Question

a) Apply the Gram–Schmidt process to find an orthogonal basis for S.

S=span{[110−1],[1301],[4220]}

b) Find projSu.

S = subspace in Exercise 14; u=[1010]

c) Find an orthonormal basis for S.

S= subspace in Exercise 14.

Answer 1

We presume that S=span{(1,1,0,−1),(1,3,0,1),(4,2,2,0)} = span {v₁,v₂,v₃} (say).

1. Let u₁ = v₁= (1,1,0,−1), u₂ = v₂-proj_u1(v₂) = v₂-[(v₂.u₁)/(u₁.u₁)]u₁ = v₂-[(1+3+0-1)/(1+1+0+1)]u₁= (1,3,0,1)- (1,1,0,−1) = (0,2,0,2)and u₃ = v₃- proj_u1(v₃) - proj_u2(v₃) = v₃ - [(v₃.u₁)/(u₁.u₁)]u₁ -[(v₃.u₂)/(u₂.u₂)]u₂ = v₃ –[(4+2+0+0)/(1+1+0+1)]u₁ – [(0+4+0+0)/(0+4+0+4)]u₂ = (4,2,2,0)- 2(1,1,0,−1)-(1/2)( 0,2,0,2) = (4,2,2,0)-(2,2,0,-2)-(0,1,0,1) = (2,-1,2,1). Then { u₁,u₂,u₃}= {(1,1,0,−1), (0,2,0,2), (2,-1,2,1)} is an orthogonal basis for S.
2. u is not described.
3. Let e₁ = u₁/||u₁||= u₁/√[1²+1²+0²+(-1)²]= (1,1,0,−1)/√3 = (1/√3, 1/√3,0,- 1/√3),
4. √[0²+2²+0²+2²]= (0,2,0,2)/√8 = (0,2/√8,0,2/√8)
5. √[2²+(-1)²+2²+1²]= (2,-1,2,1)/ √10 = (2/√10,-1/√10,2/√10,1/√10)

Then {e₁,e₂,e₃} = {(1/√3, 1/√3,0,- 1/√3), (0,2/√8,0,2/√8), (2/√10,-1/√10,2/√10,1/√10)} is an orthonormal basis for S.