###### Example 3.2.5

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

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

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

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

To illustrate, if the coordinates of \(\xvec\) in the basis \(\bcal\) are

then

We conclude that

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

where

or

This linear system for the weights defines an augmented matrix

Therefore,