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Unformatted text preview: Solutions to Homework #8, MAT1341A Question # 1: (4+4+2=10 points) Let U = span 1 1 , 2 1 1 1 and x = 3 2 3 . (a) Find the projection of x onto U . Solution: We have two choices: (1) use the method that says to choose A so that U = Col ( A ) and then calculate A T A u = A T x and the projection is A u ; (2) use GramSchmidt to get an orthogonal basis for U and then use the formula for projection onto an orthogonal basis. Method (1): We set A = 1 2 1 1 1 1 ; then A T A = 1 1 2 1 1 1 1 2 1 1 1 1 = 2 3 3 7 and A T x = 1 1 2 1 1 1 3 2 3 = 5 11 so if A T A u = A T x then u = ( A T A ) 1 A T x = 1 5 7 3 3 2 5 11 = 1 5 2 7 so the projection of x onto U = Col ( A ) is proj U ( x ) = A u = 1 5 1 2 1 1 1 1 2 7 = 1 5 16 7 9 7 Method (2): Let v 1 = (1 , , 1 , 0) then by GramSchmidt v 2 = u 2 u 2 · v 1 v 1 · v 1 v 1 = 2 1 1 1  3 2 1 1 = 1 2 1 1 2 1 1 2 So an orthogonal basis for U is { v 1 , v 2 } ; then the projection of x onto U is given by the formula: proj U ( x ) = x · v 1 v 1 · v 1 v 1 + x · v 2 v 2 · v 2 v 2 = 5 2 1 1 + 7 / 2 5 / 2 1 2 1 1 2 1 = 25 10 1 1 + 7 10 1 2 1 2 = 32 / 10 14 / 10 18 / 10 14 // 10 = 1 5 16 7 9 7 Marking: If they use the formula on the nonorthogonal basis, then 0 points and please write : the basis isn’t orthogonal. Otherwise, 2 points for a logical approach (including any I didn’t think of, and including at least one point if they correctly described what properties the projection has) and 2 points for the answer. (b) Find a basis of U ⊥ . Solution: Again, there are several approaches, but in this case most come out the same. You could carefully write out the conditions on a vector ( a, b, c, d ) for it to be orthogonal to both vectors in the spanning set for U ; or you could recall that whenever U = Row ( B ) then U ⊥ = Null ( B ). Both of these lines of reasoning lead to the same linear system to solve. So let B = 1 1 2 1 1 1 ∼ 1 1 1 1 1 then the nullspace of B is Null ( B ) = U ⊥ = { ( r, r s, r, s )  r, s ∈ R } which has as basis the set of basic solutions { ( 1 , 1 , 1 , 0) , (0 , 1 , , 1) } We verify that indeed these two vectors are orthogonal to the vectors in the spanning set for U ; and then, we notice that they are linearly independent and thus span a 2dimensional subspace. Since dim( U ) = 2 and U ⊂ R 4 , we know that U ⊥ has dimension 4 2 = 2. Thus we conclude that our vectors span U ⊥...
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 Winter '10
 DAIGLE
 Linear Algebra, Orthonormal basis, orthogonal basis

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