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Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 4 . 1 4 ) If AB = 0 then the columns of B are in the nullspace of A . The rows of A are in the left nullspace of B . Say A and B are both 3 3 of rank 2. Then the nullspace of A has dimension 3 2 = 1. But this nullspace contains the columns of B , whose span has dimension 2, a contradiction. 11 ) For A = 1 2 3 6 the columns are visibly dependent (as are the rows) so the vector 1 3 is a basis for the column space and (1 , 2) is a basis for the row space. The row reduced echelon form of A is 1 2 0 0 so a basis for the nullspace is 2 1 . Finally, A T = 1 3 2 6 1 3 0 0 so the nullspace of A T has basis 3 1 . Note the orthogonality between the column space of A and the nullspace of A T . And between the nullspace of A and the rowspace of A . For B = 1 0 3 0 the first column forms a basis for the column space and the rows are visibly dependent so the vector 1 3 is a basis for the column space and (1 , 0) is a basis for the row space. The row reduced echelon form ofthe row space....
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 '05
 HUI
 Math, Linear Algebra, Algebra, Dot Product, column space

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