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The usual method I see for checking if vectors are LD is to set them as column vectors and perform RREF. However, why can't one just set them as row vectors and perform RREF? If there are no rows of all 0s, then they are LI, otherwise they are LD.

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    $\begingroup$ Of course. Why can't you do that? It's perfectly fine. $\endgroup$ – Teresa Lisbon Sep 26 '16 at 4:10
  • $\begingroup$ Idk I guess I just never saw it anywhere so I thought there was something obvious I was missing. $\endgroup$ – sir_thursday Sep 26 '16 at 4:11
  • $\begingroup$ It's not obvious, the above fact. Just follows from row rank = column rank. $\endgroup$ – Teresa Lisbon Sep 26 '16 at 4:13
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    $\begingroup$ Actually sticking them in the rows will not only tell you if the rows are linearly independent, but will get you a (generally) simpler set of basis vectors for the space regardless of whether the vectors are linearly independent or not. So I will usually do it that way (in the rare circumstance where I do it by hand). $\endgroup$ – user137731 Sep 26 '16 at 4:13

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