# AC = BC implies A = B when C has full row rank?

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?

Since $$C$$ has full row rank, every vector in $$\Bbb R^m$$ can be written as $$Cx$$ for some $$x\in\Bbb R^n$$. In particular, each standard basis vector $$e_j\in\Bbb R^m$$ can be written as $$e_j = Cx_j$$ for some $$x_j\in\Bbb R^n$$. So suppose that $$(A-B)C=0$$. It follows that $$0=(A-B)Cx_j = (A-B)e_j$$ for all $$j=1,\dots,m$$. This means that the standard matrix of $$A-B$$ is the $$0$$-matrix.