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I understand why it wouldn't change the dependence relation between rows, but that isn't intuitively clear for the columns (for me at least).

I realize that non-trivial solutions to the equation Ax = 0 implies a dependence relation among the columns of A, but what I don't understand is why elementary matrix operations wouldn't change the solution set.

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    $\begingroup$ In fact, row reduction does change dependencies among the rows. In the RREF of a matrix, all of the zero rows—the ones that are linearly dependent—always come last. $\endgroup$ – amd Dec 4 '18 at 17:21
  • $\begingroup$ See lem.ma/6wy and the subsequent lessons. $\endgroup$ – Lemma Dec 4 '18 at 18:14
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What you call "elementary matrix operations" amount to left-multiplication by an invertible matrix; therefore you are replacing $A$ with $BA$ for some invertible $B$.

But then $BAx=0$ iff $Ax=0$, so the solution set hasn't changed.

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  • $\begingroup$ Oh that's also a great way of looking at it, forgot that elementary matrix operations is just left multiplying by an identity matrix. Thank you! $\endgroup$ – James Ronald Dec 4 '18 at 20:44
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There are many technical explanations that prove it, but don't explain why it's true. What convinces me is to imagine two columns where one is two times the other.

Does swapping 1st and 3rd entries in each column change that relationship?
Obviously no.

Does multiplying 2nd entry by, say, 7 in each column change that relationship?
Obviously no.

Does adding a multiple of the 3rd entry to the 4th entry in each column change that relationship? 
Obviously no.

That gives me an insight that convinces me and I hope that you find it helpful.

For more, see https://lem.ma/6wy and the subsequent lessons.

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  • $\begingroup$ This helps a lot, thank you! $\endgroup$ – James Ronald Dec 4 '18 at 20:44

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