Number of generators of a $2\times 2$ real orthonomal matrix inconsistent with visualization Let $ A $ be a real orthonormal matrix. From the definition $ A^T A = I $, it is easy to see that $ \det(A) = \pm 1 $, hence a real orthonormal matrix should correspond to rotation and reflection.
If $ A $ is $ N\times N $, then number of generators is given by $ g(N) = \frac{1}{2}N(N-1) $. To prove this result, a common argument goes as following:


*

*There are $ N^2 $ entries in $ A $.

*Each column of $ A $ must form a unit vector, this provides $ N $ conditions.

*Each pair of distinct columns are orthogonal, this provides $ \frac{N!}{2!(N-2)!} = \frac{1}{2}N(N-1) $ conditions.


Therefore, the number of generators is given by
$$ g(N) = N^2 - N - \frac{1}{2}N(N-1) = \frac{1}{2}N(N-1) \ . $$
But now when I try to understand this result visually, I encountered a contradiction. In particular, $ g(2) = \frac{1}{2}(2)(2-1) = 1 $, implying that there is only one generator for a $ 2\times 2 $ real orthogonal matrix. But this result does not even explain a geometric argument I try to form as following:
A linear transformation can be visualized by plotting the two columns as vectors on a Cartesian plane. Since $ A_{2\times 2} $ is orthonormal, we must have all the columns fall on the unit circle. We are free to decide where the first column lands, say $ \begin{bmatrix} 1 \\ 0 \end{bmatrix} $. This shall exhausts the only one generator $ A_{2\times 2} $ has, leading us with no choice for all the remaining entries of $ A_{2\times 2} $.
Unfortunately, this is NOT the case. By visual inspection, it is apparent that we still have to decide between $ \begin{bmatrix} 0 \\ 1 \end{bmatrix} $ and $ \begin{bmatrix} 0 \\ -1 \end{bmatrix} $ for the second column of $ A_{2\times 2} $. This argument can also be expressed in the language of algebra - suppose we pick $ A_{11} := 1 $, it is straightforward to see that both $ A_{12} $ and $ A_{21} $ must vanish, while $ A_{22} $ remains undetermined between $ +1 $ and $ -1 $.
 A: In your example, the conditions from the orthogonality are $0 = 0$, due to the zeros in your matrix. Thus, $g(N)$ is only a lower bound on the degree of freedom you have when generating orthogonal matrices, because to compute $g(N)$, we assume that all $g(N)$ conditions are non-empty and independent; which (as you have seen in your example) is not true in general. It will be, however, true if you consider random matrices over the reals, as here the chance of having zeros is itself zero. Thus one might actually work with the number $g(N)$, even if it is not exact.
You should thus ask yourself in which field you are working. In algebra, the number $g(N)$ is only a lower bound on the degree of freedom you have, in numerics for example you can assume this to be exactly the number of variables you can choose.
edit for the comment below:
Not really, no. The freedom is more like one vector and one sign. Given that you have chosen the first vector, the second has to be orthogonal with this first one. The orthogonal space of this vector has dimension one, meaning the second vector comes from a one-dimensional vector-space $\{r \cdot v \mid r \in \mathbb{R} \}$ for some vector $v$. In this vector space, there are exactly two vectors $u_1,u_2$ having length one and these are related by $u_1 = -u_2$. Thus, as soon as you have fixed the first vector, the only choice left is the sign of the second one.
This works in general: If you have an orthogonal matrix $A \in \mathbb{R}^{n \times n}$ and multiply any column with $-1$, the matrix will stay orthogonal. So you always have a choice in the sign of a vector. Thus the $g(N)$ might be defined "up to sign" to resolve such issues, maybe you can find something about this in your book/lecture notes.
