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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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 Fall '05
 HUI
 Math, Linear Algebra, Algebra

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