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:


*

*Let $v_x = \langle c_1, c_2\rangle$ be any vector in $\mathbb R^2$ where $c_1$ and $c_2$ are in $\mathbb R$.

*$v_x = c_1\langle 1, 0\rangle + c_2\langle 0, 1\rangle$ 

*Set $v_x$ = a linear combination of $v_1$ and $v_2$? How do I proceed from here?


We want to show that any $\mathbf{v} = (x,y) \in \mathbb{R}^2$ can be written as $v = a\mathbf{v_1}+b\mathbf{v_2}$. This equation can be written more explicitly like this: $\begin{pmatrix}x \\ y\end{pmatrix} = a\begin{pmatrix}2 \\ 1\end{pmatrix} + b\begin{pmatrix}4 \\ 3\end{pmatrix}$, because $\mathbf{v_1} = \begin{pmatrix}2 \\ 1\end{pmatrix}$ and $\mathbf{v_2} = \begin{pmatrix}4 \\ 3\end{pmatrix}$. We'd like to show that no matter what $(x,y)$ is, there always are $a,b$ that satisfy the above equation.
So let's do that. This equation is a vector equation, so it is really two equations. We can separate it:
$$
\begin{align}
2a+4b &= x \\
a+3b &= y \end{align}
$$
Remember that the unknowns here are $a$ and $b$. $x$ and $y$ can be anything; in fact we want to show that no matter what $x$ and $y$ are, there are $a$ and $b$ that satisfy these equations. If we say $A = \begin{pmatrix}2 & 4 \\ 1 & 3\end{pmatrix},\ \mathbf{w} = \begin{pmatrix}a \\ b \end{pmatrix},\ \mathbf{x} = \begin{pmatrix}x \\ y \end{pmatrix}$. This system of equations can be written more succintly as $A\mathbf{w} = \mathbf{x}$. Again, remember that the unknown is $\mathbf{w}$, not $\mathbf{x}$.
We know that this system has a solution $\mathbf{w}$ for every $\mathbf{x}$ if and only if $\det(A) \neq 0$. If we can show that, we have shown that no matter which $(x,y) \in \mathbb{R}^2$ we pick, there are $a,b$ such that $\begin{pmatrix}x \\ y\end{pmatrix} = a\begin{pmatrix}2 \\ 1\end{pmatrix} + b\begin{pmatrix}4 \\ 3\end{pmatrix}$. I leave this calculation to you.
