Prove Equal Matrices and Matrix-Vector Products for a Linearly Independent Spanning Set

Prove: Suppose that CC = {u1,u2,…,upu_1,u_2,\dots,u_p} is a linearly independent spanning set for Cn\mathbb{C}^n. Suppose also that AA and BB are mm x nn matrices such that AuiAu_i = BuiBu_i for every 1≤i≤n1 \le i \le n. Then A=BA = B.

By definition we know that Au=[u]1A1+[u]2A2+⋯+[u]nAnAu = [u]_1A_1 + [u]_2A_2 + \cdots +[u]_nA_n. Also since the spanning set CC is linearly independent we know that α1u1+α2u2+⋯+αpup=0\alpha_1u_1+\alpha_2u_2+\cdots+\alpha_pu_p = 0 where α1,α2,…,αp=0\alpha_1,\alpha_2,\dots,\alpha_p = 0. I’m unsure how to use the definitions above (or maybe I’m missing a definition that I need to solve this proof) to get between Aui=BuiAu_i=Bu_i and A=BA=B.




CC cannot be a linearly independent spanning set for Cn\mathbb{C}^n unless p=np=n.
– Jack
2 days ago



Yeah that was one thing that I was confused about in the problem, why they didn’t specify pp and nn to be equal.
– Justin Merriman
2 days ago


2 Answers


Hint 1: Write down what A=BA=B really means.

Hint 2: Use the fact that CC is a basis and the linearity of AA and BB.

Let UU be the matrix of which its ii-th column is uiu_i. UU is invertible.

Since AU=BUAU=BU, we have (A−B)U=0(A-B)U=0. Since UU is invertible, we have A=BA=B.