Basis for a Vector Space Dual to a Basis Set of Linear Functionals

Let  
\[V\]
  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  
\[\{ 2-2x, \; -1+2x\]
.

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