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

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14 Remark 9 From the proof of the above theorem, we have R(MA) = R(M), which means that the preconditioned least squares problem (5.1) is a consistent problem. Thus, by Theorem 4 and Theorem 6, we finally obtain our main result for our Greville preconditioner M. Theorem 7 Let A ∈ R m×n . If Assumption 1 holds, then for any b ∈ R m , preconditioned GMRES determines a least squares solution of min ‖MAx − Mb‖ x∈R n 2 (6.24) before breakdown and this solution attains min x∈R n‖b − Ax‖ 2, where M is computed by Algorithm 1. Remark 10 Since all the proofs are based on the factorization M = (I − K)F −1 V T , the theorems hold also for Algorithm 2. 7 Implementation Consideration 7.1 Detecting Linear Dependence In Algorithm 1 and Algorithm 2, one important issue is how to recognize the columns which are linearly dependent on the previous columns. In the algorithms we check whether ‖u i ‖ 2 is small to detect the rank deficiency. Simply speaking, we can set up a tolerance τ in advance, and switch to “else” when ‖u i ‖ 2 < τ. However, is this good enough to help us detect the linear dependent columns of A when we perform numerical droppings? To address this issue, we first take a look at the RIF preconditioning algorithm. The RIF preconditioner was developed for full rank matrices. However, numerical experiments showed that it also works for rank deficient matrices. For this phenomenon, our equivalent Algorithm 3 can give a good insight into the RIF preconditioner, since the only possibility for the RIF preconditioner to breakdown is when f i = 0, which implies that u i is a zero vector. From Algorithm 1 and Algorithm 2 we have u i = a i − A i−1 k i (7.1) = a i − A i−1 A † i−1 a i (7.2) = (I − A i−1 A † i−1 )a i. (7.3) Thus, u i is the projection of a i onto R(A i−1 ) ⊥ . Hence, in exact arithmetic u i = 0 if and only if a i ∈ R(A i−1 ). Algorithm 1 and Algorithm 2 have an alternative when a i ∈ R(A i−1 ), i.e. when u i = 0, Algorithm 1 and Algorithm 2 will turn to the “else” case. However, this is not always necessary, because of the numerical droppings. With numerical droppings, the u i is actually, u i = a i − A i−1 M i−1 a i (7.4) = a i − A i−1 (A † i−1 + ∆E)a i (7.5) = a i − A i−1 A † i−1 a i − A i−1 ∆Ea i (7.6) = −A i−1 ∆Ea i (7.7) ≠ 0, (7.8)
15 where ∆E is an error matrix. Hence, even though a i ∈ R(A i−1 ), since u i will not be the exact projection of a i onto R(A i−1 ) ⊥ , the RIF algorithm will not necessarily breakdown when linear dependence happens. The RIF preconditioner does not take the rank deficient columns or nearly rank deficient columns into consideration. Hence, if we can capture the rank deficient columns, we might be able to have a better acceleration to the convergence of the preconditioned problem. Assume the M we compute from Algorithm 1 or Algorithm 2 can be viewed as an approximation to A † with error matrix E ∈ R n×m , that is M = A † + E. (7.9) First note a theoretical result about the perturbation lower bound of the generalized inverse. Theorem 8 [18] If rank(A + E) ≠ rank(A), then ‖(A + E) † − A † ‖ 2 ≥ 1 ‖E‖ 2 . (7.10) By Theorem 8, if the rank of M = A † + E from our algorithm is not equal to the rank of A † , ( or A, since they have the same rank), by the above theorem, we have, ‖M † − (A † ) † ‖ 2 ≥ 1 ‖E‖ 2 (7.11) ⇒ ‖M † − A‖ 2 ≥ 1 ‖E‖ 2 . (7.12) The above inequality says that, if we denote M † = A + ∆A, then ‖∆A‖ 2 ≥ 1 ‖E‖ 2 . Hence, when ‖E‖ 2 is small, which means M is a good approximation to A † , M can be an exact generalized inverse of another matrix which is far from A, and the smaller the ‖E‖ 2 is, the further M † is from A. In this sense, if the rank of M is not the same as that of A, M may not be a good preconditioner. Thus, it is important to maintain the rank of M to be the same as rank(A). Hence, when we perform our algorithm, we need to sparsify the preconditioner M, but at the same time we also want to capture as many rank deficient columns as possible, and maintain the rank of M. To achieve this, apparently, it is very important to decide how to detect whether the exact value u i = ‖(I − A i−1 A † i−1 )a i‖ 2 is close to zero or not based on the computed value u i = ‖(I − A i−1 M i−1 )a i ‖ 2 . From the previous discussion, we have When a i ∈ R(A i−1 ), u i = a i − A i−1 A † i−1 a i = 0. Then, u i = u i − A i−1 ∆Ea i . (7.13) u i = −A i−1 ∆Ea i , (7.14) ‖u i ‖ 2 ≤ ‖A i−1 ‖ F ‖a i ‖ 2 ‖∆E‖ F . (7.15) If we require ∆E to be small, we can use a tolerance τ s. If ‖u i ‖ 2 ≤ τ s‖A i−1 ‖ F ‖a i ‖ 2 , (7.16) we suppose we detect a column a i which is in the range space of A i−1 . From now on we call τ s the switching tolerance.
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: 13 Hence, [ ] M = (I − K)F −1 I
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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