Given a real or complex vector space $V$ and a (finite) basis $B$ of it, does it always exist an inner product on $V$ such that $B$ is an orthonormal basis with respect to it?
The question is equivalent to asking: is there always a positive definite (symmetric) matrix $A$ such that if $B=\{v_1,…,v_n\}$, then
$$v_i^tAv_j=\delta_{ij}\;?$$
Best Answer
Let $\mathbb{K}$ be $\mathbb{R}$ or $\mathbb{C}$. On the $\mathbb{K}$-linear space $V$, an inner product is a bilinear form $\varphi \, : \, V \times V \, \longrightarrow \, \mathbb{K}$ which is symmetric positive definite. Let $B = \big( \varepsilon_{1},\ldots, \varepsilon_{n}\big)$ be a basis of $V$ and let $(x,y) \in V^{2}$. We can write :
$$ x = \sum_{i=1}^{n} x_{i} \varepsilon_{i} \quad \mathrm{and} \quad y=\sum_{j=1}^{n} y_{j}\varepsilon_{j} $$
where $(x_{1},\ldots,x_{n}) \in \mathbb{K}^{n}$ and $(y_{1},\ldots,y_{n}) \in \mathbb{K}^{n}$. By bilinearity of $\varphi$, we have :
$$ \begin{align*} \varphi(x,y) &= {} \varphi \Big( \sum_{i=1}^{n} x_{i} \varepsilon_{i} \, ,\sum_{j=1}^{n} y_{j}\varepsilon_{j} \Big) \\[1mm] &= \sum_{i,j=1}^{n} x_{i}y_{j} \varphi \big( \varepsilon_{i},\varepsilon_{j} \big) \end{align*} $$
As a consequence, in order to define $\varphi$, one must specify the value of $\varphi(\varepsilon_{i},\varepsilon_{j})$ for all $i$ and $j$. Choosing : $\forall i,j, \, \varphi(\varepsilon_{i},\varepsilon_{j}) = \delta_{i,j}$, you obtain an inner product on $V$ for which $B$ is an orthonormal basis.