Orthogonal matrices only defined for standard inner product? $\newcommand{\tp}[1]{#1^\mathrm{T}} \newcommand{\Id}{\mathrm{Id}} \newcommand{\n}{\{1,\ldots,n\}} \newcommand{\siff}{\quad\Leftrightarrow\quad} \newcommand{\ijth}[2][\tp{Q}Q]{[#1]_{#2}} \newcommand{\K}{\mathbb{K}}$
Let orthogonal matrices be defined as follows.

A matrix $Q\in\mathcal{M}_{m\times n}(\mathbb{K})$, where $\mathbb{K}$
  is a field, is said to be orthogonal if $$ Q^\mathrm{T}Q = \mathrm{Id}_n$$

I'm not fully sure if I'm understanding the following fact correctly:

A matrix $Q\in\mathcal{M}_{m\times n}(\K)$ is orthogonal iff
  the columns of $Q$ form an orthonormal set in $\K^m$.
Proof  Let $q_i$ denote the $i$-th column of $Q$ for all $i\in\{1,\ldots,n\}$, and let $\ijth[A]{ij}$ denote the $(i,j)$-th
  element of $A$ for any matrix $A$. Then, $Q$ being an orthogonal
  matrix is equivalent to $$\tp{Q}Q = \Id_n \siff \ijth{ij} = \delta_{ij}\,,$$ where $\delta_{ij}$ is the Kronecker delta. On the
  other hand, by the definition of matrix multiplication, $$\ijth{ij} = \sum_{k=1}^{m} \ijth[\tp{Q}]{ik}\ijth[{Q}]{kj} = \sum_{k=1}^{m} \ijth[Q]{ki}\ijth[{Q}]{kj} \stackrel{\color{red}*}{=} \langle q_i, q_j\rangle\,.$$ Thus $Q$ is orthogonal iff $$ \langle q_i, q_j\rangle = \delta_{ij} \qquad\forall (i,j)\in\n\times\n\,, $$ which is true iff $(q_i)_{i\in\n}$ form an orthonormal set.

Particularly, I'm suspicious of the equality marked with the red asterisk. Isn't that true only for the standard inner product (i.e. the dot product), defined as
$
\langle u, v \rangle = \sum_i u_iv_i\ 
$? So, are orthogonal matrices only treated in the context of the standard inner product? If so, is there a "generalization" of orthogonal matrices for general inner product spaces?
 A: It might be instructive here to start with the corresponding invariant (i.e., basis-free) description of orthogonality:

On a finite-dimensional inner product space $(\Bbb V, \langle\,\cdot\,,\,\cdot\,\rangle)$, a linear transformation $T : \Bbb V \to \Bbb V$ is said to be orthogonal if it preserves the inner product, that is, if $\langle T({\bf x}), T({\bf y}) \rangle = \langle {\bf x}, {\bf y}\rangle$.

Fixing a basis $({\bf e}_a)$ of $\Bbb V$ determines matrix representations $[T]$ of $T$ and $[\Sigma]$ of the inner product: These are characterized by
$$[T({\bf e}_a)] = \sum_b [T]_{ba} [{\bf e}_a], \qquad
  [\Sigma]_{ab} = \langle {\bf e}_a, {\bf e}_b \rangle .$$
Unwinding all of this, we see that $T$ is orthogonal if
$$[T]^{\top} [\Sigma] [T] = [\Sigma] .$$
In the special case that the basis $({\bf e}_a)$ is orthonormal, it follows immediately from the definitions that $[\Sigma] = I$, in which case the condition simplifies to the familiar definition of orthogonal matrix:
$$[T]^{\top} [T] = I .$$
Over a real inner product space, we can always choose an orthonormal basis, so the more general construction might seem like an unnecessary formalism. But such bases are not always the most convenient in applications, and if we extend our attention to nondegenerate, symmetric bilinear forms (so, drop the condition of positive-definiteness from the definition of inner product), orthogonal bases don't exist, but we still care about the notion of orthogonality.
A: Yep, just as you expected, however, all of this heavily depends on your choice of basis. Since every basis $B$ of a vector space $V$ is actually a choice of isomorphism $\varphi_B : k^{\dim(V)} \to V$, we may interpret any homomorphism $f: V \to W$ as a matrix (heavily dependent on the choice of basis). Picking an orthonormal basis w.r.t an arbitrary inner product  $\langle \_ ,\_ \rangle _V$ this isomorphism even becomes compatible with the inner product, i.e. $\langle \varphi_B (x) , \varphi_B (y)\rangle _V =\langle x , y \rangle _{\textrm{eucl}}$. So you can carry over all constructions.   However in general you can define an orthogonal matrix as commuting with the inner product. This does, using the above identification become equivalent to the definition you know.
However, I prefer doing the general case first (i.e. commuting with inner product) and then specializing using basis.
I hope this is a satisfactory answer.
