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Department of Mathematics Fall 2007 MATH 6370, Section 10583 MW 5:30-7:00PM A. Caboussat
Answers Homework Chapter 8
HOMEWORK FROM THE TEXT
Problem 1 (Chapter 8 - Problem 1) 8 1
Assume limi→∞ Ai^ = 0. Therefore
lim i→∞
Aix = 0, ∀x ∈ Rn.
Define ϕk^ the eigenvector of A associated to the eigenvalue λk. Thus:
0 = lim i→∞ Aiϕk = lim i→∞ λik ϕk, ∀k = 1,... , n.
Since ||ϕk|| = 1, this implies |λk| < 1 in order for the sequence to converge. Reciprocally, assume ρ(A) < 1. Let ε > 0 such that ρ(A) + ε < 1. Therefore, there exists a norm ||·||⋆ such that:
|||A|||⋆ 6 ρ(A) + ε < 1 ,
where |||·|||⋆ is the induced matrix norm. Therefore, for all x ∈ Rn, x 6 = 0:
0 6 lim i→∞
∣Aix
⋆ (^6) ilim→∞ |||A|||
i ⋆ ||x||⋆ 6 (ρ(A) +^ ε)
i (^) ||x|| ⋆ = 0.
Therefore limi→∞ |||A|||i⋆ = 0 and, since |||·|||⋆ is a norm, limi→∞ Ai^ = 0.
ADDITIONAL HOMEWORK
Problem 1 15
First show that the series
k=0 A
k (^) converges, then ρ(A) < 1. If the series ∑∞ k=0 A
k (^) converges, then the sequence
of the difference between partial sums Sn =
∑n k=0 A
k (^) converges to zero:
lim n→∞ Sn − Sn− 1 = lim n→∞ An^ = 0.
If limn→∞ An^ = 0, then we know that ρ(A) < 1 from the previous exercise. Reciprocally, if ρ(A) < 1, then 0 does not belong to the eigenvalues of I − A and I − A is nonsingular and therefore invertible. Moreover:
I − Ak+1^ = (I − A)(I + A + A^2 +... + Ak)
and therefore (I − A)−^1 (I − Ak+1) = (I + A + A^2 +... + Ak).
Under the assumption ρ(A) < 1, we have limk→∞Ak^ = 0. By taking the limit of the previous relation, we obtain the conclusion.
Problem 3 17
Let A, M ∈ MN and set N = M − A. To solve numerically the linear system A~x = ~b, we use the iterative method:
M~xi+1 = N~xi + ~b, i = 0, 1 , 2 ,... , ~x 0 given. let us assume that both A and M + N T^ are symmetric positive definite.
3.a) Starting from the definition, we have M + N T^ = M + M T^ − A. Therefore, ∀x ∈ RN^ ,
xT^ (M + N T^ )x = 2xT^ M x − xT^ Ax.
If x 6 = 0, 2 xT^ M x = xT^ (M + N T^ )x + xT^ Ax > 0. (1)
Therefore M is positive definite (not necessarily symmetric!), and then invertible.
3.b) Let us remark that, if A = M − N then M + N T^ = M + M T^ − A. Since A is symmetric, M + N T^ is symmetric. Let λ ∈ C an eigenvalue of M −^1 N and x ∈ CN^ the corresponding eigenvector. We have
M −^1 N x = λx ⇒ N x = λM x ⇒ (M − N )x = (1 − λ)M x ⇒ Ax = (1 − λ)M x.
Therefore x⋆Ax = (1 − λ)x⋆^ M x and
(1 − λ¯)x⋆Ax = | 1 − λ|^2 x⋆M x. (2) By taking the hermitian conjugate of (2), we obtain
(1 − λ)x⋆Ax = | 1 − λ| 2 x⋆M T^ x. (3)
By adding (2) and (3),
(2 − λ¯ − λ)x⋆^ Ax = | 1 − λ|^2 x⋆(M + M T^ )x (2 − ¯λ − λ − | 1 − λ|^2 )x⋆Ax = | 1 − λ|^2 x⋆(M + M T^ − A)x.
Since A and M + N T^ are positive definite, we have
(1 − |λ|^2 )x⋆Ax = | 1 − λ|^2 x⋆(M + N T^ )x.
Here we distinguish two cases:
- If λ 6 = 1, then | 1 − λ|^2 > 0, and thus 1 − |λ|^2 > 0 and |λ| < 1.
- If λ = 1, then Ax = 0, and leads to a contradiction.
