Numerical Methods Contents - SAM

✬
✩
➣
➤
Matrix A ∈ R n,n arising from nodal analysis satisfies
A is s.p.d., see Lemma 2.7.4 below.
• A = A T , a kk > 0 , a kj ≤ 0 for k ≠ j , (2.7.1)
n∑
• a kj ≥ 0 , k = 1,...,n , (2.7.2)
j=1
• A is regular. (2.7.3)
All these properties are obvious except for the fact that A is regular.
Proof of (2.7.3): By Thm. 2.0.3 it suffices to show that the nullspace of A is trivial: Ax = 0 ⇒ x =
0.
Pick x ∈ R n , Ax = 0, and i ∈ {1,...,n} so that
|x i | = max{|x j |, j = 1,...,n} .
Intermediate goal: show that all entries of x are the same
Ax = 0 ⇒
x i = ∑ j≠i
By (2.7.2) and the sign condition from (2.7.1) we conclude
∑
j≠i
a ij
x
a j ⇒ |x i | ≤ ∑ |a ij |
ii
j≠i
Ôº½½ ¾º
|a ii | |x j| . (2.7.4)
|a ij |
|a ii | ≤ 1 . (2.7.5)
Hence, (2.7.5) combined with the above estimate (2.7.4) that tells us that the maximum is smaller
equal than a mean implies |x j | = |x i | for all j = 1, ...,n. Finally, the sign condition a kj ≤ 0 for
k ≠ j enforces the same sign of all x i .
✸
Lemma 2.7.4. A diagonally dominant Hermitian/symmetric matrix with non-negative diagonal
entries is positive semi-definite.
A strictly diagonally dominant Hermitian/symmetric matrix with positive diagonal entries is positive
definite.
✫
Proof. For A = A H diagonally dominant, use inequality between arithmetic and geometric mean
(AGM) ab ≤ 1 2 (a2 + b 2 ):
✬
✫
n∑
x H Ax = a ii |x i | 2 + ∑ n∑
a ij¯x i x j ≥ a ii |x i | 2 − ∑ |a ij ||x i ||x j |
i=1 i≠j i=1 i≠j
n∑
≥ a ii |x i | 2 − 1 ∑
2 |a ij |(|x i | 2 + |x j | 2 )
i=1 i≠j
( n
≥ 2
1 ∑
ii |x i |
i=1{a 2 − ∑ ) (
|a ij ||x i | 2 } + 1 ∑ n
2 ii |x j |
j≠i
j=1{a 2 − ∑ )
|a ij ||x j | 2 }
i≠j
n∑
≥ |x i | 2( a ii − ∑ )
|a ij | ≥ 0 .
i=1 j≠i
Theorem 2.7.5 (Gaussian elimination for s.p.d. matrices).
Every symmetric/Hermitian positive definite matrix (→ Def. 2.7.1) possesses an LUdecomposition
(→ Sect. 2.2).
Equivalent to assertion of theorem: Gaussian elimination feasible without pivoting
In fact, this theorem is a corollary of Lemma 2.2.3, because all principal minors of an s.p.d. matrix
are s.p.d. themselves.
Sketch of alternative self-contained proof.
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Ôº½¿ ¾º
Definition 2.7.3 (Diagonally dominant matrix).
A ∈ K n,n is diagonally dominant, if
∑
∀k ∈ {1, ...,n}:
j≠k |a kj| ≤ |a kk | .
The matrix A is called strictly diagonally dominant, if
∑
∀k ∈ {1, . ..,n}:
j≠k |a kj| < |a kk | .
Ôº½¾ ¾º
Proof by induction: consider first step of elimination
( a11 b
A =
T )
1. step
b à −−−−−−−−−−−−→
Gaussian elimination
(
a11 b T )
0 Ã − bbT .
a 11
Ôº½ ¾º
➣ to show: Ã − bbT
a s.p.d. (→ step of induction argument) 11