# What is the dual basis of $F^n$?

Lemma Let $$V$$ is a finite-dimensional vector space (over the filed $$F$$), say $$n=\dim V$$. If $$x_1,\ldots,x_n$$ is a basis of $$V$$, then there exists a basis $$f_1,\ldots,f_n$$ of the dual space $$V'$$ such that $$f_i(x_j)=\delta_{ij}$$, $$i,j=1,\ldots,n$$, where $$\delta_{ii}=1$$ and $$\delta_{ij}=0$$ for $$i\neq j$$.

Example Let $$e_1,\ldots,e_n$$ be the canonical basis of $$F^n$$, then the dual basis of $$(F^n)'$$ is $$f_1,\ldots,f_n$$, where $$f_i(a_1,\ldots,a_n)=a_i$$.

My question is: Is there any other $$f_i(a_1,\ldots,a_n)$$ such that $$f_i(e_j)=\delta_{ij}$$?

Note: $$a_i\in F$$.

You have $$f_j(a_1,\ldots,a_n) = f_j(\sum_{i=1}^na_ie_i) = \sum_{i=1}^na_if_j(e_i) = \sum_{i=1}^n a_i\delta_{ji} = a_j.$$
Since $$f_i(a_1,\ldots,a_n)=a_i$$, you have:
• $$f(e_i)=1$$, since $$e_i=a_1e_1+\cdots+a_ne_n$$, eith $$a_j=0$$ if $$j\ne i$$ and $$a_i=1$$;
• $$f(e_j)=0$$ if $$j\ne i$$, for the same reason
and therefore $$f_i(e_j)=\delta_{ij}$$.