返回
1
11
1.,|| || 1,|| || || ||
|| || || ||,
P r,,
.
|| || || || || || || ||,|| ||,|| ||
|| ||,|| || || || || ||,
a
b
a
aa
P P A P A
A AP
oof
AB P AB P AP P B P A P P B
P A P B A B
设 可逆 且 则 或均为自相容的矩阵范数容易证明所定义的映射都是矩阵范数下面证明它们是相容的
11| | | | | | | | | | | | | | | |,| | | |,| | | |
| | | |,| | | | | | | | | | | |,
b
bb
A B A B P A P P B P A P P B P
A P B P A B
返回
2 1 2,| | | | | | | | | | | | | | | |,HA A A A A n A2,设则
2
2 m a x 1
2
1 1 1 2 1
| | | | ( ) ( ) | | | |
| | | | | | | | | | | |,| | | | | | | |,
H H H
H
A r A A A A A A
A A A A A
故
11:,| | | | | | | | | | | |HHA A A A A证明 由于 所以返回
2
2
2 m ax
2
2 2
1
22
11
22
2
22
|| || ( ) ( )
m ax | ||| ||
||
m ax | | m ax ( | |)
|| ||
,|| || || ||,
iHH
n
i ij
m ij j
nn
i ij i ij
jj
A r A A A A
n
aA
a
n n n
n a a
nn
A
A n A
n
故返回
4.1 特征值界的估计证,nnCA HU T UA?
n
i
ii
n
i
i t
1
2
1
2 ||||
n
ji
ij
n
i
ii tt
2
1
2 |||| )( TTtr H?
定理 1 (Shur不等式 ) 的特征值为设 nnCA
2
1 1
2
1
2 ||||||||
F
n
i
n
j
ij
n
i
i Aa
则,,,,21 n
.为正规矩阵且等号成立当且仅当 A
返回
HU T UA? HHH UTTUAA )(?
n
i
i
1
2||? )( TTtr H?
)()( TTtrAAtr HH?
)( AAtr H? 2|||| FA?
返回
11( ),( )
22
HHB A A C A A
},,,,{,,21 nCBA的特征值分别为且满足},,,,{},,,,{ 2121 nn iii
|,||||| 21 n
12,n
12,n
返回定理 2 (Hirsch) 的特征值为设 nnCA
则,,,2 n
,1?
|,|m a x||)1
,ijjii
an |,|m a x|Re|)2
,ijjii
bn
|,|m a x|Im|)3
,ijjii
cn
返回证:
nnCA)2 HHHH TUAUTAUU,
)(21)(21 TTUAAUBUU HHHH
)(21)(21 HHHH TTUAAUCUU
n
i
ii
1
22 ||||)1
n
i
n
j
ija
1 1
2|| 2
,
2 ||m a x
ijji an?
||m a x||
,ijjii
an
返回
n
i
n
j
ij
n
i
i
n
i
n
j
ij
n
i
i
c
b
1 1
2
1
2
1 1
2
1
2
|||Im|
|||Re|
2
,
2 ||m a x
ijji bn?
2
,
2 ||m a x
ijji cn?
n
i
n
j
ij
n
j
j
i
ij
n
i
ii
n
i
n
j
ij
n
j
j
i
ij
n
i
ii
c
t
b
t
1 1
2
1
1
1
2
1
2
1 1
2
1
1
1
2
1
2
||
2
||
|
2
|
||
2
||
|
2
|
返回
||m a x|Im|)3
,ijjii
cn
||m a x|Re|)2
,ijjii
bn
定理 3 (Bendixson) 的任一特则设 ARA nn,
满足值 i?
||m a x
2
)1(|Im|
,
ij
ji
i c
nn
返回证:
n
i
n
j
ij
n
i
i c
1 1
2
1
2 |||Im|? 2
,
||ij
ij
ij
c
2
,
||m a x)1( ij
ji
cnn
s
i
ii
1
22 |Im|2|Im|2
n
i
i
1
2|Im|?
2
,
||m a x)1( ij
ji
cnn
返回定理 4 则定义同上设,,,,,,
iiinn CBCA
11 Im,Re inin
证:
)1||( || 2 xxAx i?
),( Axx ),( xx i ),( xxi i
iH Axx iHH xAx
为正规矩阵B
Dd i a gBUU nH ),,,( 21
返回
xUDUx HH?
i?Re )2,( x
AAx H ),( Bxx? ),( xUDUx H?
Dyy H
n
i
ii y
1
2||?
n
i
ii
n
i
in yy
1
2
1
1
2 ||Re||
1Re in
返回定理 4(Browne),的特征值为设 nnCA
则奇异值为,,,,212 nn
,1?
),,2,1(|| 1 niin
证,HA A H e r m ite为 矩阵
2 2 212(,,,)HH nU A A U d i a g D
2( | | | | 1 )iA x x x H H Hix A x 2||HH ix A A x
2||HH ix U D U x
2||HH ix U D U x
返回
xBx ),,2,1(
1
nixxb i
n
t
tit
),,2,1(
1
nixxb i
n
t
tit
n
i
ii
n
ts
n
i
tsisit xxxxbb
11,1
)(
n
i
ii
n
i
iii xxxx
11
2
2122 || in
返回定理 6 (Hadamard不等式 ) 则设,nnCA
2/1
1 1
2
1
)]||([|d e t||)(|
n
j
n
i
ij
n
i
i aAA?
或的某一列全为且等号成立当且仅当,0A
.的列向量彼此正交A
证,),,,( 21 naaaA设线性相关naaa,,,)1 21? 0|d e t|?A
结论成立返回线性无关naaa,,,)2 21?正交化
11,11
23213133
12122
11
nnnnnn
bpbpba
bpbpba
bpba
ba
返回
100
10
1
),,,(
2
121
21
n
n
n
p
pp
bbbA
BbbbA n d e t),,,d e t (d e t 21
211,112 |||||||| iiiiii bpbpba?
211,2112 |||||||||||||||| iiiii bpbpb?
2|||| ib?
返回
BBB H d e td e t|d e t| 2 BB He t?
n
i
ib
1
2||||
n
i
ib
1
2||)||(
n
i
ia
1
2||)||(
2/1
1 1
2
1
)]||([|d e t||)(|
n
j
n
i
ij
n
i
i aAA?
1
11
1.,|| || 1,|| || || ||
|| || || ||,
P r,,
.
|| || || || || || || ||,|| ||,|| ||
|| ||,|| || || || || ||,
a
b
a
aa
P P A P A
A AP
oof
AB P AB P AP P B P A P P B
P A P B A B
设 可逆 且 则 或均为自相容的矩阵范数容易证明所定义的映射都是矩阵范数下面证明它们是相容的
11| | | | | | | | | | | | | | | |,| | | |,| | | |
| | | |,| | | | | | | | | | | |,
b
bb
A B A B P A P P B P A P P B P
A P B P A B
返回
2 1 2,| | | | | | | | | | | | | | | |,HA A A A A n A2,设则
2
2 m a x 1
2
1 1 1 2 1
| | | | ( ) ( ) | | | |
| | | | | | | | | | | |,| | | | | | | |,
H H H
H
A r A A A A A A
A A A A A
故
11:,| | | | | | | | | | | |HHA A A A A证明 由于 所以返回
2
2
2 m ax
2
2 2
1
22
11
22
2
22
|| || ( ) ( )
m ax | ||| ||
||
m ax | | m ax ( | |)
|| ||
,|| || || ||,
iHH
n
i ij
m ij j
nn
i ij i ij
jj
A r A A A A
n
aA
a
n n n
n a a
nn
A
A n A
n
故返回
4.1 特征值界的估计证,nnCA HU T UA?
n
i
ii
n
i
i t
1
2
1
2 ||||
n
ji
ij
n
i
ii tt
2
1
2 |||| )( TTtr H?
定理 1 (Shur不等式 ) 的特征值为设 nnCA
2
1 1
2
1
2 ||||||||
F
n
i
n
j
ij
n
i
i Aa
则,,,,21 n
.为正规矩阵且等号成立当且仅当 A
返回
HU T UA? HHH UTTUAA )(?
n
i
i
1
2||? )( TTtr H?
)()( TTtrAAtr HH?
)( AAtr H? 2|||| FA?
返回
11( ),( )
22
HHB A A C A A
},,,,{,,21 nCBA的特征值分别为且满足},,,,{},,,,{ 2121 nn iii
|,||||| 21 n
12,n
12,n
返回定理 2 (Hirsch) 的特征值为设 nnCA
则,,,2 n
,1?
|,|m a x||)1
,ijjii
an |,|m a x|Re|)2
,ijjii
bn
|,|m a x|Im|)3
,ijjii
cn
返回证:
nnCA)2 HHHH TUAUTAUU,
)(21)(21 TTUAAUBUU HHHH
)(21)(21 HHHH TTUAAUCUU
n
i
ii
1
22 ||||)1
n
i
n
j
ija
1 1
2|| 2
,
2 ||m a x
ijji an?
||m a x||
,ijjii
an
返回
n
i
n
j
ij
n
i
i
n
i
n
j
ij
n
i
i
c
b
1 1
2
1
2
1 1
2
1
2
|||Im|
|||Re|
2
,
2 ||m a x
ijji bn?
2
,
2 ||m a x
ijji cn?
n
i
n
j
ij
n
j
j
i
ij
n
i
ii
n
i
n
j
ij
n
j
j
i
ij
n
i
ii
c
t
b
t
1 1
2
1
1
1
2
1
2
1 1
2
1
1
1
2
1
2
||
2
||
|
2
|
||
2
||
|
2
|
返回
||m a x|Im|)3
,ijjii
cn
||m a x|Re|)2
,ijjii
bn
定理 3 (Bendixson) 的任一特则设 ARA nn,
满足值 i?
||m a x
2
)1(|Im|
,
ij
ji
i c
nn
返回证:
n
i
n
j
ij
n
i
i c
1 1
2
1
2 |||Im|? 2
,
||ij
ij
ij
c
2
,
||m a x)1( ij
ji
cnn
s
i
ii
1
22 |Im|2|Im|2
n
i
i
1
2|Im|?
2
,
||m a x)1( ij
ji
cnn
返回定理 4 则定义同上设,,,,,,
iiinn CBCA
11 Im,Re inin
证:
)1||( || 2 xxAx i?
),( Axx ),( xx i ),( xxi i
iH Axx iHH xAx
为正规矩阵B
Dd i a gBUU nH ),,,( 21
返回
xUDUx HH?
i?Re )2,( x
AAx H ),( Bxx? ),( xUDUx H?
Dyy H
n
i
ii y
1
2||?
n
i
ii
n
i
in yy
1
2
1
1
2 ||Re||
1Re in
返回定理 4(Browne),的特征值为设 nnCA
则奇异值为,,,,212 nn
,1?
),,2,1(|| 1 niin
证,HA A H e r m ite为 矩阵
2 2 212(,,,)HH nU A A U d i a g D
2( | | | | 1 )iA x x x H H Hix A x 2||HH ix A A x
2||HH ix U D U x
2||HH ix U D U x
返回
xBx ),,2,1(
1
nixxb i
n
t
tit
),,2,1(
1
nixxb i
n
t
tit
n
i
ii
n
ts
n
i
tsisit xxxxbb
11,1
)(
n
i
ii
n
i
iii xxxx
11
2
2122 || in
返回定理 6 (Hadamard不等式 ) 则设,nnCA
2/1
1 1
2
1
)]||([|d e t||)(|
n
j
n
i
ij
n
i
i aAA?
或的某一列全为且等号成立当且仅当,0A
.的列向量彼此正交A
证,),,,( 21 naaaA设线性相关naaa,,,)1 21? 0|d e t|?A
结论成立返回线性无关naaa,,,)2 21?正交化
11,11
23213133
12122
11
nnnnnn
bpbpba
bpbpba
bpba
ba
返回
100
10
1
),,,(
2
121
21
n
n
n
p
pp
bbbA
BbbbA n d e t),,,d e t (d e t 21
211,112 |||||||| iiiiii bpbpba?
211,2112 |||||||||||||||| iiiii bpbpb?
2|||| ib?
返回
BBB H d e td e t|d e t| 2 BB He t?
n
i
ib
1
2||||
n
i
ib
1
2||)||(
n
i
ia
1
2||)||(
2/1
1 1
2
1
)]||([|d e t||)(|
n
j
n
i
ij
n
i
i aAA?