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Let $V=P_2(\mathbb{R})$ with basis $B= \{1-2X,3+X,8X^2\}$. Let $T:V\rightarrow V$ be given by $T(aX^2+bX+c)=bX^2-aX+1$. Assume $T$ is linear. Find the dual basis $B^*=\{f_1,f_2,f_3\}$ by finding for each $i$ a formula for $f_i(aX^2+bX+c)$, and find $T^t(f_3)$

I'm having trouble doing this problem. So far I know that I have to set up each $f_i=0$ and solve for it, but I am getting elements that are in $V$ and not $V^*$. Thanks in advance for any help.

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1 Answer 1

Putting $\,B:=\{u_1:=1-2x\,,\,u_2=3+x\,,\,u_3:=8x^2\}\,$ , define

$$f_i(u_j):=\delta_{ij}=\begin{cases}1&,\;\;\text{if}\;\;i=j\\{}\\0&,\;\;\text{if}\;\;i\neq j\end{cases}$$

and extend the definition of each $\,f_i\,$ by linearity. Then $\,B^*:=\{f_1,f_2,f_3\}\,$ is the dual basis of $\,B\,$

Now, for example:

$$f_2(ax^2+bx+c)=f_2\left(\frac{a}{8}u_3+\left(\frac{b}{7}+\frac{2c}{7}\right)\,u_2+\left(-\frac{3b}{7}+\frac{c}{7}\right)u_1\right)=$$

$$0+\frac{b}{7}+\frac{2c}{7}+0=\frac{b}{7}+\frac{2c}{7}$$

and etc.

Unfortunately, I don't understand what you mean by $\,T^t(f_3)\,$ since $\,T\,$ is defined on $\,V\,$ , not on $\,V^*\,$ and, for me, $\,T^t\,$ is just the transpose of $\,T\,$ , so...

I'm almost sure that somehow you want this $\,T^t\,$ to work on $\,V^*\,$ but I'm not sure how you mean to do this, and anyway it is more than likely that with the above you can manage to end the exercise.

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