Bases Transformation

Consider two bases \((\mathbf{e}_{1},\mathbf{e}_{2})\) and \((\mathbf{\tilde{e}}_{1},\mathbf{\tilde{e}}_{2})\), where we consider the former the old basis and the latter the new basis.

Each vector \((\mathbf{\tilde{e}}_{1},\mathbf{\tilde{e}}_{2})\) can be expressed as a linear combination of \((\mathbf{e}_{1},\mathbf{e}_{2})\):

\begin{equation} \mathbf{\tilde{e}}_{1}\,=\,\mathbf{e}_{1}S^{1}_{1}\,+\,\mathbf{e}_{2}S^{2}_{1}\\\
\tag{1.0}\\\
\mathbf{\tilde{e}}_{2}\,=\,\mathbf{e}_{1}S^{1}_{2}\,+\,\mathbf{e}_{2}S^{2}_{2} \end{equation}

(1.0) is the basis transformation formula, and the object \(S\) is the direct transformation \(\{S^{j}_{i},\,1\,\leq\,i,\,j\,\leq\,2\}\), (assuming a \(2x2\) matrix) which can also be written in matrix form:

\begin{equation} \begin{bmatrix} \mathbf{\tilde{e}}_{1} & \mathbf{\tilde{e}}_{2} \end{bmatrix}\,=\, \begin{bmatrix} \mathbf{e}_{1} & \mathbf{e}_{2}\, \end{bmatrix} \begin{bmatrix} S^{1}_{1} & S^{1}_{2}\\\
S^{2}_{1} & S^{2}_{2} \end{bmatrix}\\\
=\, \begin{bmatrix} \mathbf{e}_{1} & \mathbf{e}_{2} \end{bmatrix}\,S\tag{1.1} \end{equation}

The resulting matrix is the direct transformation matrix from the old basis to the new basis. Note that the rows of \(S\) appear as superscripts and the columns appear as subscripts.

Converting in the reverse direction (new to old) requires doing \((1.1)\), but with the inverse matrix

\begin{equation} \begin{bmatrix} \mathbf{e}_{1} & \mathbf{e}_{2} \end{bmatrix}\,=\, \begin{bmatrix} \mathbf{\tilde{e}}_{1} & \mathbf{\tilde{e}}_{2}\, \end{bmatrix} \begin{bmatrix} S^{1}_{1} & S^{1}_{2}\\\
S^{2}_{1} & S^{2}_{2} \end{bmatrix}^{-1}\\\
=\, \begin{bmatrix} \mathbf{\tilde{e}}_{1} & \mathbf{\tilde{e}}_{2}\, \end{bmatrix} \begin{bmatrix} T^{1}_{1} & T^{1}_{2}\\\
T^{2}_{1} & T^{2}_{2} \end{bmatrix}\\\
=\, \begin{bmatrix} \mathbf{\tilde{e}}_{1} & \mathbf{\tilde{e}}_{2} \end{bmatrix}\,T\tag{1.2} \end{equation}

where \(T\) is the inverse transformation matrix, or \(T\,=\,S^{-1}\).