Question

If R3 has the Euclidean inner product, how can I find the orthogonal basis for the subspace spanned by

(2,1,2) , (-1,0,2) , (-1,1,2)? Do I need to use Gram-Schmidt?

Answer 1

Firstly, we need to check whether the given vectors are linearly independent.

Let A be the matrix with the given vectors as columns. The RREF of A is I, which implies that the given vectors are linearly independent.

Now, we can use the Gram-Schmidt process. Let v₁ = (2,1,2),v₂ = (-1,0,2) and v₃ = (-1,1,2). Also, let u₁ = v₁ = (2,1,2), u₂ = v₂-proj_u1(v₂) = v₂-[(v₂.u₁)/(u₁.u₁)]u₁ = v₂-[(-2+0+4)/(4+1+4)]u₁ =(-1,0,2)-(2/9)(2,1,2)=(-13/9,-2/9,14/9), and u₃=v₃-proj_u1(v₃)-proj_u2(v₃)=v₃–[(v₃.u₁)/(u₁.u₁)]u₁-[(v₃.u₂)/(u₂.u₂)]u₂=v₃-[(-2+1+4)/(4+1+4)]u₁-[(13/9-2/9+28/9)/(169/81+4/81+196/81)]u₂ = (-1,1,2)- (1/3)(2,1,2)- (39/41)(-13/9,-2/9,14/9)= (-1,1,2)-(2/3,1/3,2/3)-(-169/123,-26/123, 182/123 )= (-36/123, 108/123,-18/123).

Thus, { u₁, u₂, u₃} = {(2,1,2), (-13/9,-2/9,14/9), (-36/123, 108/123,-18/123) } is an orthogonal basis for the subspace spanned by (2,1,2) , (-1,0,2) , (-1,1,2).