Greville's Method for Preconditioning Least Squares ... - Projects

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where C is a nonsingular matrix. Hence, we have
R(M T ) = R(A). (5.4)
For the general case, K and F are still nonsingular. However, the expression for V is
not straightforward. To simplify the problem, we make the following assumption.
Assumption 1
1. There is no zero column in A.
2. Our preconditioning algorithm can detect all the linear independence in A.
The definition of v i can be rewritten as follows.
v i =
{ ai − Ak i ∈ R(A i ) ∉ R(A i−1 ) if a i ∉ R(A i−1 )
(A † i−1 )T k i ∈ R(A i−1 ) = R(A i ) if a i ∈ R(A i−1 )
(5.5)
Hence, we have
span{v 1 , v 2 , . . . , v i } = span{a 1 , a 2 , . . . , a i }. (5.6)
On the other hand, note the 12-th line of Algorithm 1
M i = M i−1 + 1 f i
(e i − k i )v T i , (5.7)
which implies that every row of M i is a linear combination of the vectors v T k , 1 ≤ k ≤ i,
i.e.,
R(M T i ) = span{v 1 , . . . , v i }. (5.8)
Based on the above discussions, we obtain the following theorem.
Theorem 3 Let A ∈ R m×n , m ≥ n. If Assumption 1 holds, then we have the following
relationships, where M is the approximate Moore-Penrose inverse constructed by
Algorithm 1.
R(M T ) = R(V ) = R(A) (5.9)
Remark 8 In Assumption 1, we assume that our algorithms can detect all the linear
independence in the columns of A. Hence, we allow such mistakes that a linearly
dependent column is recognized as a linearly independent column. An extreme case is
that we recognize all the columns of A as linearly independent, i.e., we take A as a full
column rank matrix. In this sense, our assumption can always be satisfied.
Hence, we proved that for any matrix A ∈ R m×n , R(A) = R(M T ). We have the
following theorem.
Theorem 4 If Assumption 1 holds, then for all b ∈ R m , the preconditioned least
squares problem (5.1), where M is constructed by Algorithm 1, is equivalent to the
original least squares problem (1.1).