Assume that W Span v1 v2 v3 where v1 1 1 0 0 v2 1 1 1 1 a

Assume that W = Span {v_1, v_2, v_3}, where v_1 = [1 1 0 0], v_2 = [1 1 1 1], and v_3 = [0 1 0 1]. (a) Find an orthogonal basis for W. (b) Find the projection of the vector y = [1 2 3 4] on the subpace W.

Solution

(a). The vectors v₁,v₂,v₃ are apparently linearly independent so that the set { v₁,v₂,v₃} is a basis for W.

Let u₁ = v₁ = (1,1,0,0)^T, u₂ = v₂-proj_u1(v₂) = v₂ –[(v₂.u₁)/(u₁.u₁)]u₁ = v₂ –[(1+1+0+0)/((1+1+0+0)]u₁ = v₂ –u₁ = (1,1,1,1)^T- (1,1,0,0)^T = (0,0,1,1)^T and u₃ = v₃-proj_u1(v₃)-proj_u2(v₃)= v₃–[(v₃.u₁)/(u₁.u₁)]u₁-[(v₃.u₂)/(u₂.u₂)]u₂ = v₃–[((0+1+0+0)/(1+1+0+0)]u₁-[(0+0+0+1)/(0+0+1+1)]u₂ = (0,1,0,1)^T-(1/2)(1,1,0,0)^T-(1/2)(0,0,1,1)^T = (-1/2, 1/2, -1/2,1/2)^T.Then { u₁,u₂,u₃} is an orthogonal basis for W.

(b). W#e have proj_v1(y)= [(y.v₁)/(v₁.v₁)]v₁ = [(1+2+0+0)/(1+1+0+0)]v₁= (3/2)(1,1,0,0)^T= (3/2,3/2,0,0)^T, proj_v2(y)= [(y.v₂)/(v₂.v₂)]v₂ = [(1+2+3+4)/(1+1+1+1)]v₂= 5/2(1,1,1,1)^T= (5/2,5/2,5/2,5/2)^T, and proj_v3(y)= [(y.v₃)/(v₃.v₃)]v₃= [(0+2+0+4)/(0+1+0+1)]v₃ = 3(0,1,0,1)^T = (0,3,0,3)^T. Hence proj_W(y) = (3/2,3/2,0,0)^T+(5/2,5/2,5/2,5/2)^T+(0,3,0,3)^T= (4,7,5/2,11/2)^T.