Unformatted text preview: 16 (a) (Statement of the problem.) Proof 1. Take x N (A). Then A x = 0 by definition of N (A). So (BA) x = B(A x) = B 0 = 0. Hence x N (BA) by definition of null space. It means that N (A) N (BA). (1) Take x N (BA). Then BA x = 0 by definition of null space. So B 1 BA x = B 1 0. That is, A x = 0. Hence x N (A). It means that N (BA) N (A). By (1) and (2), we have N (BA) = N (A). Since N (BA) = N (A), dim N (BA) = dim N (A). By the RankNullity Theorem, rank(A) + dim N (A) = n, and rank(BA) + dim N (BA) = n. (2) So we get rank(A) = rank(BA), i.e. the rank of A is equal to the rank of BA. Proof 2. Since B is nonsigular, it is a product of elementary matrices E1 , , Ek , i.e. B = E1 Ek . So BA = E1 Ek A. That is, A is row equivalent ot BA. So the system BA x = 0 can be obtained from the system A x = 0 by elementary row operations. Hence, they have the same solutions set. We get N (BA) = N (A). Also, since row equivalent matrices have the same row space, A and BA have the same row space, and therefore have the same rank. 1 ...
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 Spring '08
 samiabdul
 Linear Algebra, Algebra

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