##### Example3.2.6

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*}