Let W x y z w R4 x y z 2 0 Find out an orthonormal basi

Let W = {(x y z w) R^4: x + y + z + 2 = 0}. Find out an orthonormal basic for W.

Solution

Let y = r and z = s, and w = t. Then x = -2-y-z = -2-r-s so that (x,y,z,w)^T = ( -2-r-s, r, s, t )^T = (-2,0,0,0)^T + r (-1,1,0,0)^T + s (-1,0,1,0)^T+ t (0,0,0,1)^T. Thus a basis for W is { (-2,0,0,0)^T,(-1,1,0,0)^T, (-1,0,1,0)^T,(0,0,0,1)^T} = {v₁, v₂ ,v₃ , v₄ } (say).

Let u₁ = v₁ = (-2,0,0,0)^T. Also, let u₂ = v₁ –Proj_u1 (v₂) = v₂ –[(v₂.u₁)/(u₁.u₁)]u₁= v₂ –(1/2)u₁ = (-1,1,0,0)^T -     (-1,0,0,0)^T = ( 0,1,0,0)^T.

u₃= v₃ –Proj_u1 (v₃ )–Proj_u2(v₃) =v₃–[(v₃.u₁)/(u₁.u₁)]u₁ -[(v₃.u₂)/(u₂.u₂)]u₂ = v₃–1/2u₁ =(-1,0,1,0)^T-( -1,0,0,0)^T = (0,0,1,0)^T

u₄ = v₄ –Proj_u1 (v₄) - Proj_u2 (v₄)- Proj_u3 (v₄) = v₄ –[(v₄.u₁)/(u₁.u₁)]u₁ –[(v₄.u₂)/(u₂.u₂)]u₂–[(v₄.u₃)/(u₃.u₃)]u₃ =

(0,0,0,1)^T

Then { u₁, u₂, u₃, u₄} is an orthogonal basis for W. On normalizing u₁, the set {(1,0,0,0)^T,(0,1,0,0)^T,(0,0,1,0)^T, (0,0,0,1)^T} is an orthonotmal basis for W.