Let $V = <1,x,x^2>$ be the vector space of polynomial functions $p:\Bbb R\rightarrow \Bbb R$, I am trying to show that
$f_1(p)= \int_0^{1} p(x)dx$, $f_2(p)= \int_0^{2} p(x)dx$, and
$f_3(p) = \int_0^{3} p(x)dx$
is a basis of $V^*$, and to find its dual basis in $V$.
Thoughts:
It seems if I show that they are linearly independent, then automatically they are generating since $V^*$ has dimension $3$.
First Part (which I am more interested in):
Taking the anti derivative of a general polynomial of degree $2$ say
$a_1+a_2x+a_3x^2$, then considering a linear combination set to 0, we have by antiderivatives:
$k_1 (a_1 + a_2/2+a_3/3) + $
$ k_2 (2a_1 + 2a_2+8a_3/3) +$
$ k_3 (3a_1 + 9/2a_2+9a_3) = 0$ for all values off $a_1, a_2, a_3$ which now amounts to showing that $k_1 = k_2 = k_3 = 0$ ?
Second Part: I have got the general idea for the second part "and to find its dual basis in V". I consider a general form $a_1+a_2x+a_3x^2$ and integrate with each function solving for coefficients where $f_1(p) = 1, f_2(p) = 0, f_3(p) = 0$ for the first dual basis of $V^*$ and switching where the $1$ value goes for the other two vectors.
Best Answer
Try and find the matrices of the three linear maps: $$ f_1(1)=\int_0^1 1\,dx=1, \quad f_1(x)=\int_0^1 x\,dx=1/2, \quad f_1(x^2)=\int_0^1 x^2\,dx=1/3 $$ Thus the matrix of $f_1$ is $[1\;1/2\;1/3]$.
Similarly, the matrices of $f_2$ and $f_3$ are, respectively, $$ [2\;1\;8/3] \qquad [3\;9/2\;9] $$ and these three row vectors are linearly independent.
A polynomial such that $f_1(p)=1$, $f_2(p)=0$ and $f_3(p)=0$ must have coordinates $[a\;b\;c]^T$ such that $$ \begin{bmatrix} 1&1/2&1/3 \\ 2&1&8/3\\ 3&9/2&9 \end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix}= \begin{bmatrix}1\\0\\0\end{bmatrix} $$ and then $p=a+bx+cx^2$.
You see you just have to find the inverse of $$ \begin{bmatrix} 1&1/2&1/3 \\ 2&1&8/3\\ 3&9/2&9 \end{bmatrix} $$