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This is because A*Ap = 0 implies that p is an eigenvector of A * A corresponding to the zero eigenvalue. Hence, the determinant of A* A is necessarily zero. Finally, let us show that A'A is positive-definite. For this purpose, take any nonzero vector I and consider ~ . the product I'A'Az is necessarily positive; it cannot the product x*A*Ax,which evaluates to l l A ~ l /Then, be zero since the nullspace of A , in view of A being full rank, contains only the zero vector. 1) To see this, recall from Sec.

Morgan (Lucent Technologies), V. H. Nascimento (University of Silo Paulo, Brazil), M. R. Petraglia (Federal University of KOde Janeiro, Brazil), P. A. Regalia (The Catholic University of America), M. Rupp (Tech. Unisersitaet Wien, Austria), S. Theodoridis (University of Athens, Greece), and M. Verhaegen (Delft University, The Netherlands). I am also thankful to the many colleagues whose kind words of praise on the earlier publication have encouraged me to pursue the current project. I wish to thank L.

N - l}. which span the same linear subspace as the {hi}, by appealing to the classical Gram-Schmidt procedure. This is an iterative procedure that operates as follows. It starts with 40 = ho//(hoIl,where ilholl denotes the Euclidean norm of ho, and then repeats for i > 0: i-1 Thus, observe, for example, that T I = hl - (qGhl)qo, which is simply the residual vector that results from projecting hl onto qo. Clearly, by the orthogonality property of least-squares solutions, this residual vector is orthogonal to qo.

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