For the set of vectors S 1 0 1 1 0 2 0 3 3 1 1 5 Use the Gr

For the set of vectors S = {(1, 0, 1, 1), (0, 2, 0, 3), (-3, -1, 1, 5)} Use the Graham-Schrader procedure to find an orthonormal basis for the vector space spanned by S. Using the basis found in part la. find the projection of the vector (13, -5, -21, 11) onto the vector space spanned by S.

Solution

Let A =

1

0

-3

0

2

-1

1

0

1

1

3

5

The RREF of A is

1

0

0

0

1

0

0

0

1

0

0

0

Thus, the three vectors in S are linearly independent. Let v₁ = (1,0,1,1), v₂ = (0,2,0,3) and v₃ = (-3,-1,1,5).

(a). Let u₁= v₁. Also, let u₂=v₂–proj_u1(v₂)= v₂–[(v₂.u₁)/(u₁.u₁)]u₁= v₂–(3/3)u₁ = (0,2,0,3)-(1,0,1,1)=(-1,2,-1,2) and u₃ = v₃ - proj_u1(v₃)- proj_u2(v₃)= v₃ -[(v₃.u₁)/(u₁.u₁)]u₁-(v₃.u₂)/(u₂.u₂)]u₂ = v₃ –(3/3)u₁-(10/10) u₂ =                v₃ –u₁- u₂ = (-3,-1,1,5)- (1,0,1,1)- (-1,2,-1,2) = ( -3,-3,1,2). Then {u₁,u₂,u₃} is an orthogonal basis for span{S} . Further, let e₁ = u₁/|| u₁|| = (1/3,0,1/3,1/3), e₂ = u₂/|| u₂|| = (-1/10,2/10,-1/10,2/10) and e₃ = u₃/|| u₃||= ( -3/23,-3/23,1/23,2/23). Then {e₁,e₂,e₃} is an orthonormal basis for span{S}.

(b) Let w = (13,-5,-21,11). Then proj_u1(w) = [(w.u₁)/(u₁.u₁)]u₁ = [3/3]u₁ = (1,0,1,1)

Also, proj_u2(w) = [(w.u₂)/(u₂.u₂)]u₂ = (20/10) u₂ =2(-1,2,-1,2) = (-2,4,-2,4) and proj_u3(w) = [(w.u₃)/(u₃.u₃)]u₃ = ( -23/23)u₃ = -u₃ = -( -3,-3,1,2) = (3,3,-1,-2).

Then, the projection of W onto span{S} = (1,0,1,1)+ (-2,4,-2,4)+ (3,3,-1,-2) = (2,7,-2,3)

1	0	-3
0	2	-1
1	0	1
1	3	5