Bases Transformation

Bases Transformation

Updated: October 27, 2020

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 $$(\mathbf{e}_{1},\mathbf{e}_{2})$$:

\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}

(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{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}

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{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}

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