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Given the vector space $\begin{equation} \text{P}_2[t] = \{ a + b\,\, t + c \,\,t^2 \mid a, b, c \in \mathbb R\,\,\,\, \} \end{equation}$ and linear functionals $\phi_1, \phi_2, \phi_3 \in \left( \text{P}_2[t] \right) ^* $ defined as below;

$ \begin{equation} \phi_1\left(\,\,p\,\,\left(\,\,t\right)\,\,\right) = \int_0^1 \! p\,\,\left(\,\,t\,\,\right)\,\,\mathrm{d}t \end{equation}$

\begin{equation} \phi_2\left(\,\,p\,\,\left(\,\,t\right)\,\,\right) = p\,\,\,\,^'\,\,\left(\,\,1\right) \end{equation}

$\begin{equation} \phi_3\left(\,\,p\,\,\left(\,\,t\right)\,\,\right) = p\,\,\left(\,\,0\right) \end{equation}$

Show that $\beta^* = \{ \phi_1, \phi_2, \phi_3 \}$ is linearly independent. And determine the $\beta\,\,\,\,\,$ base which has the dual base $\beta^*$.

I really don't know how to proceed with this question, any help would be appreciated.

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    Thanks for all the comments.2011-03-27

1 Answers 1

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we have $\phi_1(p)=a+b/2+c/3, \phi_2(p)=b+2c, \phi_3(p)=a$. from this you can easily see that they are independent. if you want $p_1$ st $\phi_1(p_1)=1, \phi_2(p_1)=0, \phi_3(p_1)=0$ you just need to solve a few linear equations: $a+b/2+c/3=1, b+2c=0, a=0$. do the same to find $p_2$, $p_3$.

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    You mean $\phi_1(p_1) = 1$, $\phi_2(p_1) = 0$, $\phi_3(p_1) = 0$, right?2011-03-27