"Uniqueness of the Reduced Echelon Form
Each matrix is row equivalent to one and only one reduced echelon matrix."
The proof is in an appendix, which is a bummer, because this class feels like it could build from first principles nicely up to all its glory. The proof involves material from chapter 4, and I have to fight my way through it. Isn't he worried about being circular?
I was thinking out loud in class. I said (more or less):
If the system is consistent, it has a particular solution set. You can read the solution off from the reduced echelon form, so it can only give you one answer. [In class I wasn't thinking about free variables, and whether those could be different somehow. I was just thinking about problems with one unique solution.] We know it gives the right answer because
we've already shown that elementary row operations create row equivalent matrices, which have the same solution set.
What about an inconsistent system? I'm not sure about that. If you can break his theorem, I'll give you extra credit.
Well, I just broke his theorem, I think. (I hope none of my students are reading my blog yet.) Given the system
Have I broken his theorem? Should he have said this instead?
"Each matrix representing a consistent system of equations is row equivalent to one and only one reduced echelon matrix."