Show CA=CB iff A=B

Question

I got partial credit on this homework problem and would like a solution so I can study for my test.

The question is "Show CA=CB iff A=B, C is full column rank"

I got the "if" part which is pretty basic.

The "only if" part I got wrong though, can anyone tell me how to do it?

I put...

assume $A\neq B $

$$CA=CB$$ $$CA-CB=0$$ $$C(A-B) =0 $$ which is a contradiction as C is full row rank. I see now why this is wrong, but can anyone give me a proof for the "only if" part?

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does this work? assume $A \neq B$ $$CA=CB$$ $$C'CA=C'CB$$ $$(C'C)^{-1}(C'C)A=(C'C)^{-1}(C'C)B$$ $$A=B$$ which is a contradiction

Answer 1

If $C$ is of full column rank, it is left-invertible with some left-inverse $D$, and $CA = CB$ implies $DCA = DCB$.

To see that a matrix of full column rank is left-invertible, you can use Gaussian elimination.