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Say you have a rotation matrix $R$ and a translation matrix $t$, you can trivially have a single matrix $[\;R\;|\;t\;]$. Now say you have another matrix $R'$, can you write $R'[\;R\;|\;t\;]$ as $[\;R''\;|\;t'\;]$? I would think yes, but I would like to be sure.

Edit: $[\;R\;|\;t\;]$ is a rotation matrix with the right most colomn a translation. It's the same as applying the rotation $R$ first followed by the translation $t$. This might indeed be somewhat of a funky notation but it is used in the following book: http://szeliski.org/Book/. For example on page 50.

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What do you mean by $[R\mid t]$? Do you literally mean the $n\times 2n$ matrix $$\left[\!\!\begin{array}{ccc|ccc} R_{11} & \cdots & R_{1n} & t_{11} & \cdots t_{1n}\\ \vdots & \ddots & \vdots & \vdots & \ddots \vdots\\ R_{n1} & \cdots & R_{nn} & t_{n1} & \cdots t_{nn}\\ \end{array}\!\!\right]$$ –  Zev Chonoles Apr 24 '13 at 22:28
    
@ZevChonoles: Given from context, I would assume he means regular matrix multiplication. But I do admit, the notation is very funky... –  Tomas Lycken Apr 24 '13 at 22:30
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Welcome to Math.SE! Thank you for your question. We will be able to better answer it if you share the context, explain your notation and definitions, and let us know what you've tried so far. –  vadim123 Apr 24 '13 at 23:00
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1 Answer 1

Yes, you can. I don't know.


Originally, I assumed you meant applying the operations as rotation-rotation-translation, but since you mean rotation-translation-rotation, my reasoning does not apply to your problem. I'll leave it up here anyway, for history.

Think of it in terms of general matrix multiplication: the order of the multiplications on the left hand side here is unimportant, since matrix multiplication is associative.

In other words, if you are transforming a vector $v$, you have the first transformation as $v'=Rtv$. Applying the second rotation, $R'$, you obtain $R'v' = R'(Rtv) = (R'R)tv = R''tv$, again because of the associativity of matrix multiplication.

Note, though, that the order in which you apply these transformations is important - matrix multiplication is in general not commutative, and rotational matrices are no exception.

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I have claryfied the notation I used in my question. My exact question is that knowing that multiplication does not commute, can you still rewrite a rotation followed by translation followed by rotation as a rotation followd by a translation. –  Silverrocker Apr 25 '13 at 6:46
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OK - no, then I don't know. I've searched to find if there are any special properties of transformation matrices like these that you could use, but without result. I think that if this is possible, it is built upon the idea that any general transformation can be separated into components, so you construct the matrix for the total transformation and then separate it again - but I don't know if the assumption is valid or not, so I don't know the answer to your question. –  Tomas Lycken Apr 25 '13 at 15:03
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