Let $ V $ be a $ \mathbb {R} $ vector space, ''$ \cdot $'' as a inner product of $ V $, and $ \{e_1, \cdots, e_n \} $ as the basis for $ V $.
Is there always another basis $ \{e ^ 1, \cdots, e ^ n \} $ such that
$$ e_i \cdot e ^ j = \delta_i ^ j $$
holds for each $ i, j $ ?
The right side is Kronecker's $ \delta $.
I think $ \{e ^ 1, \cdots, e ^ n \} $ corresponds to the basis of the dual space $ V ^ * $ of $ V $. $ V \cong V ^ * $ holds, but I was wondering if the dual basis could be taken directly as a subset of $ V $.
Best Answer
Your idea is right. We can use the Riesz representation theorem isomorphism to identify the dual basis with a basis of $V$: Let $\omega_1, \dots, \omega_n \in V^{*}$ be the dual basis to $\{e_1, \dots, e_n\}$, i.e. $\omega_i(e_j) = \delta_{i, j}$. By the Riesz representation theorem, for any $\omega \in V^{*}$, there exists $v(\omega) \in V$ such that for all $x \in V$, $\omega(x) = x \cdot v(\omega)$. Moreover, $v \colon V^{*} \to V$ is an isomorphism. Hence $\{v(\omega_1), \dots, v(\omega_n)\}$ is a basis for $V$ and $e_i \cdot v(\omega_j) = \omega_j(e_i) = \delta_{i, j}$.