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

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

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

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

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

and

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

Therefore, we have

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

and

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

Hence,

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

With similar computations we get

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

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

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

The transposed matrix of this is

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

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

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

and

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

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