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

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10 Remark 7 In Algorithm 4, the definition of θ can be rewritten as θ = (u i, a j ) d i (4.2) = (Az i, Ae j ) (Az i , Az i ) (4.3) = (z i, e j ) A T A . (4.4) (z i , z i ) A T A When A is full column rank, A T A is symmetric positive definite. Hence, (·, ·) A T A defines an inner product. Hence, Algorithm 4 includes an A T A-orthogonalization procedure. Based on this observation, we conclude that for the special case in which A is full column rank, our algorithm is very similar to the RIF preconditioner which was proposed by Benzi and Tůma [3]. The difference lies in the implementation. Our algorithm looks for an incomplete factorization of A † , while RIF looks for an LDL T decomposition of A T A. 5 Equivalence Condition Consider solving the least squares problem (1.1) by transforming it into the left preconditioned form, min ‖Mb − MAx‖ 2, (5.1) x∈R n where A ∈ R m×n , M ∈ R n×m , and b is a right-hand-side vector b ∈ R m . When preconditioning a least squares problem, one important issue is whether the solution of the preconditioned problem is the solution of the original problem. For nonsingular linear systems, the condition for this equivalence is that the preconditioner M should be nonsingular. Since we are dealing with general rectangular matrices, we need some other conditions to ensure that the preconditioned problem (5.1) is equivalent to the original least squares problem (1.1). First, note the following lemma [14]. Lemma 1 and ‖b − Ax‖ 2 = min x∈R n ‖b − Ax‖ 2 ‖Mb − MAx‖ 2 = min x∈R n ‖Mb − MAx‖ 2 are equivalent for all b ∈ R m , if and only if R(A) = R(M T MA). To analyze the equivalence between the original problem (1.1) and the preconditioned problem (5.1), where M is from any of our algorithms, we first consider the simple case in which A is a full column rank matrix. After numerical droppings, we have, A † ≈ M = ZF −1 Z T A T . (5.2) Note that Z is a unit upper triangular matrix and F is a diagonal matrix with positive elements. Hence, we have M = CA T , (5.3)
11 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).
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: 9 6. k j = k j + uT i aj f i (e i
Page 13 and 14: 13 Hence, [ ] M = (I − K)F −1 I
Page 15 and 16: 15 where ∆E is an error matrix. H
Page 17 and 18: 17 Theorem 10 Let A ∈ R m×n . If
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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