with diagonal square block *R*_{ii}. Direct computations give
$$\begin{array}{rl}& \rho \left({H}_{FSOR}\right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\underset{1\le i\le s}{max}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\rho \left({H}_{FSOR}^{{R}_{ii}}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\\ & \rho \left({H}_{BSOR}\right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\underset{1\le i\le s}{max}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\rho \left({H}_{BSOR}^{{R}_{ii}}\right).\end{array}$$(41)

Since *R*_{ii} ∈ *GDE*_{n} is irreducible, Theorem 3.9 shows that for *ω* ∈ (0, 1), if $\rho ({H}_{FSOR(\omega )})=\underset{i\le i\le s}{max}\phantom{\rule{thinmathspace}{0ex}}\rho ({H}_{FSOR}^{{R}_{ii}})<1$ and $\rho ({H}_{BSOR(\omega )})=\underset{i\le i\le s}{max}\phantom{\rule{thinmathspace}{0ex}}\rho ({H}_{BSOR}^{{R}_{ii}})<1$, 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 ${R}_{ii}=A({\alpha}_{i})\notin {\mathcal{R}}_{|{a}_{i}|}^{\pi}$. Again, *R*_{ii} ∈ *GDE*_{n} but ${R}_{ii}\notin {\mathcal{R}}_{|{a}_{i}|}^{\pi}$. 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 $\rho ({H}_{FSOR(\omega )}^{{R}_{jj}})=\underset{i\le i\le s}{max}\phantom{\rule{thinmathspace}{0ex}}\rho ({H}_{FSOR}^{{R}_{ii}})<1$ and $\rho ({H}_{BSOR(\omega )}^{{R}_{jj}})=\underset{i\le i\le s}{max}\phantom{\rule{thinmathspace}{0ex}}\rho ({H}_{BSOR}^{{R}_{ii}})<1$. 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 $A\in {H}_{n}^{W}$, 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 ${R}_{ii}=A({\alpha}_{i})\phantom{\rule{thinmathspace}{0ex}}\notin {\mathcal{R}}_{|{\alpha}_{i}|}^{\pi}$. Then it follows from Theorem 3. 1, Theorem 3.2 and Theorem 3.9 that $\rho ({H}_{FSOR(\omega )})=\underset{1<i<s}{max}\phantom{\rule{thinmathspace}{0ex}}\rho ({H}_{FSOR}^{{R}_{ii}})<1$ and $\rho ({H}_{BSOR(\omega )})=\underset{1<i<s}{max}\phantom{\rule{thinmathspace}{0ex}}\rho ({H}_{BSOR}^{{R}_{ii}})<1$, 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. □

$A=I-L-U=({a}_{ij})\in {H}_{n}^{W}$ *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*.

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. □

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