Why is an orthogonal matrix called orthogonal? I know a square matrix is called orthogonal if its rows (and columns) are pairwise orthonormal
But is there a deeper reason for this, or is it only an historical reason? I find it is very confusing and the term would let me assume, that a matrix is called orthogonal if its rows (and columns) are orthogonal and that it is called orthonormal if its rows (and columns) are orthonormal but apparently that's not conventional.
I know that square matrices with orthogonal columns have no special interest, but thats not the point. If I read the term orthogonal matrix my first assumption is, that its rows (and columns) are orthogonal what is correct of course, but the more important property is that they are also orthonormal

So, Question:
Why do you call an orthogonal matrix orthogonal and not orthonormal?
Wouldn't this be more precisely and clearly?
 A: A affine transformation which preserves the dot-product on $\mathbb{R}^n$ is called an isometry of Euclidean $n$-space. In fact, one can begin without the assumption of an affine map and derive it as a necessary consequence of dot-product preservation. See the Mazur Ulam Theorem which shows this result holds for maps between finite dimensional normed spaces where the notion of isometry is that the map preserves the norm. In particular, an isometry of $\mathbb{R}^n$ can be expressed as $T(v)=Rv+b$ where $R^TR=I$. The significance of such a transformation is that it provides rigid motions of Euclidean $n$-space. Two objects are congruent in the sense of highschool geometry if and only if some rigid motion carries one object to the other.
My Point? this is the context from which orthogonal matrices gain their name. They correspond to orthogonal transformations. Of course, these break into reflections and rotations according to $\text{det}(R)= -1,1$ respective.
Likely thought: we should just call such transformations orthonormal transformations. I suppose that would be a choice of vernacular. However, we don't, so... they're not orthonormal matrices. But, I totally agree, this is just a choice of terminologies. Here's another: since $R^TR=I$ implies $R$ is either a rotation or reflection let's call the set of such matrices rotoreflective matrices. In any event, I would advocate the change of terminology you advocate, but, the current terminology is pretty well set at this point, so, good luck if you wish to change this culture.
A: Preservation of Structure
First of all note that any matrix represents a linear transformation:
$$T(x):=Ax\quad$$
(That is what is of most interest.)
Now a quick calculation shows:
$$\langle T(x),T(y)\rangle=\langle Ax,Ay\rangle=(Ax)^\intercal(Ay)=x^\intercal A^\intercal Ay$$
So a linear operator preserves scalar product iff it's adjoint is a left inverse:
$$\langle T(x),T(y)\rangle\equiv\langle x,y\rangle\iff A^\intercal A=1$$
That is it is linear and preserves angles and lengths, especially orthogonality and normalization. These transformation are the morphisms between scalar product spaces and we call them orthogonal (see orthogonal transformations).
Unfortunately I guess that is not where the name comes from historically. But one should keep in mind that a statement about column and row vectors is fancy but also special and hides what is really happening...
Invertibility
First of all note that if it bijective, so invertible, its inverse will be linear too and also preserves scalar product automatically. Thus it is an isomorphism then.
Now, a linear transformation that preserves scalar product is necessarily injective:
$$T(x)=0\implies\|T(x)\|=\langle T(x),T(x)\rangle=\langle 0,0\rangle\implies x=0$$
However it might fail to be surjective in general - take as an example the rightshift operator.
If it happens to be surjective too, so bijective, then it has an inverse matrix:
$$A^{-1}A=1\text{ and }AA^{-1}=1$$
But since the inverse is unique we have in this case:
$$A^\intercal=A^{-1}$$
Concluding that the isomorphisms are given by matrices that satisfy:
$$A^\intercal A=1\text{ and }AA^\intercal=1$$
In the finite dimensional case surjectivity directly follows by the rank nullity theorem. Thus it is enough then to check that the transpose matrix is either left inverse or right inverse instead of checking both. That check goes for any matrix to conclude injectivity by surjectivity or vice versa.
Annotation
The rank nullity theorem states that:
$$\dim\operatorname{dom}T=\dim\ker T+\dim\operatorname{ran}T$$
A: I would like to explain orthogonality in terms of vectors which explains why orthogonal matrix is called orthogonal matrix.
Dot product of two orthogonal vectors is $0.$
Assume vectors, $\ \vec a= \left[
  \begin{matrix}
  a_1 \\
  a_2 \\
  a_3 \\
  \vdots \\
  a_n
  \end{matrix}
  \right]
\text{ and } 
 \vec b = \left[
  \begin{matrix}
  b_1 \\
  b_2 \\
  b_3 \\
  \vdots \\
  b_n
  \end{matrix}
  \right]
$.
Now dot product between the vectors can be computed as:
$ (\vec a)^T\cdot(\vec b) = [a_1 b_1+a_2 b_2+a_3 b_3 + \cdots + a_n b_n] $
Assume each of the columns in matrix $Q$ as a vector.
A matrix $Q$ is an orthogonal matrix if each column vector is orthogonal to the other column vectors in the matrix $Q.$
So, for every column i and column j in matrix $Q,$ if they have to be orthogonal to each other, the dot product across every column $i$ with every column $j$ should be $0,$ when i is not equal to $j.$
Also, the magnitude of every column needs to be same.
So, let's say that when $i=j,$ the value is constant $1.$
Therefore dot product between column $i$ of matrix $Q$ and column $j$ of matrix $Q$ can be written as
$$
  \langle Q_i,Q_j\rangle = \begin{cases}
     0, & i \ne j \\
     1, & i = j
 \end{cases}
$$
So, computing dot products across all columns can be done simultaneously using $ Q^T\cdot Q$
The result will be $ I_n. $
A: A matrix $V$ with mutually orthogonal columns is called orthogonal because it maps each of the standard orthogonal coordinate directions to a new set of mutually orthogonal coordinate directions. For example, spherical coordinates are orthogonal because
$$
             x=r\sin\phi\cos\theta,\;\; y=r\sin\phi\sin\theta,\;\; z = r\cos\phi
$$
maps the coordinate lines where $r$ alone varies, $\phi$ alone varies, and $\theta$ alone varies into mutually orthogonal curves in $\mathbb{R}^{3}$. This is evidenced in the Jacobian matrix whose columns are mutually orthogonal:
$$
             \frac{\partial(x,y,z)}{\partial(r,\phi,\theta)}
                       = \left[\begin{array}{ccc}
    \sin\phi\cos\theta & r\cos\phi\cos\theta & -r\sin\phi\sin\theta \\
    \sin\phi\sin\theta & r\cos\phi\sin\theta & r\sin\phi\cos\theta \\
    \cos\phi & -r\sin\phi & 0
                               \end{array}\right]
$$
It is well known that the spherical coordinate system is orthogonal because the three different coordinate lines always intersection orthogonally. The determinant of the Jacobian matrix for this orthogonal transformation is easy to determine: it is the product of the lengths of the three columns of the Jacobian matrix, which is the product of the standard spherical distance scale factors $\frac{dl}{dr}=1$, $\frac{dl}{d\phi}=r$, $\frac{dl}{d\theta}=r\sin\phi$. The standard spherical coordinate volume element is the the product of these $dV = r^{2}\sin\phi\,dr\,d\phi\,d\theta$.
A non-zero unitary matrix goes one step further: any set of orthogonal vectors is mapped to another, which is much stronger than the standard coordinate directions being mapped to mutually orthogonal directions. This condition implies that there is a constant $C$ such that the unitary matrix maps all unit vectors to vectors of length $C$. After renormalizing by dividing by $C$, the unitary matrix is distance preserving and preserve angles: $(Vx,Vy)/\|Vx\|\|Vy\|=(x,y)/\|x\|\|y\|$.
