\[V\]
be a vector space in \[\mathbb{R}^3\]
with basis \[\left\{ \mathbf{e}_1 , \; \mathbf{e}_2, \; \mathbf{e}_3 \right\}= \left\{ \begin{pmatrix}1\\0\\1\end{pmatrix} , \begin{pmatrix}0\\1\\-2\end{pmatrix} , \begin{pmatrix}-1\\-1\\0\end{pmatrix} \right\}\]
.Find a linear functional
\[f\]
is such that\[f( \mathbf{e}_1)=0, \; f( \mathbf{e}_2)=0, \; f( \mathbf{e}_3) \neq 0\]
Since we have a basis for
\[V\]
for the dual space \[V^*\]
consisting of unique functionals \[f_i, \; i=1, \; 2, \; 3\]
such that \[f_i(\mathbf{e}_j)= \delta_{ij}\]
where \[\delta_{ij} =1, \; i=j, \; 0 \; i \neq j\]
.Then
\[f_1( \mathbf{e}_1)=f_1 \begin{pmatrix}1\\0\\1\end{pmatrix} =1, \; f_1 ( \mathbf{e}_2)=f_1 \begin{pmatrix}0\\1\\-2\end{pmatrix} =0, \; f_1( \mathbf{e}_3)=f_1 \begin{pmatrix}-1\\-1\\0\end{pmatrix} =0 \]
\[f_2( \mathbf{e}_1)=f_2 \begin{pmatrix}1\\0\\1\end{pmatrix} =0, \; f_2( \mathbf{e}_2)=f_2 \begin{pmatrix}0\\1\\-2\end{pmatrix} =0, \; f_3( \mathbf{e}_3)=f_3 \begin{pmatrix}-1\\-1\\0\end{pmatrix} =1 \]
\[f_3\]
is the required function. \[f_3\]
is of the form \[f(x,y,z+=c_1x+c_2y+c_3z\]
.Hence
\[f_3 \begin{pmatrix}1\\0\\1\end{pmatrix} =c_1+c_3=0\]
\[f_3 \begin{pmatrix}0\\1\\-2\end{pmatrix} =c_2-2c_3=0\]
\[f_3 \begin{pmatrix}-1\\-1\\0\end{pmatrix} =-c_1-c_3=1\]
These simultaneous equations have the solution
\[c_1=1, \; c_2=-2, \; c_3=-1\]
.Hence
\[f_3(x,y,z)=x_1-2x_2-x_3\]
.