# Dual basis in $V^{**}$

(All vector spaces mentioned here are finite.)

Suppose we're given the isomorphism $$\Phi:V\to V^{**}: v \mapsto (\Phi_v:V^*\to \mathbb{K}:\phi \mapsto \phi(v))$$. Consider a basis $$(\varepsilon_i)_{i\in I}$$ of $$V^*$$. Claim that there exists a $$(e_i)_{i\in I}$$ of $$V$$ such that $$(\varepsilon_i)_{i\in I}$$ is its dual basis.

My book says that this claim can be shown by considering the isomorphism $$\Phi$$ and by looking at the dual basis of $$(\varepsilon_i)_{i\in I}$$ in $$V^{**}$$. I'm not really sure what this last bit means. How do I find this dual basis in $$V^{**}$$. How can I use it to show the claim?

Thanks!

• What do you understand by a dual basis for a given basis? – k.stm May 29 '19 at 14:19
• Given $\beta=\{v_1,\dots,v_n\}$, its dual is $\beta^*=\{f_1,\dots,f_n\}$ such that $f_i(v_j)=\delta_{ij}$. – Zachary May 29 '19 at 14:23

Given a basis $$w_1,\ldots, w_n$$ of (finite-dimensional) $$W$$, we can find a basis $$\phi_1,\ldots,\phi_n$$ of $$W^*$$ such that $$\phi_i(w_j)=\delta_{ij}$$ (the dual basis). Now let $$W=V^*$$ and take the dual basis $$\phi_1,\ldots,\phi_n$$ of $$\epsilon_1,\ldots,\epsilon_n$$ in $$W^*=V^{**}$$, transport it to $$V$$ via the canonical isomorphism (producing $$e_1,\ldots,e_n$$) and verify that $$\epsilon_i(e_j)=\psi_j(\epsilon_i)=\delta_{ji}=\delta_{ij}.$$