# fixing dual basis to find basis of vector space

Given a finite-dimensional vector space $$V$$ with basis $$\{v_1,...,v_n\} \subset V$$, we also fix a basis for dual space $$V^*$$ by $$\{v_1^*,...,v_n^*\} \subset V^*$$ where: $$v_i^*(v_j) = 1$$ if $$j = i$$ and $$0$$ otherwise. If on the other hand, we are given the vector space $$V$$ and dual space $$V^*$$ and a basis $$\{f_1,...,f_n\}$$ of $$V^*$$, is it possible to find $$v_1,...,v_n \in V$$ such that $$v_i^* = f_i$$?

• Yes, $\{ v_i \}$ will be the dual basis to $\{ f_i \}$. Dec 4, 2020 at 3:03
• Sorry, can you elaborate? I am only given $f_1,...,f_n$ basis of $V^*$, how should I construct the $v_i$? Dec 4, 2020 at 3:06
• Do you know what is the canonical isomorphism $V \to V^{**}$? Dec 4, 2020 at 3:06
• Yes, taking $v$ to $E_v$ Dec 4, 2020 at 3:07

Yes, it is. The key point is that for finite-dimensional spaces, there is a natural isomorphism $$\theta:V\to V^{**}$$, given by $$\theta(v)=\text{evaluation at v} = \left(f\mapsto f(v)\right)$$. If you want, I can elaborate more on this, but this is a common isomorphism, so you should be able to read up more on this easily.
Now, taking this for granted, consider the dual basis $$\{f_1^*,\dots, f_n^*\}$$ of $$V^{**}$$. Then, because $$\theta:V\to V^{**}$$ is an isomorphism, we can consider the vectors $$v_i:= \theta^{-1}(f_i^*)$$. Now, it's a matter of unwinding definitions to see that $$\{v_1,\dots, v_n\}$$ is a basis for $$V$$, whose dual is $$\{v_1^*,\dots, v_n^*\}=\{f_1,\dots, f_n\}$$.
Yes. Consider the dual basis $$\phi_1,\dots,\phi_n$$ of $$f_1,\dots,f_n$$. Now, since $$v \mapsto E_v$$ is surjective, we have that $$\phi_i = E_{v_i}$$ for some $$v_i \in V$$ (for $$i=1,\dots,n$$). Finally, observe that $$\forall i, j \in \{1,\dots,n\} : \quad f_i(v_j) = E_{v_j}(f_i) = \phi_j(f_i) = \delta_{ij},$$ which means that $$v_1,\dots,v_n$$ is the desired basis for $$V$$.