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 $T\,=\,S^{-1}$.