"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."
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I think echelon implies leading ones and zeroes above leading 1s, but I don't have Lay handy.
ReplyDeleteYou've got that right, John.
ReplyDeleteAre those matrices not row equivalent to each other? You can multiply the second row by any arbitrary value (so 0, 0, -3 can be changed into 0, 0, 2.5) and you can then add that to the first row to change the 3rd column value to any arbitrary value.
ReplyDeleteThey are row equivalent, but they are not the same. I mistakenly thought they were in reduced echelon form. Silly me.
DeleteOh! So, in an augmented matrix (last column represents the constant on the right side of the =), I still need to finish up, and I'll get 0 1 on the last column of both of these. Duh. (Since we always stop as soon as we know a system is inconsistent, I hadn't thought about including the last column in my process.)
ReplyDeleteI was going to post in math stack exchange. Thank goodness I didn't. Feeling silly...
So now my question becomes whether we can consider the three cases (one, many, or no solutions), and build a simpler proof than Lay's. I'll keep thinking about that.