The Transition Matrix

Suppose we have a vector space
$V$
with bases
$\{ \mathbf{e}_i \}$
and
$\{ \mathbf{f}_i \}$
.
The transition matrix from
$\{ \mathbf{e}_i \}$
to
$\{ \mathbf{f}_i \}$
.
is the matrix
$B$
satisfying
$\mathbf{e}_i = B \mathbf{f}_j$
where
$B= \left( \begin{array}{cc} b_{11} & b_{21} \\ b_{12} & b_{22} \end{array} \right)$

Solving to find this matrix is equivalent to expressing the
$\{ \mathbf{e}_i \}$
in terms of the
$\{ \mathbf{f}_i \}$
.
Suppose
$\{ \mathbf{e}_i \} = \left\{ \begin{pmatrix}1\\0\end{pmatrix} , \begin{pmatrix}0\\1\end{pmatrix} \right\} \: \{ \mathbf{f}_i \} = \left\{ \begin{pmatrix}1\\1\end{pmatrix} , \begin{pmatrix}-1\\0\end{pmatrix} \right\}$

We solve
$\begin{pmatrix}1\\0\end{pmatrix} = b_{11} \begin{pmatrix}1\\1\end{pmatrix} +b_{12} \begin{pmatrix}-1\\0\end{pmatrix}$

and
$\begin{pmatrix}0\\1\end{pmatrix} = b_{21} \begin{pmatrix}1\\1\end{pmatrix} +b_{22} \begin{pmatrix}-1\\0\end{pmatrix}$

The first of these gives us the simultaneous equations
$1=b_{11}-b_{12}$
and
$0=b_{11}$

Hence
$b_{11}=0, \: b_{12}=-1$
The second returns the simultaneous equations
$0=b_{21}-b_{22}, \: 1=b_{21}$

Hence
$b_{21}=b_{22}=1$
.
The transition matrix is
$\left( \begin{array}{cc} 0 & 1 \\ -1 & 1 \end{array} \right)$