Orthonormality preserved when dualing?

Question

Let $V$ be an $n$ dimensional vector space with orthonormal basis $\{e_1, \dots, e_n\}$. Is it true that the dual base on $V^\ast$ is also orthonormal?

I think this should be a true statement, but I'm lost in the definitions. If $\{e_1, \dots, e_n\}$ is an orthonormal base for $V$, then $$\langle e_i, e_j \rangle = \delta_{ij}.$$

Now in order for this to property preserve when we dualize we should have that $$\langle \varepsilon_i, \varepsilon_j \rangle = \delta_{ij}$$ (if $\{\varepsilon_1, \dots, \varepsilon_n\}$ is the dual basis) but the issue is that I have no idea what this inner product should be on the dual $V^\ast$. If it's the usual dot product on $V$ does it immediatly induce something on $V^\ast$ or is it possible even to equip the dual space with an inner product?

Answer 1

Define $ \langle \sum\limits_{k=1}^{n}a_k\epsilon_k, \sum\limits_{k=1}^{n}b_k\epsilon_k \rangle=\langle \sum\limits_{k=1}^{n}a_ke_k, \sum\limits_{k=1}^{n}b_ke_k \rangle$ and verify that this is an inner roduct on $V^{*}$ which makes $(\epsilon_i)$ orthonormal.

Answer 2

$ \newcommand\form[1]{\langle#1\rangle} $I assume that $\form{\cdot,\cdot}$ is some kind of symmetric bilinear form. If the form $\form{\cdot,\cdot}$ is non-degenerate, i.e. $$ (\forall w.\:\form{v,w} = 0) \implies v = 0 $$ then the map $v \mapsto v^\flat$ defined by $v^\flat(w) = \form{v, w}$ is an isomorphism $V \to V^*$. So it has an inverse $\sharp : V^* \to V$, and we use this to define a form on $\epsilon,\delta \in V^*$ via $\form{\epsilon,\delta} = \form{\epsilon^\sharp,\delta^\sharp}$.

Now show that $e_i^\flat = \epsilon_i$ and the result you want follows.