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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?

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up vote 4 down vote accepted

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.

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I understand! Thank you very much. – Froskoy May 7 '12 at 16:23
@Froskoy: You're welcome! – joriki May 7 '12 at 16:23

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