R^2/(2,1),(-1,3)/Dual basis/Standard dual basis/Example

We consider ${}\mathbb {R} ^{2}$ with the standard basis ${}e_{1},e_{2}$, its dual basis ${}e_{1}^{*},e_{2}^{*}$, and the basis consisting in ${}u_{1}={\begin{pmatrix}2\\1\end{pmatrix}}$ and ${}u_{2}={\begin{pmatrix}-1\\3\end{pmatrix}}$. We want to express the dual basis ${}u_{1}^{*}$ and ${}u_{2}^{*}$ as a linear combination of the standard dual basis, that is, we want to determine the coefficients ${}a$ and ${}b$ (and ${}c$ and ${}d$) in

${}u_{1}^{*}=ae_{1}^{*}+be_{2}^{*}\,$

(and in ${}u_{2}^{*}=ce_{1}^{*}+de_{2}^{*}$). Here, ${}a=u_{1}^{*}(e_{1})$ and ${}b=u_{1}^{*}(e_{2})$. In order to compute this, we have to express ${}e_{1}$ and ${}e_{2}$ as a linear combination of ${}u_{1}$ and ${}u_{2}$. This is

${}e_{1}={\frac {3}{7}}{\begin{pmatrix}2\\1\end{pmatrix}}-{\frac {1}{7}}{\begin{pmatrix}-1\\3\end{pmatrix}}\,$

and

${}e_{2}={\frac {1}{7}}{\begin{pmatrix}2\\1\end{pmatrix}}+{\frac {2}{7}}{\begin{pmatrix}-1\\3\end{pmatrix}}\,.$

Therefore, we have

${}a=u_{1}^{*}(e_{1})=u_{1}^{*}{\left({\frac {3}{7}}u_{1}-{\frac {1}{7}}u_{2}\right)}={\frac {3}{7}}\,$

and

${}b=u_{1}^{*}(e_{2})=u_{1}^{*}{\left({\frac {1}{7}}u_{1}+{\frac {2}{7}}u_{2}\right)}={\frac {1}{7}}\,.$

Hence,

${}u_{1}^{*}={\frac {3}{7}}e_{1}^{*}+{\frac {1}{7}}e_{2}^{*}\,.$

With similar computations we get

${}u_{2}^{*}=-{\frac {1}{7}}e_{1}^{*}+{\frac {2}{7}}e_{2}^{*}\,.$

The transformation matrix from ${}u^{*}$ to ${}e^{*}$ is thus

${}M_{\mathfrak {e^{*}}}^{\mathfrak {u^{*}}}={\begin{pmatrix}{\frac {3}{7}}&-{\frac {1}{7}}\\{\frac {1}{7}}&{\frac {2}{7}}\end{pmatrix}}\,.$

The transposed matrix of this is

${}{{\left(M_{\mathfrak {e^{*}}}^{\mathfrak {u^{*}}}\right)}^{\text{tr}}}={\begin{pmatrix}{\frac {3}{7}}&{\frac {1}{7}}\\-{\frac {1}{7}}&{\frac {2}{7}}\end{pmatrix}}={\begin{pmatrix}2&-1\\1&3\end{pmatrix}}^{-1}={\left(M_{\mathfrak {e}}^{\mathfrak {u}}\right)}^{-1}\,.$

The inverse task to express the standard dual basis with ${}u_{1}^{*}$ and ${}u_{2}^{*}$, is easier to solve, because we can read of directly the representations of the ${}u_{i}$ with respect to the standard basis. We have

${}e_{1}^{*}=2u_{1}^{*}+u_{2}^{*}\,$

and

${}e_{2}^{*}=-u_{1}^{*}+3u_{2}^{*}\,,$

as becomes clear by evaluation ${}u_{1},u_{2}$ on both sides.