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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?

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    $\begingroup$ So your real question is in the last sentence. $\endgroup$
    – Paul Frost
    Commented Jan 19, 2023 at 23:02
  • $\begingroup$ @PaulFrost The main question is the one entitled, but I need the result of the last sentence for it presumeably. $\endgroup$
    – Walker
    Commented Jan 19, 2023 at 23:08

2 Answers 2

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$ \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.

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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.

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  • $\begingroup$ What is $\delta_k$? @geetha290krm $\endgroup$
    – Walker
    Commented Jan 19, 2023 at 23:24
  • $\begingroup$ @Walker Sorry, I typed $\delta_k$ for $e_k$. $\endgroup$ Commented Jan 19, 2023 at 23:26

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