Thus, the quest for the smallest error is actually equivalent to the quest for the largest(in absolute value) inner product between the residual rk−1 and the normalized vectors of the matrix A.In the Update Provisional Solution stage, we minimize the term kAx − bk2 2 withrespect to x, such that its support is Sk. We denote ASk as a matrix of size n × jSkjthat contains the columns from A that belong to this support. Thus, the problem tobe solved is a minimization of kASkxSk − bk2 2, where xSk is the non-zero portion ofthe vector x. The solution is given by zeroing the derivative of this quadratic form
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