Page 170 - Matrices theory and applications

9.3. Two Cases of Convergence
Proof
If the method is convergent, we have ρ(L ω ) < 1. However,
det((1 − ω)D)
det((1 − ω)D + ωF)
n
=
det L ω =
det(D − ωE)
Hence
1/n det D =(1 − ω) . 153
ρ(L ω ) ≥| det L ω | = |1 − ω|.
9.3 Two Cases of Convergence
In this section and the following one we show that simple and natural
hypotheses on A imply the convergence of the classical methods. We also
compare their efficiencies.
9.3.1 The Diagonally Dominant Case
We assume here that one of the following two properties is satisfied:
1. A is strictly diagonally dominant,
2. A is irreducible and strongly diagonally dominant.

Proposition 9.3.1 Under one or the other of the hypotheses (1) and (2),
the Jacobi method converges, as well as the relaxation method, with ω ∈
(0, 1].
Proof
Jacobi method: The matrix J = D −1 (E + F) is clearly irreducible if A
is. Furthermore,
n

|J ij |≤ 1, i =1,... ,n,
j=1
in which all inequalities are strict if (1) holds, and at least one in-
equality is strict under the hypothesis (2). Then either Gershgorin’s
theorem (Theorem 4.5.1) or its improvement, Proposition 4.5.2 for
irreducible matrices, yields ρ(J) < 1.
Relaxation method: We assume that ω ∈ (0, 1]. Let λ ∈ CC be a nonzero
eigenvalue of L ω . It is a root of
det((1 − ω − λ)D + λωE + ωF)= 0.

Hence, λ+ω−1 is an eigenvalue of A := ωD −1 (λE+F). This matrix
is irreducible when A is. Then Gershgorin’s theorem (Theorem 4.5.1)