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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
length is given by
x = A� (AA� )−1 y ,
and from part (b), A� (AA� )−1 = A+ . Therefore
x = A+ y .
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
�Σz − c�2 = r
(σi zi − ci )2 +
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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