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Why not orthogonal residuals?

Since the direction vectors have not yet been specified, we still have some degree of freedom that we may use to help choose an optimum method. We might suppose that it could be possible to choose the residuals themselves in a way so that they are orthogonal from one step to the next. But, we soon find this is wrong for, if we were to try this, we would find that the condition

(r_n,r_n-1) = 0   implies that

_n = (r_n-1,r_n-1) (r_n-1,Mp_n-1) = (r_n-1,Mp_n-1) ||Mp_n-1||^2.   But, this condition (impossible) is actually impossible because the Cauchy-Schwartz inequality for vectors states that

(r_n-1,Mp_n-1)^2 ||r_n-1||^2||Mp_n-1||^2,   where the equality in (CchySchwartz) occurs only when the two vectors are proportional which will virtually never be true for such an iteration scheme.

Since orthogonality of the residuals is impossible, the next best condition we might try to impose is conjugacy of the residual vectors. This condition can be stated as

(M^Tr_n,M^Tr_j) = (r_n,MM^Tr_j) = 0     for     j = 1,2,...,n-1.   Thus, if this condition is met, the residuals will be orthogonal relative to the inner product $(\cdot,{\bf MM}^T\cdot)$.The set of conditions (conjugateres) is not in conflict with (optimumalpha) so we assume it holds and construct the directions ${\bf p}_n$ that satisfy both (conjugacy) and (conjugateres).


previous up next print clean
Next: Conjugate directions Up: CONJUGATE DIRECTIONS AND CONJUGATE Previous: Linear iteration
Stanford Exploration Project
11/11/1997