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

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An appropriate switching tolerance τ s is important to the efficiency of the preconditioner.
When we choose a small τ s, e.g. 10 −8 , we have less chance to detect the
linearly dependent columns. On the other hand, when a relatively large τ s, e.g. 10 −3 , is
chosen, the preconditioning algorithm tends to find more linearly dependent columns.
When a large τ s is taken, many linearly dependent columns can be found. However, it
is also possible that Assumption 1 is violated so that the original least squares problem
can not be solved. When a relatively small τ s is taken, it is possible that less or no
linearly dependent columns can be found. Hence, the preconditioner can loose some
efficiency. When τ s is fixed, a larger dropping tolerance τ d results in less possibility
to detect linearly dependent columns, and a smaller τ d results in larger possibility to
detect linearly dependent columns. Choosing optimal τ s and the dropping tolerance τ d
to achieve the balance between them remains a problem.
Another issue related with the dropping tolerance τ d is how much memory is needed
to store the preconditioner. From the previous discussion, we know that R(M T ) =
span{V }. When numerical droppings is performed on V , the relation may not hold.
Hence, in Algorithm 2 we only perform numerical droppings on the columns of K. In
this way, as long as Assumption 1 holds, we have R(M T ) = span{V }. However, on
the other hand, we cannot control the sparsity of V directly. We will discuss this issue
again at the end of this section.
7.2 Right-Preconditioning
So far we assumed A ∈ R m×n , m ≥ n, and only discussed left-preconditioning. When
m ≥ n, it is better to perform left-preconditioning since the size of the preconditioned
problem will be smaller. When m ≤ n, right-preconditioning will be better, i.e, solving
the preconditioned problem
min ‖b − ABy‖ 2. (7.17)
y∈R m
When m ≤ n, Theorem 3 and Theorem 5 still hold, since in the proof of these two
theorems we did not refer to the fact that m ≥ n. Then, according to the following
lemma from [14], it is easy to obtain similar conclusions for right-preconditioning.
Lemma 2 min ‖b − Ax‖ x∈R n 2 = min ‖b − ABy‖ y∈R m 2 holds for any b ∈ R m if and only if
R(A) = R(AB).
Theorem 9 Let A ∈ R m×n . If Assumption 1 holds, then for any b ∈ R m we have
min ‖b − Ax‖ x∈R n 2 = min ‖b − AMy‖ 2, where M is a preconditioner constructed by Algorithm
y∈R m
1.
Proof From Theorem 5, we know that R(M) = R(A T ), which implies that there exists
a nonsingular matrix C ∈ R m×m such that M = A T C. Hence,
R(AM) = R(AA T C) (7.18)
= R(AA T ) (7.19)
= R(A). (7.20)
Using Theorem 2 we complete the proof.
✷