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4.1 - Math 2940 Solutions Fall 2011 Section 4.1 4 If AB = 0...

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Math 2940 Solutions, Fall 2011 Section 4 . 1 4 )I f 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 )F o r A = ° 12 36 ± 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 ° 00 ± so a basis for the nullspace is ° 2 1 ± . Finally, A T = ° 13 26 ± ° ± 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 = ° 10 30 ± the ±rst 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 of
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4.1 - Math 2940 Solutions Fall 2011 Section 4.1 4 If AB = 0...

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