Let A, B, C be matrices, Is it true that AC = BC implies A = B when C has full row rank, or alternatively if we take the transpose, CA=CB and C has full column rank implies A=B?
I think I have come up with proof for the above statement. I'll explain the main idea for simplicity. Let C be a matrix with m rows and n columns since it has full row rank, $n \geq m$. Therefore, we can separate C into two submatrices, each containing columns of C. The former submatrices $C_{0}$ contain any m columns that are linearly independent, while the latter contains the rest. Thus we have $AC_{0} = BC_{0}$, multiplying by $C_{0}^{-1}$ on both side, we arrive at $A = B$, which is what we want to prove.
I was thinking about this while learning linear programming, and it's pretty odd that I did not find a post asking for this. Would you mind letting me know if the above statement is correct?