Finding a Linear Functional defined By its Value on Basis Vectors

Let  
\[V\]
  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  
\[f_3(x,y,z)=x_1-2x_2-x_3\]
.

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