Let $V$, a vector space over $\mathbb{C}$ from a finite dimension with an inner product $\beta \langle .,. \rangle$. Let $B=\{v_1,\ldots, v_n\}$, a basis for $V$. If $v_1,\ldots,v_n$ is an orthonormal basis, we have learn in class (apparently. I didn't) that:
For $v,w\in V$: $$ \beta(v,w) = \overline{([v]_B)^T} \cdot [w]_B $$
I need to prove the generalization (for any basis, not necessarily an orthonormal) of this, for:
$$ \beta(v,w) = \overline{([v]_B)^T} \cdot G \cdot [w]_B $$
Where $G\in M_n(\mathbb{C})$ such that $G_{ij} = \beta(v_i,v_j)$
Take $\;v,w\in V\;$ and express them as linear combinations of the given orthonormal basis:
$$\begin{cases}v=\sum\limits_{k=1}^na_iv_i\\{}\\w=\sum\limits_{k=1}^nb_iv_i\end{cases}$$
so using the basic properties of a complex inner product:
$$\beta(v,w)=\beta\left(\sum\limits_{k=1}^na_iv_i\,,\,\,\sum\limits_{k=1}^nb_iv_i\right)=\sum_{i,j=1}^na_i\overline {b_j}\,\overbrace{\beta(v_i,v_j)}^{\delta_{ij}}=\sum_{i=1}^na_i\overline{b_i}=\left([v]_B\right)^t\overline{[w]_B}$$
where $\;[v]_B\;$ means column vector .
The above is what you say you apparently didn't learn, but now you have.
Now, if $\;B=\{v_1,...,v_n\}\;$ is not precisely an orthonormal basis, can you see how the matrix $\;G=(\beta(v_i,v_j))\;$ fits in the formula? Complete the argument.