Finally at 00 1 13 13 3 t so the nullspace of a has

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Unformatted text preview: basis for the nullspace is . Finally, AT = 00 1 ￿ ￿ ￿ ￿ ￿ ￿ 13 13 −3 T → so the nullspace of A has basis . Note the orthogonality 26 00 1 between the column space of A and the nullspace of AT . And between the nullspace of A and the rowspace of ￿ . A ￿ 10 For B = the first column forms a basis for the column space and the rows are 30 ￿￿ 1 visibly dependent so the vector is a basis for the column space and (1, 0) is a basis for 3 ￿ ￿ 10 the row space. The row reduced echelon form of B is so a basis for the nullspace 00 ￿ ￿ ￿￿ 0 13 T is . F...
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