##### Example3.2.5

In this section's preview activity, we considered the vectors

\begin{equation*} \vvec_1 = \twovec{2}{1}, \vvec_2 = \twovec{1}{2} \text{,} \end{equation*}which form a basis \(\bcal=\{\vvec_1,\vvec_2\}\) for \(\real^2\text{.}\)

In the standard coordinate system, the point \((2,-3)\) is found by moving 2 units to the right and 3 units down. We would like to define a new coordinate system where we interpret \((2,-3)\) to mean we move two times along \(\vvec_1\) and 3 times along \(-\vvec_2\text{.}\) As we see in the figure, doing so leaves us at the point \((1,-4)\text{,}\) expressed in the usual coordinate system.

We have seen that

\begin{equation*} \xvec = \twovec{1}{-4} = 2\vvec_1 - 3\vvec_2 \text{.} \end{equation*}The coordinates of the vector \(\xvec\) in the new coordinate system are the weights that we use to create \(\xvec\) as a linear combination of \(\vvec_1\) and \(\vvec_2\text{.}\)

Since we now have two descriptions of the vector \(\xvec\text{,}\) we need some notation to keep track of which coordinate system we are using. Because \(\twovec{1}{-4} = 2\vvec_1 - 3\vvec_2\text{,}\) we will write

\begin{equation*} \coords{\twovec{1}{-4}}{\bcal} = \twovec{2}{-3} \text{.} \end{equation*}More generally, \(\coords{\xvec}{\bcal}\) will denote the coordinates of \(\xvec\) in the basis \(\bcal\text{;}\) that is, \(\coords{\xvec}{\bcal}\) is the vector \(\twovec{c_1}{c_2}\) of weights such that

\begin{equation*} \xvec = c_1\vvec_1 + c_2 \vvec_2 \text{.} \end{equation*}To illustrate, if the coordinates of \(\xvec\) in the basis \(\bcal\) are

\begin{equation*} \coords{\xvec}{\bcal} = \twovec{5}{-2} \text{,} \end{equation*}then

\begin{equation*} \xvec = 5\vvec_1 - 2\vvec_2 = 5\twovec{2}{1}-2\twovec{1}{2} = \twovec{8}{3} \text{.} \end{equation*}We conclude that

\begin{equation*} \coords{\twovec{8}{3}}{\bcal} = \twovec{5}{-2} \text{.} \end{equation*}This demonstrates how we can translate coordinates in the basis \(\bcal\) into standard coordinates.

Suppose we know the expression of a vector \(\xvec\) in standard coordinates. How can we find its coordinates in the basis \(\bcal\text{?}\) For instance, suppose \(\xvec=\twovec{-8}{2}\) and that we would like to find \(\coords{\xvec}{\bcal}\text{.}\) We have

\begin{equation*} \coords{\twovec{-8}{2}}{\bcal}=\twovec{c_1}{c_2} \end{equation*}where

\begin{equation*} \twovec{-8}{2} = c_1\vvec_1 + c_2\vvec_2 \end{equation*}or

\begin{equation*} c_1 \twovec{2}{1} + c_2 \twovec{1}{2} = \twovec{-8}{2}. \end{equation*}This linear system for the weights defines an augmented matrix

\begin{equation*} \left[\begin{array}{rr|r} 2 \amp 1 \amp -8 \\ 1 \amp 2 \amp 2 \\ \end{array}\right] \sim \left[\begin{array}{rr|r} 1 \amp 0 \amp -6 \\ 0 \amp 1 \amp 4 \\ \end{array}\right] \text{.} \end{equation*}Therefore,

\begin{equation*} \coords{\twovec{-8}{2}}{\bcal} = \twovec{-6}{4} \text{.} \end{equation*}