\[V\]
be the vector space of polynomials of degree 2 and let \[\phi_1 , \phi+2 : V \rightarrow \mathbb{R}\]
be linear functionals on \[V|\]
defined by\[\phi_1 (\mathbf{v})= \int^1_0 (a+bx)dx\]
\[\phi_2 (\mathbf{v})= \int^2_0 (a+bx)dx\]
We want to find a basis
\[\{ \mathbf{v}_1, \mathbf{v}_2 \}\]
of \[V\]
dual to \[\{ \phi_1 , \phi_2 \}\]
.The vector space
\[V\]
has dimension 2 since \[\{1,x\}\]
forms a basis. The set of all linear functions from \[V\]
to \[\mathbb{R}\]
forms a vector space of dimension 2. The dual vector space \[V^*\]
spanned by \[\{ \phi_1 , \phi_2 \}\]
has dimension 2 since \[ \phi_1 , \phi_2 \]
are linearly independent so form a basis for \[V^*\]
.We can use this basis to find the dual basis for
\[V\]
using the following theorem.Let
\[\{ f_1, \; f_2 \}\]
be a basis for \[V^*\]
and let \[\{ \mathbf{e}_1, \; \mathbf{e}_2 \}\]
be a basis for \[V\]
then\[f_i (\mathbf{e}_j)= \left\{ \begin{array}{cc} 1 & i=j \\ 0 & i \neq j \end{array} \right. \]
Let
\[\mathbf{e}_1=a_1+b_1x, \; \mathbf{e}_2=a_2+b_2x \]
. Then\[\phi_1 (\mathbf{e}_1)= \int^1_0 (a_1+b_1x)dx=a_1+\frac{b_1}{2}=1\]
\[\phi_2 (\mathbf{e}_1)= \int^2_0 (a_1+b_1x)dx=2a_1+2b_1=0\]
\[\phi_1 (\mathbf{e}_2)= \int^1_0 (a_2+b_2x)dx=a_2+\frac{b_2}{2}=0\]
\[\phi_2 (\mathbf{e}_2)= \int^2_0 (a_2+b_2x)dx=2a_2+2b_2=1\]
Solving these equations for
\[a_1, \; b_1. \; a_2, \; b_2\]
gives \[a_1=2, \; b_1=-2, \; a_2=-1, \; b_2=2\]
. Hence the basis dual to \[\{ \phi_1, \; \phi_2 \}\]
is \[\{ 2-2x, \; -1+2x\]
.