# Linear algebra - Dual, functionals

Given three functionals

• $f_1(p) = \int_0^1 p(t)\,dt$
• $f_2(p) = \int_0^2 p(t)\, dt$
• $f_3(p) = \int_0^{-1} p(t)\, dt$

defined on $V = P_2$, the space of all polynomials over $\mathbb R$ of degree not greater than 2. Asking to prove that $\{f_1, f_2, f_3\}$ is a basis for $V^*$ by finding the basis for $V$ of which it is the dual.

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Please, check your input and correct the mistakes. The question is obscure. – Siminore Apr 22 '12 at 16:49
Are $f_1$ and $f_3$ the same? – Davide Giraudo Apr 22 '12 at 16:53
f1 and f3 are not the same. f3' integral is from 0 to -1. – Megan Apr 22 '12 at 19:44
I wanted add tags: dual or dual space, functionals, linear algebra. But I was not allowed to create new tags. Sorry. – Megan Apr 22 '12 at 19:45
Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. – Arturo Magidin Apr 22 '12 at 21:17

You want to find polynomials $p_1,p_2,p_3$ of degree at most $2$ with the property that $f_i(p_j) = 1$ if $i=j$, and $f_i(p_j) = 0$ if $i\neq j$.
For instance, you want to find $p_1(x) = a+bx+cx^2$ such that \begin{align*} 1=f_1(p_1) &=\int_0^1(a+bx+cx^2)\,dx \\ &= \left.\left( ax + \frac{b}{2}x^2 + \frac{c}{3}x^3\right)\right|_{0}^1\\ &= a + \frac{b}{2}+\frac{c}{3}.\\ 0 =f_2(p_1) &= \int_0^2(a+bx+cx^2)\,dx\\ &= \left.\left( ax + \frac{b}{2}x^2 + \frac{c}{3}x^3\right)\right|_{0}^2\\ &= 2a + 2b + \frac{8c}{3}.\\ 0 = f_3(p_1) &= \int_0^{-1}(a+bx+cx^2)\,dx\\ &= \left.\left( ax + \frac{b}{2}x^2 + \frac{c}{3}x^3\right)\right|_{0}^{-1}\\ &= -a + \frac{b}{2} -\frac{c}{3}. \end{align*} So this gives us a system of three linear equations in three unknowns: $$\begin{array}{rcccccl} a & + & \frac{1}{2}b & + & \frac{1}{3}c & = & 1\\ 2a & + & 2b & + & \frac{8}{3}c & = & 0\\ -a & + & \frac{1}{2}b & - & \frac{1}{3}c & = & 0. \end{array}$$ Solving it will give the value of $a$, $b$, and $c$.
Replacing the solution vector with $(0,1,0)^T$ gives $p_2$; replacing the solution vector with $(0,0,1)^T$ gives $p_3$.
So, in essence, we are trying to find the inverse of a matrix; that matrix is related to the functions $f_1$, $f_2$, $f_3$. In fact, it is the inverse of the matrix corresponding to the linear transformtion $T\colon V\to V$ given by $T(p) = (f_1(p),f_2(p),f_3(p))$.