Find a basis of the subspace of R4

Question

Find a basis of the subspace of R⁴ defined by the equation −5x₁+3x₂−4x₃−4x₄=0

Answer 1

Solution: v₁=[3 5 0 0], v₂=[0 4 3 0], v₃=[0 0 4 -4] is an obvious choice of basis. How to see this:

Your subspace (let's denote it by Π) has dimension 3 (just one linear equation in R⁴). Write the equation of Π as (n,x)=0, with n=[-5 3 4 4], and x=[x₁ x₂ x₃ x₄] - any point of Π. Scalar product (n,x) equals 0 implies that n is orthogonal complement of Π (R⁴=Π⊕n). You can choose v₁ orthogonal to n in the intersection of Π with plane given by x₃=x₄=0: v₁=[3 5 0 0] gives (v_1,n)=3*(-5)+5*(3)=0. Similarly choose v₂ and v₃. And v_1,v_2,v_3, are linearly independent: compose matrix B from these vectors, B =

[3 5 0 0]

[0 4 3 0]

[0 0 4 -4],

B is of full rank (first 3 columns are linearly independent because they form upper triangular square matrix with nonzero entries on the diagonal).