Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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
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 $\mathbb{R}^2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.