# Vector Space of polynomials and the dual space with integrals

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.

• Try taking the anti derivative for a general second degree polynomial that might help. – A.Riesen Dec 19 '16 at 22:44

## 2 Answers

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 $$\begin{bmatrix} 1&1/2&1/3 \\ 2&1&8/3\\ 3&9/2&9 \end{bmatrix}$$

• I understand everything until you say to find the inverse, why are you finding the inverse? Is it so that you multiply the inverse by [1,0,0]? – IntegrateThis Dec 19 '16 at 23:09
• @JamesDickens The system I wrote is “find the first column of the inverse of the matrix”, the other two are finding the second and third columns. – egreg Dec 19 '16 at 23:11
• AH YES ok I get it now. – IntegrateThis Dec 19 '16 at 23:12

To show it is a basis, it suffices you show that such functionals are linearly independent. Suppose then that $\sum \lambda^i f_i= 0$. If you evaluate this at the basis elements $1,2x,3x^2$ you obtain the equations

$$\lambda^1 +2\lambda^2 + 3\lambda^3=0$$ $$\lambda^1 +4\lambda^2 + 9\lambda^3=0$$ $$\lambda^1 +8\lambda^2 + 27\lambda^3=0$$

and this is $V(1,2,3)\vec \lambda =0$ where $V(1,2,3)$ is a Vandermonde matrix. Since $V(1,2,3)$ is invertible, $\vec\lambda=0$, so the set $\{f_1,f_2,f_3\}$ is linearly independent, and hence a basis because $\dim V^\ast=3$.

To obtain a dual basis, you have to solve three systems of equations. To illustrate, if $p_i=a_0+2a_1x+3a_2x^2$ is dual to $f_i$ then

$$a_0+a_1+a_2=\delta_{1i}$$ $$a_0+4a_1+9a_2=\delta_{2i}$$ $$a_0+8a_1+27a_2=\delta_{3i}$$

so you have to invert $V(1,2,3)$.