Suppose we have: $$ E= \mathbb{R^2}[t]=\{p(t)=a_0+a_1t+a_2t^2\} $$ $$m_0, m_1, m_2$$ Where $m_0, m_1, m_2$ are different real numbers.
We define $$F_i:E\rightarrow \mathbb{R}, i = 0,1,2$$ By $$F_i(p)= \int_0^{m_i}{p(t)dt} $$
1) Show that $B^*=\{F_0,F_1, F_2\}$ is a basis of $E^*$
2) Give the anti dual basis of $B^*$
So, we find $F_i$:
$F_0(p)=a_0m_0+\frac{a_0 m_0^2}{2}+\frac{a_0 m_0^3}{3} $
$ F_1(p)=a_1m_1+\frac{a_1 m_1^2}{2}+\frac{a_1 m_1^3}{3}$
$F_2(p)=a_2m_2+\frac{a_2 m_2^2}{2}+\frac{a_2 m_2^3}{3}$
And we find the vectors of $B^*$, by defining $B$ as the canonical basis $B=\{v_0,v_1,v_2\}$:
$$F_0(V_0)=m_0=1 \hspace{1cm} F_1(V_0)=m_1=0 \hspace{1cm} F_2(V_0)=m_2=0$$ $$F_0(V_1)=\frac{m_0^2}{2}=0 \hspace{1cm}F_1(V_1)=\frac{m_1^2}{2}=1 \hspace{1cm}F_2(V_1)=\frac{m_2^2}{2}=0 $$ $$F_0(V_2)=\frac{m_0^3}{3}=0 \hspace{1cm}F_1(V_2)=\frac{m_1^3}{3}=0 \hspace{1cm}F_2(V_2)=\frac{m_2^3}{3}=1 $$ And from that point I have no idea what to do. How do I prove linear independence?