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In a paper I'm reading I've come accross this notation several times now and can't figure out what it's supposed to represent. Sometimes it is applied to just one term or a product of terms like this and sometimes it seems to be applied to a list of arguments like the occurrence in this matrix.

Can anyone help me understand the meaning of this notation?

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  • $\begingroup$ @MishaLavrov This is the paper: uu.diva-portal.org/smash/get/diva2:827236/FULLTEXT01.pdf $\endgroup$ – Sean Duffy Apr 17 '17 at 23:36
  • $\begingroup$ Looks like p.10 says "$[\boldsymbol{a}]_\times$ is the skew-symmetric matrix representation of the vector cross product" which is a more concise and less helpful version of the answer you've already received. But given that the paper has a section for symbols where they define things like transpose I'm annoyed that it's not included there. $\endgroup$ – Misha Lavrov Apr 17 '17 at 23:44
  • $\begingroup$ Oh and Appendix A spells it out, except that due to an unfortunate typesetting failure it defines $[\boldsymbol{a}]_x$ instead. $\endgroup$ – Misha Lavrov Apr 17 '17 at 23:46
  • $\begingroup$ I usually write it as $ [ {\bf r} \times ]$ to indicage that the vector $\bf r$ is transformed into the matrix for the cross product operator. $\endgroup$ – ja72 Apr 23 '17 at 23:18
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Let $V=\begin{pmatrix}a\\b\\c\end{pmatrix}$.

$[V]_{\times}$ denotes the antisymmetrical matrix, associated with the linear operation "taking the cross product with $V$", i.e.,

$$[V]_{\times}X:=V \times X=\begin{pmatrix} -cy+bz\\ \ \ \ cx-az\\-bx+ay \end{pmatrix},$$

which means that $$[V]_{\times}=\begin{pmatrix}\ \ \ 0&-c& \ \ \ b\\ \ \ \ c& \ \ \ 0&-a\\-b& \ \ \ a& \ \ \ 0\end{pmatrix}.$$

(this notation is described in (https://en.wikipedia.org/wiki/Cross_product)).

Remarks:

1) Very often $V$ is assumed to be unitary ($a^2+b^2+c^2=1$).

2) Here is a very interesting paper (https://arxiv.org/pdf/1312.0788v1.pdf) where this notation is used with many of its properties grouped in its Appendix A.

3) This notation doesn't look very old ; maybe one should post the question of its origin (robotics ?) on the "history of mathematics" section of SE.

Edit:

4) I have had a glance at the source paper you have indicated after I wrote my answer.

In the block-defined matrix, $\hat R^n_{b,k}f_k^b$ has to be a vector. There are two ways to consider this notation:

  • a) either as a single matrix $\hat R^n_{b,k}$ or $\left(\hat R^n_{b}\right)_k$applied to a vector $f_k^b$, yielding a vector.

  • b) or (much less likely) a linear combination of matrices applied to vectors, with a hidden summation sign according to Einstein notation (https://en.wikipedia.org/wiki/Einstein_notation), $\sum_{b=1}^m \hat R^n_{b,k}f_k^b$ where the summation is on the index that is present once in a lower, once in a upper position, here index $b$.

5) See also the recent document (https://people.eecs.berkeley.edu/~wkahan/MathH110/Cross.pdf) and the older one (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.825.1726&rep=rep1&type=pdf) (1989)

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    $\begingroup$ Thanks, this does seem to be what the notation means. However I'm still a little unclear what the computation taking place in the matrix in my second example is. My first thought would be that the vectors are being multiplied and then the anti-symmetrical matrix is taken, but with the comma I'm not so sure. Is it possible this is just a notational error? $\endgroup$ – Sean Duffy Apr 18 '17 at 0:16
  • $\begingroup$ Ah yes of course, apologies, I was pretty tired when I wrote that comment and it didn't occur to me that the comma was within the subscript. Of course R is a matrix being applied to the vector f, yielding a vector result from which the anti-symmetric matrix is taken. $\endgroup$ – Sean Duffy Apr 18 '17 at 13:51

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