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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.2011-01-25
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    The word you want is *uple*.2011-01-25
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    No it's not, it's *tuple*.2011-01-25
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    @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.)2011-01-25
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    @Mariano, thanks. I just edited the question with the correct spelling.2011-01-25
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    Grrr, I missed the $t$ in the comment above. @Hendrik: it is spelt *tuple*.2011-01-25
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    @Mariano: Thanks a lot! In principle I know this, but from my mistake you can now try and infer my mother tongue.2011-01-25
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    @Hendrik: as far as my experience with lots of American and British people... your mother tongue is English! :)2011-01-25

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