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There's a question in Jim Hefferon's Linear algebra book (chapter 1 section 1 problem 3.22), that asks what "happens" when a homogenous system has only 0=0 equations after applying Gaussian operations. By what happens, I'm guessing the question is about what the solution set looks like.

Let's say that the system is over $m$ unknowns. At the moment, I'm thinking that the solution set will look like $\{ (c_1, c_2 \ldots c_m) | c_1, c_2 \ldots c_m \in \mathbb R \}$; none of the variables are leading variables, so they can all be parameters to describing the solution set. This seems like kind of weak reasoning to me, though, does anyone have a better way of explaining why this is true? I guess I'm also assuming that I'm correct, so if I'm wrong please explain why as well.

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    $\begingroup$ Something to consider: how can one end up with only $0=0$ equations after applying row operations? (e.g. what must you start with?) $\endgroup$ – apnorton Jun 27 '13 at 2:35
  • $\begingroup$ anorton:I guess if you had a system like $x + 3y = 4, 2x + 6y = 8$, you could simultaneously perform the Gaussian operations $-2(\rho_1) + \rho_2$ and $-\frac{1}{2}(\rho_2) + \rho_1$ to make both equations 0 = 0. In this case, however, the system isn't underspecified; you're just "throwing away" both equations. So, the solution set you come up with won't be correct. Not sure how else you'd end up with all 0 = 0 equations though. $\endgroup$ – cer0 Jun 27 '13 at 19:11
  • $\begingroup$ Actually, I don't think doing two row operations simultaneously is a correct way to end up with all $0=0$ equations. Doing multiple operations at each step is just a notation convenience. The theorem for Gauss's method states that the solution sets are identical for a system before and after the application of a single Gaussian operation, so doing two operations at once violates the guarantee of this theorem. $\endgroup$ – cer0 Jul 11 '13 at 17:47
  • $\begingroup$ @cero Your most recent message is correct. (As a side note: I'm still trying to think of a system other than $0=0$ that will row reduce to a zero matrix... :)) $\endgroup$ – apnorton Jul 11 '13 at 17:58
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Yes, you are correct. The system is underconstrained, so it makes sense that there are free variables. In fact, the system has no constraints (since each equation is a tautology that is always true, regardless of the value of any of the variables), so all variables are free variables.

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