Lunear Functional of a Arbitrary Vector Given Values of the Functional on a Basis
\[V\]
be a vector space in \[\mathbb{R}^3\]
with basis \[\left\{ \mathbf{e}_1 , \; \mathbf{e}_2, \; \mathbf{e}_3 \right\}= \left\{ \begin{pmatrix}1\\0\\1\end{pmatrix} , \begin{pmatrix}0\\1\\-2\end{pmatrix} , \begin{pmatrix}-1\\-1\\0\end{pmatrix} \right\}\]
.The linear functional
\[f\]
is such that\[f( \mathbf{e}_1)=1, \; f( \mathbf{e}_2 )=2, \; f( \mathbf{e}_3)=3, \;\]
Find
\[f \begin{pmatrix}a\\b\\c\end{pmatrix} \]
.We can write
\[ \begin{pmatrix}a\\b\\c\end{pmatrix} =c_1 \begin{pmatrix}1\\0\\1\end{pmatrix} +c_2 \begin{pmatrix}0\\1\\-2\end{pmatrix} +c_3 \begin{pmatrix}-1\\-1\\0\end{pmatrix} = \begin{pmatrix}c_1-c_3\\c_2-c_3\\c_1-2c_2\end{pmatrix}\]
Solving for
\[c_1, \; c_2, \;c_3\]
in terms of \[a, \; b, \; c\]
gives \[c_1=2a-2b-c, \; c_2=a-b-c, \; c_3=a-2b-c\]
.Hence
\[ \begin{pmatrix}a\\b\\c\end{pmatrix} =(2a-2b-c) \begin{pmatrix}1\\0\\1\end{pmatrix} +(a-b-c) \begin{pmatrix}0\\1\\-2\end{pmatrix} +(a-2b-c) \begin{pmatrix}-1\\-1\\0\end{pmatrix} \]
Therefore
\[\begin{equation} \begin{aligned} f(a,b,c) &= (2a-2b-c) f \begin{pmatrix}1\\0\\1\end{pmatrix} +(a-b-c) \begin{pmatrix}0\\1\\-2\end{pmatrix} +(a-2b-c) \begin{pmatrix}-1\\-1\\0\end{pmatrix} \\ &= (2a-2b-c) \times 1 + (a-b-c) \times 2 + (a-2b-c) \times 3 \\ &= 4a-7b-3c \end{aligned} \end{equation}\]
