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

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16 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. ✷
17 Theorem 10 Let A ∈ R m×n . If Assumption 1 holds, then GMRES determines a solution to the right preconditioned problem before breakdown happens. min ‖b − AMy‖ y∈R m 2 (7.21) Proof The proof is the same as Theorem 6. ✷ Theorem 11 Let A ∈ R m×n . If Assumption 1 holds, then for any b ∈ R m , GMRES determines a least squares solution of min ‖b − AMy‖ y∈R m 2 (7.22) before breakdown and this solution attains min x∈R n‖b − Ax‖ 2, where M is computed by Algorithm 1. We would like to remark that it is preferable to perform Algorithm 1 and Algorithm 2 to A T rather than A when m < n, based on the following three reasons. By doing so, we construct ˆM, an approximate generalized inverse of A T . Then, we can use ˆM T as the preconditioner to the original least squares problem. 1. In Algorithm 1 and Algorithm 2, the approximate generalized inverse is constructed row by row. Hence, we perform a loop which goes through all the columns of A once. When m ≥ n, this loop is relatively short. However, when m < n, this loop could become very long, and the preconditioning will be more time-consuming. 2. Another reason is that, linear dependence will always happen in this case even though matrix A is full row rank. If m τ s ∗ ‖A i−1 ‖ F ∗ ‖a i ‖ 2
Page 1 and 2: Noname manuscript No. (will be inse
Page 3 and 4: 3 Greville’s method. In Section 3
Page 5 and 6: 5 we can express A † i in a unifi
Page 7 and 8: 7 Remark 4 If k i is sparse, when w
Page 9 and 10: 9 6. k j = k j + uT i aj f i (e i
Page 11 and 12: 11 where C is a nonsingular matrix.
Page 13 and 14: 13 Hence, [ ] M = (I − K)F −1 I
Page 15: 15 where ∆E is an error matrix. H
Page 19 and 20: 19 Table 1 Information on the test
Page 21 and 22: 21 preconditioner is usually much d
Page 23: 23 problems are equivalent to the o

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