It can be shown that any (complex) inner product on $V=\mathbb{C}^n$ can be written as $\langle v,w\rangle=\bar v^{T}Aw$ for some $n\times n$ matrix $A$.
Conversely suppose we consider $F:V\times V\to\mathbb{C}$ such that $F(v,w)=\bar{v}^{T}Aw$, when is $F$ an (complex) inner product? (In other words, we need sesquilinear form, conjugate symmetry and positive-definiteness)
Now imposing conjugate symmetry and non-degeneracy, we can deduce $A=\overline{A^t}$ and $\det(A)\not =0.$ However this is not enough as I considered a few examples that do not satisfy positive-definiteness. Hence what matrix would guarantee such positive-definiteness? Do we need the determinant to be greater than $0?$
Many thanks in advance!