# 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?
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$v_1=$ what? Similarly, $v_2=$ what? – Alex Becker Sep 20 '12 at 1:32
I am having an error rendering the <> symbols. Any suggestions? It shows up fine when I am editing – CodeKingPlusPlus Sep 20 '12 at 1:34
Fixed it, thanks! – CodeKingPlusPlus Sep 20 '12 at 1:36
I need to explicitly show that the vectors $v_1$ and $v_2$ span $\mathbb R^2$ – CodeKingPlusPlus Sep 20 '12 at 1:43

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.

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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.

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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.

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I need to only use the definition of Span. – CodeKingPlusPlus Sep 20 '12 at 1:47
This indeed uses only the defintion. I'll edit to make it more explicit. – Javier Sep 20 '12 at 1:48
The above answer, and also mine, use the definition of span, but you need to show a solution exists and for this nothing as clear, simple and basic as linear equations...what else?! – DonAntonio Sep 20 '12 at 1:56

Well, suppose $e_1=\langle 1,0\rangle$ and $e_2=\langle 0,1\rangle$--that is, $\{e_1,e_2\}$ is the standard ordered basis for $\Bbb R^2$. It's readily seen that $e_2=v_2-2v_1$, and that $e_1=\frac12(v_1-e_2)=\frac12(3v_1-v_2)$. At that point, you can write every vector in $\Bbb R^2$ uniquely as a linear combination of $v_1,v_2$, pretty much exactly as you described.

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