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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?

Froskoy
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2 Answers2

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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.

joriki
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  • I understand! Thank you very much. – Froskoy May 07 '12 at 16:23
  • @Froskoy: You're welcome! – joriki May 07 '12 at 16:23
  • @joriki, Is it true that If $A$ is $3x3$ matrix such that $\forall_{u \in R^3} \space u'Au = 0$ then A is skew symmetric. My reasoning goes as follow: $u'Au = u'v$ where $(Au=v)$, and $u'v = 0$ that means $A$ transforms $u$ in such a way that it has angle of $90$ degree with itself after transform. So It has something to do with skew symmetric matrix. I am not able find whether my statement is correct and if it is correct then please give some idea why that statement is correct? – Vimal Patel Jan 01 '20 at 03:24
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    @VimalPatel: That's correct; in fact it's true in any number of dimensions. You can prove it concretely by first using $u=(1,0,0),(0,1,0),(0,0,1)$ to show that the diagonal elements are zero, then using $u=(1,1,0),(1,0,1),(0,1,1)$ to show that the off-diagonal elements are skew-symmetric. I'm sure there's also a nice more abstract proof based on universal properties instead of coordinates. – joriki Jan 01 '20 at 04:08
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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!

HelloGoodbye
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