$\newcommand{\tp}[1]{#1^\mathrm{T}} \newcommand{\Id}{\mathrm{Id}} \newcommand{\n}{\{1,\ldots,n\}} \newcommand{\siff}{\quad\Leftrightarrow\quad} \newcommand{\ijth}[2][\tp{Q}Q]{[#1]_{#2}} \newcommand{\K}{\mathbb{K}}$
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?
Best Answer
It might be instructive here to start with the corresponding invariant (i.e., basis-free) description of orthogonality:
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.