Partial Solutions to Homework IV
Y. Zhou
Section 4.5 (16)
Proof.
kA−1 − Bk
= kA−1 (I − AB)k ≤ kA−1 kkI − ABk = kA−1 kkAB − Ik
∞
X
(I − AB)k k (using Theorem 2 in page 200)
≤ kB
k=0
∞
X
≤ kBkk
(I − AB)k k ≤ kBk
k=0
≤ kBk
∞
X
∞
X
k(I − AB)k k
k=0
kI − ABkk = kBk
k=0
∞
X
k
k=0
≤ kBk
1−
(21) In order to show that 2B − BAB approximates A−1 closer than B, we need to prove that
kI − A(2B − BAB)k ≤ kI − AB|.
Indeed,
kI − A(2B − BAB)k
= kI − 2AB − ABABk = k(I − AB)2 k
≤
kI − ABk2
≤
kI − ABk
since kI − ABk < 1.
Section 4.6 (2)
Proof. The iteration matrix of the Richardson method is (I − A), thus we only need to show that
ρ(I − A) < 1 for A with given property.
Since
n
X
kaij k,
aii = 1 >
j=1,j6=i
(I − A) has zeros in its main diagonal and
kI − Ak∞ = max
1≤i≤n
n
X
By definition, ρ(I − A) = inf kI − Ak ≤ kI − Ak∞ < 1.
k·k
|aij | < 1.
j=1
(21) ρ(AB) ≤ ρ(A)ρ(B) may not true for the all pairs of n × n matrix since the infinimums of
kAk, kBk and kABk may not be obtained at the same type of norm.
If both A and B are upper triangular matrices, then all the eigenvalues of A and B are located
in their main diagonals, and the eigenvalue of AB are the products of the eigenvalues of A and B,
i.e.,
λi (AB) = λi (A) · λi (B),
and thus
max ikλi (AB)k = max ikλi (AB)k ≤ max ikλi (A)k · max ikλi (B)k,
suggesting that ρ(AB) ≤ ρ(A)ρ(B).
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