###### 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*}
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