Page 35 - Compact Numerical Methods For Computers

Formal problems in linear algebra 25
In other words, we have
+
A A = 1 . (2.34)
n
When A has only k linearly independent columns, it will be satisfactory if

+
A A =
(2.35)

but in this case x is not defined uniquely since it can contain arbitrary components
from the orthogonal complement of the space spanned by the columns of A. That
is, we have
+ +
x = A b + (1 – A A) g (2.36)
n
where g is any vector of order n.
The normal equations (2.22) must still be satisfied. Thus in the full-rank case, it
is straightforward to identify
+ T -l T
A = (A A) A . (2.37)
In the rank-deficient case, the normal equations (2.22) imply by substitution of
(2.36) that
T T + T T +
A Ax = A AA b+(A A – A AA A)g (2.38)
T
= A b .
If
T
A AA + = A T (2.39)
+
then equation (2.38) is obviously true. By requiring A to satisfy
+
AA A = A (2.40)
and
+ T +
(AA ) = AA (2.41)
this can indeed be made to happen. The proposed solution (2.36) is therefore a
+
least-squares solution under the conditions (2.40) and (2.41) on A . In order that
T
(2.36) gives the minimum-length least-squares solution, it is necessary that x x be
minimal also. But from equation (2.36) we find
T T + T + T + T + T + T +
x x = b (A ) A b + g (1 – A A) (1 – A A)g + 2g (1 – A A) A b (2.42)
which can be seen to have a minimum at
g = 0 (2.43)
if
+
(1 – A A) T