###### Example 3.2.6

Let's consider the basis of \(\real^3\text{:}\)

\begin{equation*}
\bcal = \left\{
\threevec{1}{0}{-2},
\threevec{-2}{1}{0},
\threevec{1}{1}{2}
\right\}\text{.}
\end{equation*}

It is relatively straightforward to convert a vector's representation in this basis to the standard basis, using the matrix whose columns are the basis vectors:

\begin{equation*}
C_{\bcal} =
\left[\begin{array}{rrr}
1 \amp -2 \amp 1 \\
0 \amp 1 \amp 1 \\
-2 \amp 0 \amp 2 \\
\end{array}\right]\text{.}
\end{equation*}

For example, suppose that the vector \(\xvec\) is described in the coordinate system defined by the basis as \(\coords{\xvec}{\bcal} = \threevec{2}{-2}{1}\text{.}\) We then have

\begin{equation*}
\xvec = C_{\bcal}\coords{\xvec}{\bcal} =
\left[\begin{array}{rrr}
1 \amp -2 \amp 1 \\
0 \amp 1 \amp 1 \\
-2 \amp 0 \amp 2 \\
\end{array}\right]
\threevec{2}{-2}{1}
=
\threevec{7}{-1}{2}\text{.}
\end{equation*}

Consider now the vector \(\xvec=\threevec{3}{1}{-2}\text{.}\) If we would like to express \(\xvec\) in the coordinate system defined by \(\bcal\text{,}\) then we compute

\begin{equation*}
\coords{\xvec}{\bcal} = C^{-1}_{\bcal}\xvec
=
\left[\begin{array}{rrr}
\frac14 \amp \frac 12 \amp -\frac38 \\
-\frac14 \amp \frac12 \amp -\frac18 \\
\frac14 \amp \frac12 \amp \frac18 \\
\end{array}\right]
\threevec{3}{1}{-2}
=
\threevec{2}{0}{1}\text{.}
\end{equation*}