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I'm attempting to show that $\text{Col}(A) \neq \text{Col}(R)$ where $R$ is the reduced row-echelon form of $A$ in general. Now of course I could use a simple counterexample but I find those a bit boring and wanted to try to use a different method to stretch my logic a bit.

I'm attempting to structure it like a proof by contradiction. We know $R = (E_1E_2 \ldots E_k)A$ for some elementary matrices $E_n$. Thus suppose $\text{Col}(A) = \text{Col}(R)$ and there is some $\mathbf{v}$ in both sets:

\begin{align*} \exists u,p: A\mathbf{u} = R\mathbf{p} = \mathbf{v} &\implies (E_1E_2 \ldots E_k)A\mathbf{p} = A\mathbf{u} \\ &\implies A\mathbf{p} = (E_1E_2 \ldots E_k)^{-1}A\mathbf{u} \\ &\implies \mathbf{p} = (E_1E_2 \ldots E_k)^{-1}\mathbf{u} \end{align*}

But now I'm a bit stuck. How can I prove that this cannot be true for all possible $\mathbf{v}$? There does not appear to be any direct conflict with the dimensions of the elements involved, as the elementary matrices are $n \times n$ and the vectors $\mathbf{p}$ and $\mathbf{u}$ are $n \times 1$. Is such a proof possible?

Perhaps this approach would require more complicated or more general knowledge than I have at my introductory level, but a structure like this involving elementary matrices has worked for me before for similar proofs so I am a bit befuddled on what I'm missing here.

I'm aware of questions like this one but its answer uses a counterexample and the resources I can find from there are also of a similar form.

naiveai
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  • How come you don't want to just give a counterexample? – wormram Nov 07 '23 at 03:48
  • @morrowmh As I said, basically out of sheer curiosity. Which may be simply naivety speaking but now that I've gotten into the problem I find it difficult to get it out of my head. – naiveai Nov 07 '23 at 03:50
  • Please explain what RREF means. – J.-E. Pin Nov 07 '23 at 03:53
  • @J.-E.Pin Apologies, edited the question - it's the reduced row-echelon form. – naiveai Nov 07 '23 at 03:56
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    You are missing the boat; you should find all "counterexamples." Try to describe all matrices $A$ such that $\text{Col}(A) \neq \text{Col}(R).$ At the same time, try to describe all matrices $A$ such that $\text{Col}(A) = \text{Col}(R).$ If you are very lucky, the union of your two sets will be all matrices – Will Jagy Nov 07 '23 at 04:03

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If $A$ is (square and) invertible, then so is any matrix obtained from it by row operations (as these do not change the rank), including any row reduced echelon form. In particular their column space will be the whole space, just like that of $A$. So what you want to prove is not true in this case; there is no way you could prove such a statement without making any assumption about $A$ that excludes this case.

Marc van Leeuwen
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