Showing vectors span a vector space by definition

I need to show that the vectors $v_1 = \langle 2, 1\rangle$ and $v_2 = \langle 4, 3\rangle$ span $\mathbb R^2$ by definition. By definition if I can write any vector in $\mathbb R^2$ as a linear combination of $v_1$ and $v_2$ then the vectors span $\mathbb R^2$. How do I show this? Here is what I have been working with:

1. Let $v_x = \langle c_1, c_2\rangle$ be any vector in $\mathbb R^2$ where $c_1$ and $c_2$ are in $\mathbb R$.
2. $v_x = c_1\langle 1, 0\rangle + c_2\langle 0, 1\rangle$
3. Set $v_x$ = a linear combination of $v_1$ and $v_2$? How do I proceed from here?

Solution 1:

What you need to prove is the following:

$$\forall\,v:=(a,b)\in\Bbb R^2\,\,\,\exists\,x,y\in\Bbb R\,\,\,s.t.\,\, v=x(2,1)+y(4,3)\Longleftrightarrow$$

$$\Longleftrightarrow \text{the linear system}\,\,\left\{\begin{array}{}2x+4y=a\\{}\\\;\;x+3y=b\end{array}\right.$$

has a solution for any $\,a,b\in\Bbb R\,$

Now, the above system always has a solution (and, in fact, a unique one for each choice of $\,a,b\in\Bbb R\,$) since the reduced coefficients matrix's determinant is $\,2\cdot 3-1\cdot 4=2\neq 0\,$ , and voilá.

Solution 2:

If you can prove that $\langle 1,0\rangle$ and $\langle 0,1\rangle$ can be written as linear combinations of $v_1$ and $v_2$, and then combine that fact with statements 1 and 2, that would be enough.

In general, if you can show that all elements of one known basis can be written as linear combinations of a set of linearly independent vectors, then it follows that the aforementioned set is also a basis.