The orthogonalities appear again 1 17 if s 0 s

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Unformatted text preview: inally, B = is in row reduced echelon form so the nullspace of B T has 1￿ 00 ￿ −3 basis . The orthogonalities appear again! 1 17) If S = 0, S ⊥ contains all vectors that have dot product 0 with the zero vector. So S is all of R3 . If S is spanned by (1, 1, 1) then S ⊥ consists of all (a, b, c) with (a, b, c) · (1, 1, 1) = 0, that is all solutions to a + b + c = 0, a plane. A basis for this plane is {(−1, 1, 0), (−1, 0, 1)}. If S is spanned by (1, 1, 1) and (1, 1, −1), then S ⊥ consists of all (a, b, c) with (a, b, c) · (1,...
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