46 3 Orthogonality
w
Fig. 3.8. Least norm squared solution: (ATA) is not invertible and A^ = QiE'Qj
Table 3.1. How to solve Ax = b, where A G Rmx"
Caseb e -£(4)b e M(AT)b <jL 11(A)
bgAf(A)Subcaser=n=mr=m < n
A=[B\N]r=m
[A\\b
B\N
or <A^ -< n6'
0m
I\N~\
O \(ATA):
invertible
(ATA):
not
invertibleSolutionx=A~^1 bXB =
B_16-
B~^1 NxnXB =
B-'b-B^NXnmanyx=A*bmany
x=A*b
min.normTypeExact
uniqueExact
manyExactmanyTrivialLeastSquares
Unique
Least
Squares
Least
Norm
SquaresSpecial Forms
A=LU => Lc=b,Ux=c
A=QR => Rx=QTb
A = QtEQl =»
x=Q 2 Z-xQlb
B=LU => Lc=b,UxB=c
B=QR => RxB=QTb
B=Q-,EQl =>
xB=Q2S-^1 Qjb
B=LU =• Lc=b,UxB-c
B=QR => RxB=QTbB=Q 1 EQ 2 r =>xB=Q2Z~^1 QTbT
x=aVaG' -N'
I
M.n~r)A=QR => Rx=QTb
A = QxEQl =>
x=Q 2 E-lQjbA^QiEQZ =*
x=Q 2 E1iQjbInverseA^A-^1A^B-^1A* « B-^1noneA*=
(ATA)~^1 ATA*=
Q 2 E^Ql