I'm trying to understand the role that orthonormal basis have, and I come across this passage:
For a general inner product space $V$, an orthonormal basis can be used to define normalized orthogonal coordinates on $V$. Under these coordinates, the inner product becomes a dot product of vectors. Thus the presence of an orthonormal basis reduces the study of a finite-dimensional inner product space to the study of $\mathbb{R}^n$ under dot product. Every finite-dimensional inner product space has an orthonormal basis, which may be obtained from an arbitrary basis using the Gram–Schmidt process.
and I am trying to digest what this means. I'll try and break it down by the real and complex vector spaces, and see where that leads me.
So $\varphi$ is a unitary transformation, so I could see how
"...the presence of an orthonormal basis reduces the study of a [real] finite-dimensional inner product space to the study of $\mathbb{R}^n$ under dot product"
makes sense.
So $\psi$ is a unitary transformation, hence I could also see how
"...the presence of an orthonormal basis reduces the study of a [complex] finite-dimensional inner product space to the study of $\require{cancel}\cancel{\mathbb{R}^n} \ \mathbb{C}^n$ under
dot productcomplex dot product"
makes sense. So it could just be that this passage is "shorthand" for these two conclusions together. That's the simplest interpretation.
But it could also be that, they actually do mean that "the presence of an orthonormal basis reduces the study of a [complex] finite-dimensional inner product space to the study of $\mathbb{R}^n$ under dot product"; I attempted to figure out how that would possible - my intuition being that if $H \cong \mathbb{C}^n$ and $\mathbb{C}^n \cong \mathbb{R}^{2n}$ then maybe $H \cong \mathbb{R}$??
Lets say we have $\mathbb{C}^n$ over $\mathbb{C}$ with the standard (complex) basis and complex dot product $\langle \cdot, \cdot \rangle_{\mathbb{C}^n}:\mathbb{C}^n \times \mathbb{C}^n \to \mathbb{C}$. I know that $\mathbb{C}^n$ could also be viewed as a real vector space, and it's standard (real) basis would just be:
and it's (real) inner product would be:
$$
\begin{align}
\langle \cdot, \cdot \rangle_{\mathbb{C}^n(\mathbb{R})}: \mathbb{C}^n \times \mathbb{C}^n \to \mathbb{R}; &&&
(\pmb u, \pmb v) \mapsto \operatorname{Re}(\langle \pmb u, \pmb v \rangle_{\mathbb{C}^n}) .
\end{align}
$$
So when viewing $\mathbb{C}^n$ a real vector space, I could define a mapping to $\mathbb{R}^{2n}$ over $\mathbb{R}$ with the standard basis and dot product $\langle \cdot, \cdot \rangle_{\mathbb{R}^n}:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ - that map being:
$$
\newcommand{\iu}{{\mathrm{i}\mkern 1mu}}
\begin{align}
\chi: \mathbb{C}^n \to \mathbb{R}^{2n};
&&& (a_{1}+\iu b_{1}, a_{2}+\iu b_{2}, \dots, a_{n}+\iu b_{n}) \mapsto (a_{1},b_{1},a_{2},b_{2},\dots,a_{n},b_{n}).
\end{align}
$$
I know that that $\chi$ is linear and bijective (proof too long, didn't include), and also:
So $\chi$ is also a unitary transformation, meaning $\mathbb{C}^n$ and $\mathbb{R}^{2n}$ are isomorphic as real inner product spaces. But that kinda breaks the intuition I had earlier, since its not obvious that $H$ and $\mathbb{R}^{2n}$ are isomorphic as real inner product spaces. And even if they were, I don't see how the study of $H$ can be reduced to the study of $\mathbb{R}^{2n}$ under the dot product.
I think my first (and the simplest) interpretation is correct. Is this so?