# Reordering an unknown linear algebra theorem and proof into the correct order

Good day everyone,

I have got a little puzzle that needs solving. A friend has given me a theorem and its proof which has been jumbled-up with two pieces introduced into it that do not belong in the text.

I am currently trying to re-order the pieces and eliminate the two incorrect portions so that the text makes sense.

So far I've figured a starting point and two different combinations. Namely the starting point being number 6 and one combination of 13, 10, 11 and 14, 8 but I'm not too sure if these are correct.

Each portion of text has a number, written on the right-hand side of the page. The answer I am trying to achieve should simply consist of an ordered list of these numbers.

Here is the problem below:

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Here's a start: You need to start with a well formed Theorem. 6 is first. 4 would have to next to complete the Theorem (6. by itself is not a theorem as there is no result, only a hypothesis). Obviously, 2. is next to start the proof. – David Mitra Apr 27 '12 at 17:00

Clearly 6 must be the first block. However, it’s only the hypothesis of the theorem; we still need the conclusion. The only block that simply states a conclusion is 4, so it comes next, and it must be followed by 2, the beginning of the proof. You are correct about the sequences 13,10,11 and 14,8.

Now you can either try to extend the pieces that you have, try to work from the beginning of the proof, or try to work back from the end. In fact it will be easiest to do all of these.

First, there’s a symbol at the end of one block that clearly marks it as the last block of the proof; once you find that block, you can set it aside temporarily.

The proof begins by defining a function $\varphi:\Bbb R^n[x]\to\Bbb R^{n+1}$. The domain and codomain of $\varphi$ are vector spaces, so it seems likely that we want $\varphi$ to be a linear transformation, and we probably want it to have other nice properties as well. This should suggest that 1 is the next block after 2. Note that it lays out a plan of attack: the proof will continue by showing that $\varphi$ is surjective and injective. Where next? Not 13: ‘We can now show’ [emphasis added] is a fairly clear indication that some prerequisite result has already been shown. In fact 14 seems to be the only block that can follow here, just on the basis of the English, and it makes sense that 14 would come next: we’ve just said that we need to show that $\varphi$ is both surjective and injective, and that does mean that we aim to show that $\varphi$ is bijective. Thus, we’ve connected the beginning with one of your sequences to make 6,4,2,1,14,8.

Now 14 and 8 introduce the polynomials $q_{\,r}$ and calculate $q_{\,r}^{(m)}(x)$, presumably in order to do something with them. Where else do they appear? In your 13 sequence, and in 12. Note that 12 and 10 both conclude that $\varphi$ is surjective, so one of them is almost certainly one of the intrusive blocks. But you already have good evidence that one of the two is part of the proof, so that’s what probably follows 8, and the other is the intrusion.

Now perhaps you should try to extend the 13,10,11 sequence. 11 ends with an incomplete sentence about the rank-nullity theorem, so its continuation ought to be fairly easy to find. In fact the only possibilities are 5 and 7; which one states a conclusion that uses the rank-nullity theorem?

Once you get to here, there’s very little left to do, and the logic of the proof should be clear enough to let you finish the job fairly easily.

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