工程矩阵理论第章矩阵的广义逆

(3) R(A) = R(AA+) = R(AAH) = K(IAA+);
证明: X R(A) Y n s.t. X = AY X = AA+AY R(AA+),
AH(AAH)+ = AH(AH)+A+ = AH(A+)HA+ = …
(8) 利用定理4.2.6 (奇值分解), (9)() A+AB = A+AC
AB = AA+AB = AA+AC = AC.
第六章 矩阵的广义逆
定理6.1.3 设A
sn, 则
§6.1 广义逆及其性质
(1) AA+X =
X, 0,
(4) (GA)H = GA,
则称G为A的广义逆
(或Moore-Penrose逆, 简称MP-逆).
第六章 矩阵的广义逆
§6.1 广义逆及其性质
二. 存在性与唯一性
定理6.1.1 设A sn, 则A有唯一的广义逆. 证明: (存在性) 根据定理4.2.6 (奇值分解),
存在酉矩阵U与V使得
A=U
其中k
, k+ =
k1, k 0, 0, k = 0;
证明: 根据Penrose方程直接验证.
第六章 矩阵的广义逆
§6.1 广义逆及其性质
(5) AH = AHAA+ = A+AAH; (6) (AHA)+ = A+(AH)+,
(AAH)+ = (AH)+A+;
证明: (5) AHAA+ = AH(AA+)H = (AA+A)H = AH. A+AAH = (A+A)HAH = (AA+A)H = AH.
X R(A), X K(AH);
证明: X R(A) Y n s.t. X = AY
AA+X = AA+AY = AY = X.
X K(AH) AHX = 0 AA+X = (AA+)HX = (A+)HAHX
= 0.
第六章 矩阵的广义逆
§6.1 广义逆及其性质
(2) A+AX =
= A+(AA+)HAA+(AH)+ = A+AA+AA+(AH)+ = A+AA+(AH)+ = A+(AH)+;
[(AHA)A+(AH)+]H = [(AH)+]H(A+)HAH(AH)H
= A+(A+)HAHA = A+(AA+)HA = A+AA+A = A+A = (A+A)H = [A+(AA+)A]H = AH(AA+)H(A+)H
(6) 利用定理4.2.6(奇值分解),
或根据Penrose方程直接验证.
(AHA)A+(AH)+(AHA) = AHAA+(A+)HAHA
= AHAA+(AA+)HA = AHAA+AA+A = AHAA+A = AHA;
第六章 矩阵的广义逆
§6.1 广义逆及其性质
A+(AH)+(AHA)A+(AH)+ = A+(A+)HAHAA+(AH)+
= (YAXA)HY = (YA)HY = YAY = Y.
第六章 矩阵的广义逆
§6.1 广义逆及其性质
注: A的广义逆记为A+.
例1 (1) 若A为可逆阵, 则A+ = A1.
(2) O+ = OT.
例2
(1)
AO OB
+
=
A+ O
O B+
,
OA BO
+
=
O A+
B+ O
.
(2)
A O
+
= (A+, O),
X, 0,
X R(AH), X K(A);
证明: X R(AH) Y s s.t. X = AHY A+AX = A+AAHY = (A+A)HAHY = (AA+A)HY = AHY = X.
X K(A) AX = 0 A+AX = 0.
第六章 矩阵的广义逆
§6.1 广义逆及其性质
故B+ =
1/2 1/2
,
A+ =
B O
+ = (B+, O) =
1/2 1/2
0 0
.
第六章 矩阵的广义逆
§6.1 广义逆及其性质
三. A+的性质
定理6.1.2 设A sn, 则
(1) (A+)+ = A;
(2) (AH)+ = (A+)H;
(3) (AT)+ = (A+)T;
(4) (kA)+ = k+A+,
(A, O)+
=
A+ O
.
第六章 矩阵的广义逆
例3
设A =
1 0
1 0
, 求A+.
解: 令B = (1, 1), B+ =
x y
,
§6.1 广义逆及其性质
则 (x+y)(1, 1) = BB+B = B = (1, 1),
x x
y y
= (B+B)H = B+B =
x y
x y
,
由此可得x = y = 1/2.
第六章 矩阵的广义逆
第一节 广义逆及其性质
第二节 A+的求法
第三节 广义逆的一个 应用
第六章 矩阵的广义逆
§6.1 广义逆及其性质
§6.1 广义逆及其性质
一. Penrose方程与MP-逆
定义6.1.1 设A sn. 若存在G ns满足
(1) AGA = A;
(2) GAG = G; (3) (AG)H = AG; Penrose方程
(3) (AX)H = AX, (AY)H = AY;
(4) (XA)H = XA, (YA)H = YA,
则X = XAX = X(AX)H = XXHAH
= XXH(AYA)H = XXHAH(AY)H
= X(AX)H(AY)H = XAXAY = XAY
= XAYAY = (XA)H(YA)HY
D O
O O
VH,
其中D = diag(1, …, r),
1, …, r > 0为AHA的特征值.
令G = V
D1 O
O O
UH,
ns
则可直接验证G为A的广义逆.
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§6.1 广义逆及其性质
(唯一性) 设X, Y满足
(1) AXA = A = AYA;
(2) XAX = X, YAY = Y;
第六章 矩阵的广义逆
§6.1 广义逆及其性质
(7) A+ = (AHA)+AH = AH(AAH)+; (8) (UAV)+ = VHA+UH,
其中U, V为酉矩阵; (9) A+AB = A+AC AB = AC.
证明: (7) (AHA)+AH = A+(AH)+AH = A+(A+)HAH = A+(AA+)H = A+AA+ = A+.
= AHAA+(A+)H = (AHA)A+(AH)+; [A+(AH)+(AHA)]H = AH(AH)H[(AH)+]H(A+)H
= AHAA+(A+)H = AH(AA+)H(A+)H = [A+(AA+)A]H = (A+A)H = A+A = A+(AA+)A = A+(AA+)HA = A+(A+)HAHA = A+(AH)+(AHA).