Question

Is the set of all x, y, z such x+ 3y + 2z = 0 a subspace of R^3 ? If so find a basis for the space.

Answer 1

Let X₁ = (x₁,y₁,z₁)^T and X₂ = (x₂,y₂,z₂)^T be 2 arbitrary vectors in V (say), the given subset of R³ and let k be an arbitrary scalar. Then x₁+ 3y₁ + 2z₁ = 0 and x₂+ 3y₂ + 2z₂ = 0. Further, X₁+X₂ = (x₁,y₁,z₁)^T +(x₂,y₂,z₂)^T = (x₁+x₂,y₁+y₂,z₁+z₂)^T and (x₁+x₂)+ 3(y₁ +y₂)+ 2(z₁+z₂) = (x₁+ 3y₁ + 2z₁)+( x₂+ 3y₂ + 2z₂)=0+0 = 0. This implies that X₁+X₂ ∈ V so that V is closed under vector addition. Also kX₁ = k(x₁,y₁,z₁)^T = (kx₁,ky₁,kz₁)^T and kx₁+ 3ky₁ + 2kz₁ =k(x₁+ 3y₁ + 2z₁ ) =k.0 = 0. This implies that kX₁ ∈ V so that V is closed under scalar multiplication. Also, since 0+3.0+2.0 = 0, hence the zero vector (0,0,0)^T ∈ V so that V is a vector space, and , therefore, a subspace of R³.