Here is the question I want to solve:
Let $V$ be a vector space of all polynomial functions $p$ from $R$ into $R$ which has degree $2$ or less: $$p(x) = c_0 + c_1 x + c_2 x^2,$$
Define three linear functionals on $V$ by
$f_1(p) = \int_{0}^{1}p(x)dx, f_2(p) = \int_{0}^{2}p(x)dx, f_1(p) = \int_{0}^{-1}p(x)dx.$
Show that $\{f_1, f_2, f_3\}$ is a basis for $V^*$ by exhibiting the basis for $V$ of which it is the dual.
My trial:
I evaluated $f_1(p) = c_0 + c_1 / 2 + c_2/ 3, f_2(p) = 2c_0 + 2c_1 + 8c_2/ 3, f_3(p) = – c_0 + c_1 / 2 – c_2/ 3.$ But then I do not know how to complete, could someone help me please?
Best Answer
Well....looks like you've got a handful here.
Let $\{p_{1},p_{2},p_{3}\}$ be the basis for $V$.
Let $p_{i}=c_{1}^{i}+c_{2}^{i}x+c_{3}^{i}(x^{2})$ then
Then by definition of dual basis you have $f_{1}(p_{1})=1$ , $f_{1}(p_{2})=0$ and $f_{1}(p_{3})=0$. Similarly you have $f_{2}(p_{1})=0$,$f_{2}(p_{2})=1$ and $f_{2}(p_{3})=0$ and $f_{3}(p_{1})=0$, $f_{3}(p_{2})=0$ and $f_{3}(p_{3})=1$
These give you $9$ equations in $9$ variablies i.e $\{c_{j}^{i}\}_{1\leq i,j\leq 3}$.
$c_{1}^{1}+\frac{c_{2}^{1}}{2}+\frac{c_{3}^{1}}{3}=1$
$c_{1}^{2}+\frac{c_{2}^{2}}{2}+\frac{c_{3}^{2}}{3}=0$
$c_{1}^{3}+\frac{c_{2}^{3}}{2}+\frac{c_{3}^{3}}{3}=0$
$2c_{1}^{1}+\frac{4c_{2}^{1}}{2}+\frac{8c_{2}^{1}}{3}=0$
$2c_{1}^{2}+\frac{4c_{2}^{2}}{2}+\frac{8c_{3}^{2}}{3}=1$
$2c_{1}^{3}+\frac{4c_{2}^{3}}{2}+\frac{8c_{3}^{3}}{3}=0$
$c_{1}^{1}-\frac{c_{2}^{1}}{2}+\frac{c_{3}^{1}}{3}=0$
$c_{1}^{2}-\frac{c_{2}^{2}}{2}+\frac{c_{3}^{2}}{3}=0$
$c_{1}^{3}-\frac{c_{2}^{3}}{2}+\frac{c_{3}^{3}}{3}=1$
Solve them. Thank me later.
After that you will arrive at $\{p_{1},p_{2},p_{3}\}$ and once you show that they are linearly independent .Just show that the matrix whose ijth entry is $c_{i}^{j}$ has non-zero determinant . You will have that $\{f_{1},f_{2},f_{3}\}$ is the dual basis corresponding to $\{p_{1},p_{2},p_{3}\}$.
This being said. The best way to show $\{f_{1},f_{2},f_{3}\}$ is a basis is to find the representation of $f_{1},f_{2},f_{3}$ in terms of the standard dual basis and show that the matrix formed by those coefficients has non-zero determinant.
A well known theorem in dual spaces gives you that if $\{v_{1},v_{2},...v_{n}\}$ forms a basis for $V$ and $\{g_{1},g_{2},...,g_{n}\}$ is the dual basis corresponding to it then any functional $f:V\to\mathbb{F}$ can be uniquely written as $\displaystyle f=\sum_{i=1}^{n}f(v_{i})g_{i}$
That is if $g_{1},g_{2},g_{3}$ denotes the standard dual basis correspoding to $1,x,x^{2}$.
Then $f_{1}=f_{1}(1)g_{1}+f_{1}(x)g_{2}+f_{1}(x^{2})g_{3}$
Similarly we get the representations for $f_{2}$ and $f_{3}$.
Thus to show that $\{f_{1},f_{2},f_{3}\}$ is a basis , it suffices to show that the determinant:-
$$\begin{vmatrix}f_{1}(1)&f_{2}(1)&f_{3}(1)\\f_{1}(x)&f_{2}(x)&f_{3}(x)\\f_{1}(x^{2})&f_{2}(x^{2})&f_{3}(x^{2})\end{vmatrix}\neq 0$$
This proves linear independence of these $3$ vectors in a space of dimension $3$ and hence it shows that it is a basis.