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Unformatted text preview: CS 435 : Linear Optimization Fall 2008 Lecture 6: Row Rank. Convex Sets. Lecturer: Sundar Vishwanathan Computer Science & Engineering Indian Institute of Technology, Bombay 1 Row Rank of a matrix The space f x : Ax = g is called the null space of the matrix A . Having already seen that the dimension of the null space of A is equal to the n k where k is the number of linearly independent columns of the matrix A , we now examine how this relates to the number of linearly independent rows of A . Indeed, we will show that it is equal to n r where r is the number of linearly independent rows of A . This, in some sense, is not surprising. Unlike the column space, here both x and the rows of A , have the same number of coordinates. As vectors, both x and rows of A are part of R n . Recall that two vectors are perpendicular if their dot product is zero. What we are interested in is the set of all vectors x which are perpendicular to all rows of A . If the row space is the space spanned by the rows of A then what we are looking for is the orthogonal complement of the row space of A . It is hence believable that the dimension of the orthogonal complement is n r One way of proving this is to use Gaussian Elimination which we do next. We have proved that Gaussian Elimination does not change the set f x : Ax = g . From what we have done so far, it is easy to infer that Gaussian Elimination does not change the number of linearly independent columns of A . Why? We will next prove that Gaussian Elimination does not change the space spanned by the rows. This is to be expected, since, we know that it does not change the orthogonal complement! So, one way to prove this is to show that the orthogonal complement of the orthogonal complement of a subspace U of R n is the subspace U itself. We will do it di erently but the reader is encouraged to try that approach. Lemma 1 Gaussian Elimination does not change the row space of a matrix....
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