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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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From the way the vector space is given, you already have a basis for it. One way to show that the linear functionals are linearly independent would be to show that the three triples that you get when you evaluate them on that basis are linearly independent. – joriki Mar 27 '11 at 17:08
What have you tried so far? Have you tried applying the definition of linear independence? It may help to calculate the values of each $\phi_i$ on a general polynomial $a+bt+ct^2$. – Alon Amit Mar 27 '11 at 17:08
Thanks for all the comments. – hattenn Mar 27 '11 at 18:21
up vote 2 down vote accepted

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? – hattenn Mar 27 '11 at 18:16
Thank you, done :) – hattenn Mar 27 '11 at 18:20

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