Given a basis for an inner product space $V$, is there a "dual" basis for $V$? 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 $.

Consider the space $E_1=e_2^\perp\cap e_2^\perp\cap\ldots\cap e_n^\perp$. This is the intersection of $n-1$ spaces and the dimension of each of them is $n-1$. Therefore, $\dim E_1\geqslant1$ (actually, it is equal to $1$). Take $v\in E_1\setminus\{0\}$. You cannot have $e_1.v=0$, because otherwise $v$ would be orthogonal to every $e_k$ and therefore $v$ would be $0$. Now, define $e^1$ as $\frac1{e_1.v}v$. You can define $e^2,\ldots,e^n$ in the same way.
