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True/false: If $\text{span}\left\{v_1,v_2\right\}= \mathbb{R}^2$ for $v_1,v_2 \in \mathbb{R}^2$ then $(v_1,v_2)$ is a basis of $\mathbb{R}^2$.

I believe the statement is true.

Basis means we need linearly independent vectors and $\text{span}\left \{v_1,v_2 \right\}$ means the vectors $v_1,v_2$ are linearly independent and thus a basis of $\mathbb{R}^2.$

Correct or not?

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  • $\begingroup$ I think you need to expand a bit on why $v_1$ and $v_2$ are linearly independent. $\endgroup$ – amd Mar 16 '17 at 19:32
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EDIT

The statement is true because you took exactly 2 vectors.

Suppose we took for instance 3 vectors instead of 2, say $(1,0)$, $(0,1)$ and $(2,0)$.

The span of all of them is $\mathbb{R}^2$ but they are not a basis in $\mathbb{R}^2$ because they are not linearly independent.

Instead, $\text{span}\{(1,0),(0,1)\}=\mathbb{R}^2$ and these vectors are linearly independent, so they form a basis in $\mathbb{R}^2$.

In general, we define the span of some vectors as all possible linear combinations of those vectors. And what we need for a set of vectors $S$ to form a basis of a vector space $V$ is the following:

  1. $\text{span}(S)=V$.
  2. The set $S$ is linearly independent.
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  • $\begingroup$ But he wrote $\{v_1, v_2\}$ in his question and the statement is true for 2 vectors. You took 3 vectors as a counterexample. $\endgroup$ – windircurse Mar 16 '17 at 10:36
  • $\begingroup$ @windircurse You are right. I went for a counterexample too fast because I saw a misunderstanding of what $\text{span}$ is. Editing the answer now... $\endgroup$ – Edu Mar 16 '17 at 10:39
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$ \text{Span}\{v_1,v_2\}$ is a sub space of $\mathbb{R}^2$ Now both have same dimension. Now let if possible $v_1,v_2$ are linearly dependent then $v_1=av_2 $ $,a\neq0$ then $v_1\in L(v_2)$ and this gives dimension of $\mathbb{R}^2$ is 1 which is a clear cut contradiction.

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  • $\begingroup$ And what does that mean? :P The statement would still be true, no? $\endgroup$ – tenepolis Mar 16 '17 at 10:48
  • $\begingroup$ $span$ is a shortcut for stating that the two vectors span a vector space - that is any vector of that space can be created from the vectors listed as the spanning set of the space. Now you need to think whether they constitute a basis. $\endgroup$ – Nox Mar 16 '17 at 10:56

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