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Why is the following tuple of vectors not a basis of $\mathbb{R}^{2}$? $\left ( \left ( \begin{array}{c} 1\\ 0\\0 \end{array} \right ), \left ( \begin{array}{c} 0\\ 1\\0 \end{array} \right ) \right )$

I would've thought that just like $\left ( \left ( \begin{array}{c} 1\\ 0\end{array} \right ), \left ( \begin{array}{c} 0\\ 1 \end{array} \right ) \right )$ the 1st tuple would span $\mathbb{R}^{2}$ and the vectors clearly linearly independent...

Hopefully I'm not fundamentally misunderstanding something, I'm doing one of those "self-test at the end of the chapter only solutions, no explanations" things. It'd be great if anyone could offer a brief explanation so that I can move on... thanks!

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Yeah as TonyK mentions, first the basis should be an element of $\mathbb{R}^{2}$. So this is not a basis, where as if you consider $\mathbb{R}^{2}$ as a subspace of $\mathbb{R}^{3}$ then yes. –  anonymous Jan 25 '11 at 15:45
    
The word you want is uple. –  Mariano Suárez-Alvarez Jan 25 '11 at 16:07
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No it's not, it's tuple. –  TonyK Jan 25 '11 at 16:12
    
@Mariano: Am I missing some joke? I'd call this a pair. (And I see tupel a lot if one doesn't want to specify the number of entries.) –  Hendrik Vogt Jan 25 '11 at 16:12
    
@Mariano, thanks. I just edited the question with the correct spelling. –  ghshtalt Jan 25 '11 at 16:13

1 Answer 1

up vote 9 down vote accepted

Strictly speaking, it is the basis of a two-dimensional subspace of $\mathbb{R}^3$, which happens to be isomorphic to $\mathbb{R}^2$.

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