What $X$ is: The axis of rotation.
What $A$ does to $X$: $A$ is a rotation.
What $M$ requires of $X$: $X$ must be in the nullspace of $M$.
This means $A$ maps $X$ to itself:
\[
A X = X \quad \text{with eigenvalue } \lambda = 1.
\]
$A$ is an orthogonal matrix, so its transpose is its inverse: $A^T = A^{-1}$.
The inverse matrix $A^{-1}$ represents the inverse rotation (rotation by $-\theta$ around the same axis $X$).
If $X$ is unchanged by rotation $A$, it must also be unchanged by the inverse rotation $A^{-1}$.
If your vector $X$ is the axis of rotation, what happens when you rotate around it by $-\theta$ ? Nothing. It still stays exactly where it is.
This means $A^T$ maps $X$ to itself:
\[
A^T X = A^{-1} X = X.
\]
We found Result 1: $A X = X$
We found Result 2: $A^T X = X$
Hence, $A X = A^T X$.
The condition for a vector $X$ to be in the nullspace of $M = A - A^T$ is precisely $A X = A^T X$.
Therefore, the axis of rotation ($X$) is guaranteed to be in the nullspace of $M$.
Theorem: The rank of a $3 \times 3$ skew-symmetric matrix must be even (0 or 2).
The matrix $M = A - A^T$:
\[
\underbrace{M}_{\text{Skew-Symmetric}}
\xrightarrow{\text{Nonzero?}}
\underbrace{\text{Yes, because } \theta \neq 0, \pi}_{\text{Check Constraint}}
\xrightarrow{\text{3D Rule}}
\underbrace{\text{Rank } = 2}_{\text{Fixed Result}}
\]
