Question

Let {v1, ..., vk} ⊂ R n be nonzero mutually orthogonal vectors (i.e. every vector in the set is orthogonal to every other vector in the set) and define the n × k matrix A = [ v1 · · · vk ] . Show that the only solution to Ax = 0 is the trivial solution.

Answer 1

Let {v₁,v₂ ..., v_k} be non-zero mutually orthogonal vectors in Rⁿ . Then the vectors v₁,v₂ ..., v_k are linearly independent(non-zero mutually orthogonal vectors are linearly independent). Let A the n × k matrix with v₁,v₂ ..., v_k as columns and let X = (x₁,x₂,…x_k)^T be a solution to the equation AX = 0. Further, let v_i = (a_i1,a_i2,…,a_in)^T(1 ≤ i ≤k). Then, we have x₁a₁₁+x₂a₂₁+…x_ka_k1 = 0 for each i (1 ≤ i ≤k) so that x₁v₁ +x₂v₂ +…+x_kv_k = 0. However, since the vectors v₁,v₂ ..., v_k are linearly independent , therefore, x₁=x₂=…=x_k = 0 . Hence X = 0. Thus the equation AX = 0 has only a trivial solution.