The Dual Basis For a Vector Space
\[V\]
be a vector space defined over a field \[F\]
. The set of all linear functionals (a linear functional sends \[\mathbf{v} \in V\]
to elements of \[F\]
) is also a vector space. If \[\phi_1 , \; \phi_2\]
are linear functionals and \[\mathbf{v}\]
is a fixed vector, then\[( \phi_1 + \phi_2 ) \mathbf{v} = \phi_1( \mathbf{v} )+ \phi_( \mathbf{v} )\]
\[ \phi ( c\mathbf{v}) =c \phi ( \mathbf{v} ))\]
where
\[c \in F\]
.Let
\[V^*\]
be the set of all such linear functional;s. This set is closed under addition and scalar multiplication.Since
\[V^*\]
is a vector space it has a basis, and \[dim(V^*)=dim(V)\]
.Let
\[V= \mathbb{R}^3\]
with basis \[\left\{ \begin{pmatrix}1\\-1\\3\end{pmatrix} , \begin{pmatrix}0\\1\\-1\end{pmatrix}, \begin{pmatrix}0\\3\\-2\end{pmatrix} \right\}\]
.For any particular ordering of the bases, there is a unique matrix that sends the normal basis
\[\{ \mathbf{e}_i \}_i, \; i=1,2,...,n\]
for \[V\]
onto the basis for \[V^*\]
.There is a unique linear functional
\[f_i\]
satisfying \[f_i( \mathbf{e}_j)=\delta_{ij}\]
where \[\delta_{ij}=1, \;i=j, \; 0, i \neq j \]
.In this way we can find a set of
\[n\]
linearly independent linear functionals. This set forms a basis for \[V^*\]
, called the dual basis.We need so fin
\[\{ \phi_1 , \phi_2, \; \phi_3 \}\]
such that\[\phi_1 (x,y,z)=a_1x+a_2y+a_3z\]
\[\phi_2 (x,y,z)=b_1x+b_2y+b_3z\]
\[\phi_3 (x,y,z)=c_1x+c_2y+c_3z\]
and
\[\phi_1(\mathbf{v}_1)=1, \; \phi_1(\mathbf{v}_2)=0, \; \phi_1(\mathbf{v}_3)=0 \]
\[\phi_2(\mathbf{v}_1)=0, \; \phi_2(\mathbf{v}_2)=1, \; \phi_2(\mathbf{v}_3)=0 \]
\[\phi_3(\mathbf{v}_1)=0, \; \phi_3(\mathbf{v}_2)=0, \; \phi_3(\mathbf{v}_3)=1 \]
where
\[\mathbf{v}_1 = \begin{pmatrix}1\\-1\\3\end{pmatrix} , \; \mathbf{v}_2 = \begin{pmatrix}0\\1\\-1\end{pmatrix}, \; \mathbf{v}_3 = \begin{pmatrix}0\\3\\-2\end{pmatrix}\]
.We solve
\[\phi_1 (1,-1,3)=a_1-a_2+3a_3=1\]
\[\phi_1 (0,1,-1)=a_2-a_3=0\]
\[\phi_1 (0,3,-2)=3a_2-2a_3=0\]
The solution is
\[a_1=1, \; a_2=0, \; a_3=0 \rightarrow \phi_1 (x,y,z)=x\]
.\[\phi_2 (1,-1,3)=b_1-b_2+3b_3=0\]
\[\phi_2 (0,1,-1)=b_2-b_3=1\]
\[\phi_2 (0,3,-2)=3b_2-2b_3=0\]
The solution is
\[b_1=7, \; b_2=-2, \; b_3=-3 \rightarrow \phi_2 (x,y,z)=7x-2y-3z\]
.\[\phi_3 (1,-1,3)=c_1-c_2+3c_3=0\]
\[\phi_3 (0,1,-1)=c_2-c_3=0\]
\[\phi_3 (0,3,-2)=3c_2-2c_3=1\]
The solution is
\[c_1=-2, \; c_2=1, \; c_3=1 \rightarrow \phi_3 (x,y,z)=-2x+y+z\]
. 
