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In Hoffman and Kunze they have two exercises where they ask to show that if two homogeneous linear systems have the exact same solutions then they have the same row-reduced echelon form.

They first ask to prove it in the case of $2\times 2$ (Exercise 1.2.6) and then they ask to prove it in the case $2\times 3$ (Exercise 1.4.10). I am able to prove it in both of these special cases, but as far as I can tell Hoffman and Kunze never tell us whether or not this is true in general.

So that's my question, is this true in general? And if not, can anybody provide a counter-example? Thank you!

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Solutions to the homogeneous system associated with a matrix is the same as determining the null space of the relevant matrix. The row space of a matrix is complementary to the null space. This is true not only for inner product spaces, and can be proved using the theory of non-degenerate symmetric bilinear forms.

So if two matrices of the same order have exactly the same null space, they must also have exactly the same row space. In the row reduced echelon form the nonzero rows form a basis for the row space of the original matrix, and hence two matrices with the same row space will have the same row reduced echelon form.

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  • $\begingroup$ Thanks bro, I'm digesting this, will check it off as soon as I'm convinced. Not that I don't believe you, I do, I just want to understand a bit better. But I'm glad to know the answer because it was really bugging me! Thanks again, Greg $\endgroup$ – Gregory Grant Jun 19 '16 at 16:02
  • $\begingroup$ Hi @GregoryGrant, yes the theory is actually quite deep...I highly recommend you have a look at the following article: jstor.org/stable/2695466?seq=1#page_scan_tab_contents . And regarding the complementary spaces, have a look at page 7 top of this doc: researchgate.net/publication/…. $\endgroup$ – Christiaan Hattingh Jun 19 '16 at 16:06
  • $\begingroup$ Thank you for those pointers! I'm surprised H&K didn't at least tell the reader that this was true in general rather than leave us wondering. Maybe it's in there somewhere but I have yet to find it. But I must say it's a good thing, if this wasn't true then life would just be that much more complicated. $\endgroup$ – Gregory Grant Jun 19 '16 at 16:19
  • $\begingroup$ condensed answers are also available at the wiki entry under additional properties: en.wikipedia.org/wiki/Row_equivalence#Additional_properties . But it seems to refer primarily to inner product spaces (orthogonal complement) and the sources aren't really referenced...the detail you will find in my previous links. $\endgroup$ – Christiaan Hattingh Jun 19 '16 at 16:33
  • $\begingroup$ @ChristiaanHattingh Can you please explain your last sentence? Specifically, if you set the vectors of two different bases of a subspace as the row vectors of two different matrices and then take the RREF of those matrices, can you prove that they will be the same RREF? Thanks for the help! $\endgroup$ – Noble Mushtak Dec 29 '16 at 15:42

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