Question
Let $V$ be a finite dimensional vector space over $\Bbb F$ and $V^*$ it's dual space. Let $f_1 … f_n$ be a basis for $V^*$. Prove that $\exists ! e_1 … e_n$ – basis for $V$ s.t. $f_1 … f_n $ is its dual basis.
Thought:
Someone showed me a hint with an inverse of the matrix of functionals… I don't really understand how this proves the question…
Best Answer
I am assuming that the question is about $f_i=e_i^*$.
Existence: The map $ev:V\to V^{**}$ is an isomorphism (is injective and both spaces have the same dimension). Let $g_1,\cdots,g_n$ be the dual basis of $\{f_i\}$. This is easy to construct since you can define your morphisms $g_i$ in a basis. Now, put $e_i=ev^{-1}(g_i)$. The set $\{e_1,\cdots,e_n\}$ is a basis of $V$ since $ev$ is an isomorphism. Let us show that its dual basis is $\{f_1,\cdots,f_n\}$: $$ f_j(e_i)=ev(e_i)(f_j)=g_i(f_j)=\delta_{ij}. $$
Uniqueness: Let $e'_i$ be another predual basis and write $e'_i=\sum a_{ij}e_j$. Applying $f_k$ in both sides of the equation we get $$ \delta_{ik}=f_k(e'_i)=\sum a_{ij}f_k(e_j)=\sum a_{ij}\delta_{jk}=a_{ik}, $$ so that $e'_i=e_i$.