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I would like to show the following claim:

Let $l_1, l_2 \in \mathbb{R}^{n+1}$ be two distinct lines then $\exists$ a linear isomorphism $A : \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$ with $Al_1 = (\lambda, 0, 0, \dots , 0)$ and $Al_2 = (0, \lambda, 0, \dots , 0)$

I thought this $A$ needed to prove the claim is the basis transformation matrix from the standard basis of $\mathbb{R}^{n+1}$ into the basis consisting of $l_1$, $l_2$ and some linearly independent basis vectors.

My question now is: Are my thoughts correct? And if so, how can I construct the missing vectors to make $\{l_1, l_2 \}$ into a basis of dimension $n + 1$?

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I think your sentence "I thought..." is gramatically messed up somehow. :) – Rasmus Mar 4 '11 at 9:39
@Rasmus: No, I think it's OK -- it's easier to parse if you replace "this $A$" by "that this $A$ that's", but it's OK as it is. – joriki Mar 4 '11 at 10:13
Why do you write $\mathbb{R}^{n+1}$ and not $\mathbb{R}^{n}$? – joriki Mar 4 '11 at 10:20
@joriki: The actual problem I am thinking about is something related to the real projective plane and the question here is a sub-problem of my actual problem. In the actual problem I am thinking about $\mathbb{R}^{n+1}$ but the dimension doesn't matter for this sub-problem so I left it to be $n + 1$. – Rudy the Reindeer Mar 4 '11 at 10:47
@joriki: Right, thanks. – Rasmus Mar 4 '11 at 11:46
up vote 3 down vote accepted

Your thinking is basically correct, except: $l_1$, $l_2$ are vectors, not lines; and you have to exclude the case that one of them is zero.

About extending the basis, see this and this.

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many thanks, the first link is exactly what I was looking for. In fact, I googled for it for a while but somehow failed to find it even though I was certain it had to be "somewhere out there". – Rudy the Reindeer Mar 4 '11 at 10:51

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