# Vectors - Span RR^2?

## Any help here is much appreciated

May 3, 2018

A requirement for any two vectors to span ${\mathbb{R}}^{2}$ is that the vectors are linearly independent .

For convenience we normally use a natural basis for vectors based on a standard cartesian coordinate system. Thus we normally use standard vectors $\boldsymbol{\underline{\hat{i}}}$ and $\boldsymbol{\underline{\hat{j}}}$, or in column format the basis:

$\boldsymbol{{B}_{1}} = \left\{\boldsymbol{\underline{\hat{i}}} , \boldsymbol{\underline{\hat{j}}}\right\} = \left\{\begin{matrix}\begin{matrix}1 \\ 0\end{matrix} \\ \begin{matrix}0 \\ 1\end{matrix}\end{matrix}\right\}$

Using this basis $\boldsymbol{{B}_{1}}$ we can can represent any vector by just using its standard cartesian coordinates in the column vector, so that the coordinate $\left(3 , 2\right)$ is trivially represented as:

$\left(\begin{matrix}3 \\ 2\end{matrix}\right) = 3 \boldsymbol{\underline{\hat{i}}} + 2 \boldsymbol{\underline{\hat{j}}} = 3 \left(\begin{matrix}1 \\ 0\end{matrix}\right) + 2 \left(\begin{matrix}0 \\ 1\end{matrix}\right)$

However we can readily show that the vectors:

$\boldsymbol{\underline{u}} = \left(\begin{matrix}1 \\ 1\end{matrix}\right) \setminus \setminus$ and $\boldsymbol{\underline{v}} = \left(\begin{matrix}- 3 \\ 2\end{matrix}\right)$

are also linearly independent, and as such can also be used as a basis:

$\boldsymbol{{B}_{2}} = \left\{\boldsymbol{\underline{u}} , \boldsymbol{\underline{v}}\right\} = \left\{\begin{matrix}\begin{matrix}1 \\ 1\end{matrix} \\ \begin{matrix}- 3 \\ 2\end{matrix}\end{matrix}\right\}$

And now to represent the coordinate $\left(3 , 2\right)$, we seek $l a m \mathrm{da} , \mu \in \mathbb{R}$ st:

$\left(\begin{matrix}3 \\ 2\end{matrix}\right) = l a m \mathrm{da} \boldsymbol{\underline{u}} + \mu \boldsymbol{\underline{v}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = l a m \mathrm{da} \left(\begin{matrix}1 \\ 1\end{matrix}\right) + \mu \left(\begin{matrix}- 3 \\ 2\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 & - 3 \\ 1 & 2\end{matrix}\right) \left(\begin{matrix}l a m \mathrm{da} \\ \mu\end{matrix}\right)$

And solving this system, yields the solution:

$l a m \mathrm{da} = \frac{12}{5} \setminus \setminus$, and $\mu = - \frac{1}{5}$

Thus we can write the coordinate using this spanning basis $\boldsymbol{{B}_{2}}$ as:

$\left(\begin{matrix}3 \\ 2\end{matrix}\right) = \frac{12}{5} \left(\begin{matrix}1 \\ 1\end{matrix}\right) - \frac{1}{5} \left(\begin{matrix}- 3 \\ 2\end{matrix}\right)$