I am looking over Linear Algebra,and am stuck on this problem.
V is the space of real polynomials of degree 2 or less, and I want to find a dual basis to the basis $\{1,x,x^2\}$. I know the general technique, that each element of the dual basis $e_i'$evaluates to $0$ on the basis vectors $e_j$ for which $i \neq j$ and $1$ when $i=j$ but every time I try and do this on these basis vectors, I get stuck.
Any help appreciated.
Also, a more general question: what is the method for determining in general whether or not something is a dual basis? I know you have to show it to be linearly independent and spanning, and showing linear independence is generally quite easy, but how do I show spanning? If I have a dual basis I'm presuming it doesn't necessarily have to satisfy the above criterion (evaluating to 0 and 1 on different basis vectors), and this is just a way of finding one dual basis, so how would I test whether or not a given dual basis is spanning?
Thanks.
For linear functionals it is sufficient to define their behaviour on a basis.
In finite dimensional spaces, a basis $e_k$ has a uniquely defined dual basis $e^*_l$ defined by $e^*_l(e_k) = \delta_{kl}$.
It is straightforward to check that the $e^*_l$ is a basis. If $f$ is an arbitrary linear functional, then we can write $f = \sum_l f(e_l)e^*_l$. To see this, note that $f(\sum_k a_k e_k) = \sum_k a_k f(e_k)$ and $(\sum_l f(e_l)e^*_l)(\sum_k a_k e_k) = \sum_l \sum_k a_k f(e_l)e^*_l(e_k) = \sum_k a_kf(e_k)$. Hence is is spanning. To check independence, just evaluate $\sum_l b_l e^*_l = 0$ at $e_k$ to get $b_l = 0$.
To check if a collection $f_l$ is a dual basis, just evaluate on the underlying basis. It is dual iff $f_l(e_k) = \delta_{lk}$.
In the specific example you gave, we can easily find a formula for the dual basis. If we let $e_k(x) = x^k$ for $k=0,1,2$ be a basis for $P_2$ then the dual basis must satisfy $e^*_l(x \mapsto x^k) = \delta_{kl}$, or $e^*_l(x \mapsto \sum_k p_k x^k) = p_l$, that is, it 'picks out' the $p_l$ coefficient of a polynomial. One way of writing this is $e^*_l(p) = {1 \over l!} p^{(l)}(0)$.