{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# solhw8 - Solutions to Homework#8 MAT1341A Question 1(4 4...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 Gram-Schmidt 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 Gram-Schmidt 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 non-orthogonal 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 2-dimensional 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 ⊥...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

solhw8 - Solutions to Homework#8 MAT1341A Question 1(4 4...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online