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.

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

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

  • $\begingroup$ 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]? $\endgroup$ – IntegrateThis Dec 19 '16 at 23:09
  • $\begingroup$ @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. $\endgroup$ – egreg Dec 19 '16 at 23:11
  • $\begingroup$ AH YES ok I get it now. $\endgroup$ – 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 $V(1,2,3)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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