What is the geometrical action of a skew-symmetric matrix on an arbitrary vector? What is the geometrical action of a skew-symmetric matrix on an arbitrary vector?
The rotation matrix is a skew-symmetric matrix when $\theta$ is some multiple of $\frac{\pi}{2}$. But it cannot be true that every skew-symmetric matrix represents a rotation?
Also, since the leading diagonal is zero, it cannot represent a scaling nor a shear. In fact, none of the standard transformation matrices on Wikipedia seem to fit the pattern of an arbitrary skew-symmetric matrix.
So can anything be said about the geometrical action of a skew-symmetric matrix on an arbitrary vector?
 A: As explained at Wikipedia, a skew-symmetric matrix can be brought into the form
$$\begin{bmatrix}
\begin{matrix}0 & \lambda_1\\ -\lambda_1 & 0\end{matrix} &  0 & \cdots & 0 \\
0 & \begin{matrix}0 & \lambda_2\\ -\lambda_2 & 0\end{matrix} &  & 0 \\
\vdots &  & \ddots & \vdots \\
0 & 0 & \cdots & \begin{matrix}0 & \lambda_r\\ -\lambda_r & 0\end{matrix} \\
& & & & \begin{matrix}0 \\ & \ddots \\ & & 0 \end{matrix}
\end{bmatrix}$$
by an orthogonal transformation. Each of the $2\times2$ blocks on the diagonal is a rotation through $\pi/2$ times a scaling by $\lambda_i$. In other words, there is a basis such that the basis vectors form pairs (except for the ones that are annihilated) and the action in the plane formed by each pair is a rotation through $\pi/2$ and a scaling.
A: As also explained on Wikipedia, a skew-symmetric matrix can be thought of as an infinitesimal rotation. So there will be an axis around which a rotating motion will, for each point in space, give rise to a velocity that equals the skew-symmetric matrix times the position of the same point. Hope that made sense!
