Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
1  
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
add comment

1 Answer

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 $p_2$, $p_3$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.