Let $V$ be a vector space over a field $\Bbb F$ and let $(f)=\{f_1,f_2.\ldots,f_n\}$ be a basis for $V^*$. Prove there is a basis $(b)=\{b_1,\ldots,b_n\}$ for $V$ s. t. $(f)$ is its dual basis by observing the linear map $T:V\to\Bbb F^n$ defined by $Tx=(f_1(x),f_2(x),\ldots,f_n(x))$.
Note: There already are some posts (this and this) dealing with the same problem, but there it's solved in a different way by invoking the canonical isomorphism $\varphi(x):V\to V^{**},\varphi(x)=\hat x$.
Here's what I've done so far:
First, $T(x)=(0,0,\ldots,0)\iff x\in\bigcap\limits_{i=1}^n\operatorname{Ker}(f_i)$. Since $f_1,f_2,\ldots,f_n$ are linearly independent, $\bigcap\limits_{i=1}^n\operatorname{Ker}(f_i)=\{0\},$ as proven here, it follows that $Tx=(0,0,\ldots,0)\iff x=0_V.$ In turn, $\operatorname{Ker}(T)=\{0\}\land \dim V=\dim(\Bbb F^n)=n\implies T$ is an isomorphism. Now, we're wondering there is a basis $(b)=\{b_1,\ldots,b_n\}$ for $V$ s. t. the matrix representation $T(e,b).$ where $(e)$ is a canonical basis for $\Bbb F$, is an identity matrix. Let $(c)$ be an arbitrary basis for $V$. Then $$T(e,b)=T(e,c)I(c,b)=\begin{bmatrix}1&0&0&\ldots&0&0\\0&1&0&\ldots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\ldots&1&0\\0&0&0&\ldots&0&1\end{bmatrix}\implies I(c,b)=(T(e,b))^{-1}$$ and now, the columns of $I(c,b)$ are vectors from a basis $(b)$ expressed in the basis $(c)$ an $(b)$ is the basis we were looking for.
I feel my arguing isn't complete. Is there anything wrong or missing?
Thank you very much for the feedback!
Best Answer
Hint: You showed that $T$ is an isomorphism. In particular, it is surjective so you can find $b_i \in B$ with $Tb_i = e_i$ where $\{e_1, \dots, e_n\}$ is the canonical basis of $\mathbb{F}^n$. Take two minutes to convince yourself that $\{b_1, \dots, b_n\}$ is the basis of $V$ you are looking for.