Page 67 -

2. Two-Point Boundary Value Problems
54
Diagonal dominant matrices occur frequently in numerical analysis.
Example 2.6 The matrix given by (2.14), derived in the previous section,
is diagonal dominant. This follows since the desired inequality holds with
equality for all rows, except for the first and the last, while we have strict
inequality in these two rows.
Lemma 2.1 Assume that the coefficient matrix A of the triangular system
(2.19) is diagonal dominant and that β k =0 for k =2, 3,... ,n. Then the
variables δ k ,k =1, 2,... ,n determined by Algorithm 2.1 are well defined
and nonzero.
Proof: We prove by induction that
|δ k | > |γ k | for k =1, 2,... ,n.
By assumption this holds for k = 1. Assume now
|δ k−1 | > |γ k−1 | for some k such that 2 ≤ k ≤ n.
Since δ k−1 =0, m k , and hence δ k , is well defined and
β k
δ k = α k − δ k−1 γ k−1 .
By the induction hypothesis |γ k−1 /δ k−1 | < 1, and hence, since β k =0,
|β k || γ k−1 | < |β k |.
δ k−1
Therefore, by the triangle inequality and since the system is diagonal dom-
inant we obtain
|δ k |≥|α k |−|β k || γ k−1 | > |α k |−|β k |≥|γ k | .
δ k−1

Assume that the system (2.19) satisfies the assumptions given in Lemma
2.1 above. Then, if the vector b = 0, also the vector c = 0, and hence,
by tracking the system (2.23) backwards, the unique solution of (2.23) is
v = 0. However, since the systems (2.19) and (2.23) are equivalent, this
means that v = 0 is the only solution of (2.19) when b = 0. Hence, A is
nonsingular. We have therefore obtained the following result:
Proposition 2.3 Assume that the coefficient matrix A of (2.19) satisfies
the properties specified in Proposition 2.1 above. Then, the system has a
unique solution which can be computed by Algorithm 2.1.

As a direct consequence of this proposition, and the result of Example
2.6, we reach the following result: