Exercise 46 a suppose a cn has full column rank then

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Unformatted text preview: he smallest length �x�2 is given by x = A+ y. If A has full row rank, we have shown in chapter 3 that the solution with the smallest ˆ length is given by x = A� (AA� )−1 y , ˆ and from part (b), A� (AA� )−1 = A+ . Therefore x = A+ y . ˆ Similary, it can be shown that the pseudo inverse is the solution for the case when a matrix A has a full column rank (compare the results in chapter 2 with the expression you found in part (b) for A+ when A has full column rank). Now, let’s consider the case when a matrix A is rank deficient, i.e., rank (A) = r < min(m, n) where A ∈ C m×n and is thus neither full row or column rank. Suppose we have a singular value decomposition of A as A = U ΣV � , where U and V are unitary matrices. Then the norm we are minimizing is �Ax − y � = �U ΣV � x − y � = �U (ΣV � x − U � y )� = �Σz − U � y �, where z = V � x, since � · � is unaltered by the orthogonal transformation, U . Thus, x minimizes �Ax − y � if and only if z minimizes �Σz − c�, where c = U � y . Since the rank of A is r, the matrix Σ has the nonzero singular values σ1 , σ2 , · · · , σr in its diagonal entries. Then we can rewrite �Σz − c�2 as follows: �Σz − c�2 = r n � � (σi zi − ci )2 + c2 . i i=1 i=r+1 c It is clear that the minimum of the norm can be achived when zi = σii for i = 1, 2, · · · , r and the rest of the zi ’s can be chosen arbitrarily. Thus, there are infinitely many solutions z and the ˆ 6 solution with the minimum norm can be achieved when zi = 0 for i = r...
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