Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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:



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.

share|cite|improve this answer

Your Answer


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.