Example 1: The least squares method
Find the
least squares solution
for the system
x
1
-
x
2
= 2
x
1
+
x
2
= 4
2
x
1
+
x
2
= 8
Solution:
The
coefficient matrix
is
A =
1
-1
1
1
2
1
The least squares solution
satisfies
A
T
A
= A
T
.
We have
A
T
A =
1
1
2
-1
1
1
1
-1
1
1
2
1
=
6
2
2
3
and
A
T
=
1
1
2
-1
1
1
2
4
8
=
22
10
so that
6
2
2
3
=
22
10
We use the
Gaussian elimination
to solve
:
6
2
22
2
3
10
~
2
3
10
0
-7
-8
~
1
0
23/7
0
1
8/7
-Add the 2nd row multiplied by -3 to the 1st row
-Interchange the first two rows
-Add the 2nd row multiplied by 3/7 to the 1st row
-Divide the 1st row by 2
-Divide the 2nd row by -7
-We have:
x
1
= 23/7
x
2
= 8/7
We now know the
least squares solution
.
The residual vector (the error) is
=
-
A
=
2
4
8
-
1
-1
1
1
2
1
23/7
8/7
=
-1/7
-3/7
2/7
and ||
||
2
2
= 2/7
Go back to theory
satisfies
ATA = AT
.:
.
||22 = 2/7
