Consider the set of vectors S 1 0 1 23 13 23 14 1 14 a ls t

Consider the set of vectors S = {[1 0 -1], [-2/3 1/3 -2/3]. [1/4 1 1/4]}. (a) ls the set S orthogonal? (b) Is the set S orthonormal? c) If your answer to either (a) or (b) is \"no\", construct an orthonormal set of vectors {u vector_1, u vector_2, u vector_3} such that Span{u vector_1, u vector_2, u vector_3} = Span (S).

Solution

11.(a). We have (1,0,-1)^T.(-2/3,1/3,-2/3)^T= (-2/3)+0+(2/3) = 0; (1,0,-1)^T.(1/4,1,1/4)^T=(1/4)+0-(1/4) =0 and

(-2/3,1/3,-2/3)^T. (1/4,1,1/4)^T= (-1/6)+(1/3)-(1/6) =0. Hence the set S is orthogonal.

(b). Only the vector (-2/3,1/3,-2/3)^T is a unit vector. The magnitude of neither of the other 2 vectors is 1. Hence S is not orthonormal.

(c). Let v₁= (1,0,-1)^T and u₁= v₁/|| v₁|| = [1/(1+0+1)]v₁ = (1/2) (1,0,-1)^T = (1/2,0,-1/2)^T u₂ = (-2/3,1/3,-2/3)^T , v₃ = (1/4,1,1/4)^T and u₃= v₃/|| v₃|| = [1/(1/16+1+1/16)]v₃ = (22/3) (1/4,1,1/4)^T = ( 2/6, 22/3, 2/6)^T.

Then span {u₁,u₂,u₃}= span(S),and {u₁,u₂,u₃} is an orthonormal set where u₁=(1/2,0,-1/2)^T, u₂ =                      (-2/3,1/3,-2/3)^T and u₃= ( 2/6, 22/3, 2/6)^T.