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 $\mathbb{R}^2$.

Question

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?

Answer 1

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.

Answer 2

$ \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.