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Let $v \in \mathbb{R}^3$ and $R$ be a rotation matrix (probably orthogonal is enough). I think the following property is true:

$$ [R\cdot v] = R \cdot [v] \cdot R^T$$ where $[v]$ is the skew symmetric form of the vector $v$. I should verify this by direct calculation, but I am not in the mood right now, hence I want to know if the following proof is true:

proof: let $u \in \mathbb{R}^3$ hence $$ [R\cdot v] \cdot \left( R \cdot u\right) = \left(R\cdot v\right) \times \left( R\cdot u\right) = R \cdot \left( v\times u\right) = R \cdot [v] \cdot R^T \cdot \left( R \cdot u\right)$$ therefore $$ [R\cdot v] \cdot \left( R \cdot u\right) = R \cdot [v] \cdot R^T \cdot \left( R \cdot u\right) $$ hence $\left([R\cdot v] - R \cdot [v] \cdot R^T \right) \cdot w = 0$ for all $w = R \cdot u$ with $u \in \mathbb{R}^3$. It follows that $\left([R\cdot v] - R \cdot [v] \cdot R^T \right) = O_{3 \times 3}$

Edit:

Suppose $v = \begin{bmatrix} a\\b\\c\end{bmatrix}$ then $[v] = \begin{bmatrix}0 &-c &b\\ c &0 &-a\\ -b &a &0 \end{bmatrix}$

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  • $\begingroup$ by skew symmetric form of $\begin{pmatrix}a\\b\\c\end{pmatrix}$ do you mean $\begin{pmatrix}0&-c& b\\ c& 0&-a\\-b& a&0\end{pmatrix}$ ? $\endgroup$
    – Jean Marie
    Sep 5, 2017 at 21:44
  • $\begingroup$ @JeanMarie yes that is what I mean $\endgroup$
    – C Marius
    Sep 5, 2017 at 21:47
  • $\begingroup$ I'm not sure what you're looking for here. Your statement is true and your proof is valid---does that help? $\endgroup$
    – wrvb
    Sep 5, 2017 at 22:04
  • $\begingroup$ @wrvb Yes that is enough! Thank you for your attention! By the way ... is this a well known property that I have just rediscovered? $\endgroup$
    – C Marius
    Sep 5, 2017 at 22:05
  • $\begingroup$ Yes, it's pretty well known. Most people who have worked with angular velocities in three dimensions will at some point write down some form of your identity, since it allows you to relate the angular velocity in the stationary and nonstationary frames: $\frac{d}{dt}R = R[\omega]$, while $\frac{d}{dt}R^T = -R^T[R \omega]$ $\endgroup$
    – wrvb
    Sep 5, 2017 at 22:11

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