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中科院数值线性代数典型例题复习题（含答案）.pdf
5Œ5Œ6SK
'Œ˜:(K)
1. A ∈ Cn×n, K
∞
(1)m^v, k
k=0
Ak´æ ⇔ lim
k→∞ Ak = 0
∞
k=0
Ak = (I − A)−1
(1)
(2)
dØu´æ?Œ', keªO, =:3 Cn×n NŒ · , ƒ:
(I − A)−1 − m
k=0
Ak ≤ A m+1
1− A
(3)
∞
k=0
y†. (1)⇒: e
k
= n > N + 1
Ak ´æ, dCauchy´æOKk ∀ > 0, ∃N > 0, n > N ,
An+1 < ,
An < .
k→∞ Ak = 0.
ˇd,
lim
m
k→∞ Ak = 0, K ρ(A) < 1, ˇd3,NŒ · , ƒ A < 1.
⇐: ˇ lim
- Sm =
Ak, @o
k=0
Sm = I + A + A2 + ··· + Am
≤ I + A +··· + A m
1− A m+1
1− A
1− A
1
=
≤
1
(I − A)Sm = I + A + A2 + ··· + Am − A − A2 − ··· − Am+1 = I − Am+1
∞
k=0
Ak ´æ.
K Sm k., =
(2)ˇLOk
- m → ∞ k
=
@o
(I − A)−1 − m
∞
k=0
≤ A m+1
A k=
k=0
(I − A)
Ak = I,
(I − A)−1 =
Ak.
∞
k=0
∞
∞
k=0
k=m+1
A m+1
1− A
Ak =
Ak ≤ A m+1
(4)
(5)
∞
k=0
Ak
2. A B 'O m × n, n × m, K
(1) AB BA k()›Œ)"A;
(2) A, B n , _, K AB BA q.
y†. m ≤ n, K
λmdet(
) = λmdet(
Im A
B λIn
λmdet(
Im A
B λIn
)
)
Im A
B λIn
0
Im
−B In
Im
0 λIn − BA
A
)
= λmdet(
= λmdet(λIn − BA)
) = det(
λIm −A
0
In
λIm − AB 0
λIn
B
Im A
B λIn
)
= det(
= λndet(λIm − AB)
2
=, det(λIn − BA) = λn−mdet(λIm − AB).
ˇd, AB Aı“ BA Aı“"k›Œ.
(2) A _, K A−1(AB)A = BA, = AB BA q.
3.(1) A n × n Hermite, K
x∗Ax
x∗x
λ1 = max
x=0
x∗Ax
x∗x
, λn = min
x=0
(6)
(2) A, B n Hermite, B ‰,P µ1 ≥ ··· ≥ µn A ’u B Ø
A, = µi det(A − µB) , K
x∗Ax
x∗Bx
, µn = min
x=0
x∗Ax
x∗Bx
µ1 = max
x=0
(7)
.
.
y†. (1)ˇ A ·Hermite, ˇd3j Q, ƒ A = Q∗ΛQ, ¥ Λ =
diag (λ1, λ2,··· , λn), λ1 ≥ λ2 ≥ ··· ≥ λn. Ø?¿ x = 0, - y = Qx, d y = 0.
x∗Ax
x∗x
=
x∗Q∗ΛQx
x∗Q∗Qx
=
y∗Λy
y∗y
=
n
n
i=1 |yi|2
i=1 |yi|2 ≤
n
n
i=1 λi|yi|2
i=1 |yi|2 =
λn
x∗Ax
x∗x
⁄–
λn =
n
n
i=1 λi|yi|2
i=1 |yi|2 ,
n
n
i=1 |yi|2
i=1 |yi|2 = λ1.
≤ λ1
x1 A ØAu λ1 A, xn A ØAu λn A,
x∗
1Ax1
x∗
1x1
λ1x∗
1x1
x∗
1x1
=
(2)ˇ B ‰, det(B− 1
λ1 = max
x=0
2 ) = 0, K
x∗
nAxn
x∗
nxn
λnx∗
nxn
x∗
nxn
=
= λn,
, λn = min
x=0
x∗Ax
x∗x
.
= λ1,
x∗Ax
x∗x
2 ) = det(B− 1
2 AB− 1
2 − µI),
2 ) , = µ1, µ2,··· , µn B− 1
2 AB− 1
2 ⁄kA.Ø
det(B− 1
µi det(µI−B− 1
?¿ x = 0, - y = B 1
x∗Ax
x∗Bx
2 )det(A − µB)det(B− 1
2 AB− 1
2 x, d y = 0.
2 B− 1
x∗B 1
2 AB− 1
2 B 1
2 x
x∗B 1
=
2 B 1
2 x
y∗B− 1
2 AB− 1
2 y
y∗y
,
=
3
K
λ1 = max
y=0
y∗B− 1
2 AB− 1
2 y
y∗y
x∗Ax
x∗Bx
= max
x=0
, λn = min
y=0
y∗B− 1
2 AB− 1
2 y
y∗y
x∗Ax
x∗Bx
.
min
x=0
4. A, B n Hermite‰,K
κ(A + B) ≤ max{κ(A), κ(B)},
max{ κ(A)
κ(B)
,
κ(B)
κ(A)
} ≤ κ(AB) ≤ κ(A)κ(B),
(8)
(9)
¥ κ(A) L« A ^Œ.
y†. (1) A, B n Hermite‰, K A + B ·Hermite‰.
λ1(C), λn(C) 'OL« C A, K λ1(A), λn(A), λ1(B), λn(B) > 0,
d1nK,
λ1(A + B) = max
x=0
λn(A + B) = min
x=0
x∗(A + B)x
x∗x
x∗(A + B)x
x∗x
≤ max
x=0
≥ min
x=0
x∗Ax
x∗x
x∗Ax
x∗x
x∗Bx
x∗x
x∗Bx
x∗x
+ max
x=0
+ min
x=0
= λ1(A) + λ1(B)
= λn(A) + λn(B),
ˇd,
|^ a
b ≤ c
d ⇒ a+c
b+d ≤ c
κ(A + B) =
λ1(A + B)
λn(A + B)
≤ λ1(A) + λ1(B)
λn(A) + λn(B)
d, a, b, c, d > 0, i.e κ(A) ≤ κ(B), K λ1(A)
.
λn(A) ≤ λ1(B)
λn(B), =
κ(A + B) =
λ1(A) + λ1(B)
λn(A) + λn(B)
≤ λ1(B)
λn(B)
= κ(B),
ii.e κ(B) < κ(A), K λ1(B)
λn(B) < λ1(A)
λn(A), =
κ(A + B) =
λ1(A) + λ1(B)
λn(A) + λn(B)
≤ λ1(A)
λn(A)
= κ(A),
n, κ(A + B) ≤ max{κ(A), κ(B)}.
(2)dŒN5,
κ(AB) = AB (AB)−1 = AB B−1A−1 ≤ A B A−1 B−1 = κ(A)κ(B),
4
κ(A) = A A−1 = ABB−1 BB−1A ≤ AB B−1 B (AB)−1 = κ(B)κ(AB)
κ(B) = B B−1 = AA−1B B−1A−1A ≤ A−1 AB (AB)−1 A = κ(A)κ(AB),
n,
max{ κ(A)
κ(B)
,
κ(B)
κ(A)
} ≤ κ(AB) ≤ κ(A)κ(B).
5. A ∈ Cn×n n Hermite, B · A k f“, (1 ≤ k ≤ n−1),
¿ A B A'O λ1 ≥ λ2 ≥ ··· ≥ λn µ1 ≥ µ2 ≥ ··· ≥ µn, K
λi ≥ µi ≥ λn−k+i, i = 1, 2,··· , k.
(10)
y†. B · A k f“,
B ∗
∗ ∗
A =
Ik
0
∈ C n×k, K B = U∗AU .
- U =
d44‰n, K3 i fm P ⊆ Ck ƒ
x∗Bx.
µi = min
x∈P,x=1
- Q = {y = U x|x ∈ P}, · Cn i fm, k
x∈P,x=1
µi = min
≤ max
dim S=i
min
y∈S,y=1
x∗Bx = min
x∈P,x=1
y∗Ay = λi.
x∗U∗AU x = min
y∈Q,y=1
y∗Ay
(11)
λi(M ) L« M 1 i A, k
λj(−A) λj(−B) = −λk−j+1(B)
- j = k − i + 1 k
−λn−k+i = λk+1−i(−A) −λi(B) = −µi.
5
=(.
6. A ∈ Cm×n
p
, `ƒ Ak ∈ Cm×n
A − Ak F = min{ A − B F : B ∈ Cm×n
(1 ≤ k ≤ p) ƒ
p
k
XJ?b‰ A σ1 ≥ σ2 ≥ ··· ≥ σn, `y:
). A ')
A = U∗
min
rank(B)≤k
= (σ2
1
2 .
k+1 + ··· + σ2
n)
Σp 0
0
0
V
}
(12)
(13)
Kk
A − B F = U∗
Σp = diag (σ1, σ2,··· , σp), σ1 ≥ σ2 ≥ ··· ≥ σp.
Σp 0
0
0
V − B F =
Σp 0
0
0
− U BV ∗ F
- C = U BV ∗,
A − B 2
F =
i=j
c2
ij +
min{m,n}
j=p+1
c2
jj +
p
j=1
(σj − cjj)2
dd,
d,
=
min
rank(B)≤k
A − B F = (σ2
k+1 + ··· + σ2
n)
1
2 .
0,
0,
σk,
0,
cij =
i = j
i = j ≥ p + 1
i = j ≤ k
k + 1 ≤ i = j ≤ p.
6
Ak = U∗
Σk 0
0
0
V.
7. A ∈ Cn×n, λ(A) = {λi}, σ(A) = {σi}, `y: XJ |λ1| ≥ |λ2| ≥ ··· ≥
|λn|,|σ1| ≥ |σ2| ≥ ··· ≥ |σn|, K
k
σi ≥ k
|λi|, k = 1, 2,··· , n
(14)
i=1
i=1
y†. k = n ,
n
i=1
σi = det(A∗A) = |det(A)|2 =
n
i=1
|λi|
k < n , A Shur')
λ1
0
λ2
. . .
U = U∗
∗
λn
U
T11 T12
0
T22
A = U∗T U = U∗
¥ U j, T n, T11 k , ˇd, T A
A .
T 2 =
, T ∗T =
T 2
11
0
∗
T 2
22
2,··· , µ2
k.
11T11 ∗
T ∗
∗
∗
.
T ∗
11T11 A µ2
KdA'‰n, T ∗
1, µ2
11T11 · T ∗T f, ˇd,
2d det(T ∗
11T11) = |det(T 2
11)|, k
k
i=1
i ≥ µ2
σ2
i .
k
µ2
i =
|λi|2,
i=1
7
@o
n, k(14)ڮ.
k
i=1
!5§|'(K)
8.5§| Ax = b, ¥
|λi|.
i=1
σi ≥ k
, b =
2
A =
−1
−1
−1 10−10 10−10
10−10 10−10
1
2(1 + 10−10)
−10−10
10−10
(1)y: x = (1010,−1, 1) ·§|), ^Œ· κ∞(A) = 2(1010 + 1) ≈
2 × 1010;
(2)y†: XJ |E| < 10−8|A|, (A + E)y = b, Kk |x − y| < 10−7|x|. øL†, =
ƒ A ^ŒØ, A 6˜7‰‹ x ªCz;
(3)‰´ D = diag (10−5, 105, 105), y†: κ∞(DAD) ≤ 5.
y†.
2
−1
−1
−1 10−10 10−10
10−10 10−10
1
, x = (1010,−1, 1) ·§|).
0
1
2
− 1
2
A−1 =
− 1
2 − 1
1010
−1
1
=
−10−10
10−10
2(1 + 10−10)
.
1
2
1
1
4(2 − 1010)
4(2 + 1010) − 1
4(2 + 1010)
4(2 − 1010)
4(2 + 1010) = 5 × 109 + 1
2.
A = 4, A−1 = 1
κ∞(A) = 4(5 × 109 + 1
(2)P = 10−8. ˇ Ax = b, (A + E)y = b, k x− y = A−1Ey, = x = (I + A−1E)y.
5¿ |E| < 10−8|A|,
4(1010 − 2) + 1
2 + 1
2) ≈ 2 × 1010.
4 × 10−10
4 × 10−20 4 × 10−20
2 + 4 × 10−10 4 × 10−10 4 × 10−10
2 + 4 × 10−10 4 × 10−10 4 × 10−10
.
8
|A−1E| ≤ 10−8|A−1||A| = 25
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