Simple question: Euclidean versus Hermitian form This may be a basic question, but why is the inner product of bra and ket Euclidean inner product [link] and not more general Hermitian form? [link] Is there something fundamental stating that $M$ should equal the identity? 
 A: I would rather answer your question in a different way. I request you to be patient as the start diverges a bit from your question. There is a recent upsurge due to finding real eigenvalues for Non-Hermitian matrices and this has led to the idea of generalising hermiticity of operators in QM to something else, known as Pseudo-Hermiticity.
If you study them you will know that idenity as a metric is important for Hermitian matrices or else they do not form a Complete orthogonal vector space i.e. a Hilbert space. In fact Gram Schmidt orthogonalization will also not work in such a case !
I hope this answers your question.
A: *

*The inner product on a finite-dimensional "bra-ket space" is not "Euclidean", as that denotes the standard inner product on Euclidean space $\mathbb{R}^n$. It is, however, in the case of $\mathbb{C}^n$ given by
$$ x^\dagger y = \sum (x^i)^\ast y^i,$$
the Wikipedia article is just wrong about calling this Euclidean.

*Every complex inner product space $(V,\langle\dot{},\dot{}\rangle)$ of dimension $n$ is isometrically isomorphic to $\mathbb{C}^n$ with this standard inner product, so allowing arbitrary Hermitian forms is not a generalization. The isomorphism is just given by choosing an orthonormal basis of $V$ and mapping this to the standard basis in $\mathbb{C}^n$.
