Why do two linearly independent vectors span $\mathbb{R}^2$?

Question

I understand it intuitively, and I know how to prove it geometrically, but what is the the algebraic proof of this fact.

Answer 1

Take $v=(a,b)$ and $w= (c,d)$. Linear independence means that the system

$$ \begin{cases} \alpha a + \beta c= 0 \\ \alpha b + \beta d=0 \end{cases}$$

has the only solution $\alpha= \beta =0$. Assume there exists a vector $(e,f)$ not spanned by those two vectors. Then there is no solution to

$$ \begin{cases} \alpha a + \beta c= e \\ \alpha b + \beta d=f \end{cases}$$

Assume $a,b,c,d \neq 0$, otherwise the thesis would be trivial (can you see it?). Then from the equations you find

$$ \alpha= (ed-ef)/(ad-cb) \text{ and } \beta=(fa-eb)/(ad-cb)$$

The denominator is the determinant of $\begin{bmatrix} a & c\\ b & d\end{bmatrix}$, and you can see easily its being $\neq 0$ (i.e. the second system having a solution) is equivalent to being linearly independent. Indeed you can write the first system (of linear independence) as

$$ \begin{cases} \alpha = - \beta c/a \\ \beta (ad-bc)/a=0 \end{cases}$$

Now if $ad -bc \neq 0$ you're forced to choose $\beta=0$, and consequently $\alpha=0$. Otherwise choose whatever $\beta$ you like and you can find a $\alpha$ which works.

Answer 2

Two linearly independent vectors span $\mathbb{R}^2$ because any vector in $\mathbb{R}^2$ can be expressed as a linear combination of the two linearly independent vectors.

For example, consider two linearly independent vectors in $\mathbb{R}^2$, $v_1 = [1, 2]$ and $v_2 = [3, 4]$. Any vector in $\mathbb{R}^2$, such as $[7, 10]$, can be written as a linear combination of $v_1$ and $v_2$.

$[7, 10] = 1 * [1, 2]+ 2 * [3, 4] $