Let $B=\{1,x,x^2,…,x^n\}$ be the standard basis of $P(\mathbb{R})$, where $P(\mathbb{R})$ is the vector space of all polynomials with real coefficients of degree at most $n$. What is the dual basis $B^*$ of $P(\mathbb{R}$), and express this dual basis in terms of the derivatives of a polynomial.
I asked this question before, but am still confused.
This is what I have so far:
The dual space of $P(\mathbb{R})$ is the set Hom$(P(\mathbb{R}),R)$, where $R$ is the vector space of all single-tuples given by $\mathbb{R}$.
Because $B$ is a basis for $P(\mathbb{R})$, then there exists a dual basis of linear functions $B^*=\{f_0,f_1,…,f_n\}$, such that $f_i(v_i)=1$ and $f_i(v_j)=0$ for $i\neq j$ and $v_i\in B$.
More so, by linearity of $f_i$, we see for $0\leq i \leq n$, and $c_0,c_1,…,c_n\in \mathbb{R}$
$$f_i(c_0x^0+c_1x+c_2x^2+…+c_nx^n)=f_i(c_0x^0)+f_i(c_1x)+f_i(c_2x)+…+f_i(c_nx^n)=c_i$$
This is where I am stuck. The above formula implies $f_i=c_i$, so would the dual basis $B^*=\{c_0,c_1,…,c_n\}$? This doesn't seem right at all, and I think I am doing something wrong.
Any help would be much appreciated. Thanks in advance!
Best Answer
Given a basis $(e_k)_{1\leq k\leq n}$ of a vector space $V$ the elements $e_k^*$ of the dual basis compute the $e_k$-coordinates of arbitrary vectors $x\in V$: One has $$x=\sum_{k=1}^n \xi_k e_k\quad\Leftrightarrow\quad \xi_k= e_k^*(x)\qquad(1\leq k\leq n)\ .$$ Consider now a polynomial $$a(t)=\sum_{k=0}^n \alpha_k t^k$$ of degree $\leq n$. Its coordinates with respect to the basis ${\cal B}:=\bigl(t^k\bigr)_{0\leq k\leq n}$ are the coefficients $\alpha_k$ of this polynomial. The dual basis $(e_k^*)_{0\leq k\leq n}$ of ${\cal B}$ then consists of functionals (or "operations") that compute for a given polynomial function $a$ its coefficients $\alpha_k$.
If we now remember that such an $a$ is its own Taylor expansion centered at $t=0$ then it becomes clear that we can identify $e_k^*$ as $$e_k^*(a)={1\over k!}{d^k a\over dx^k}(0)\qquad(0\leq k\leq n)\ .$$