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 $.
 A: 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 $e_i \cdot v(\omega_j) = \omega_j(e_i) = \delta_{i, j}$.
A: It always exists. In this proof, I will deliberately confuse the column vector with the row vector.
Since $V$ is n-dimensional $\mathbb R$-vector space, so there must exists an inner-product preserving isomorphism $L:V\to\mathbb R^n$(It can constructed by Gram-Schmidt orthogonalization).
I'll make a matrix $A$ such that i-th column of $A$ is $Le_i$.
Then, $A$ must be non-singular matrix because all of its column vector is linear independent, so $A^{-1}$ exists.
Thus, I'll let i-th row vector of $A^{-1}$ as $f^i$, then following is truth because of matrix multiplication: $f^i\cdot Le_j=\delta_{ij}$.
Then, let $L^{-1}f^i=e^i$. Since $L$ is inner-product preserving isomorphism, so $e^i\cdot e_j=\delta_{ij}$ also holds.
A: Given any real inner product space $V$, there is a linear map $T:V\to V^*$ where $T(v)$ is the functional $V\to\mathbb{R}$ defined by $T(v)(w)=v\cdot w$.  This map $T$ is injective because if $v\neq 0$ then $T(v)(v)=v\cdot v>0$ so $T(v)\neq 0$.  If $V$ is finite dimensional, then $V$ and $V^*$ have the same finite dimension so injectivity of $T$ implies it is actually an isomorphism.
So, here is how you can get your "dual basis" within $V$.  Let $\{f^1,\dots,f^n\}\subset V^*$ be the usual dual basis to $\{e_1,\dots,e_n\}$.  Then let $e^i=T^{-1}(f^i)$.  Since $T$ is an isomorphism, $\{e^1,\dots,e^n\}$ is a basis for $V$.  We also have $T(e^i)(e_j)=f^i(e_j)=\delta_j^i$.  But by definition of $T$, this means $e^i\cdot e_j=\delta_j^i$, as desired.
A: 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.
