Retrieving angle of rotation from transformation matrixby Phong Tags: angle, matrix, retrieving, rotation, transformation 

#1
Dec1909, 02:19 AM

P: 5

Hi!
How do I calculate the angle of rotation for each axis by a given 4x4 transformation matrix? The thing is that all values are a kind of mixed up in the matrix, so I cannot get discrete values to start calculating with anymore. Thanks, Phong 



#2
Dec1909, 04:17 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

A four by four transformation matrix? Are you rotating in four dimensional space or is this a projective space?
First find the eigenvalues. A rotation matrix, in four dimensions may have two real and two complexconjugate eigenvalues or two pairs of complex eigenvalues. If there are two real eigenvalues they must be either 1 or negative one. The eigenvectors corresponding to those eigenvalues give the axes of rotation. The complex eigenvalues will have modulus 1 and are of the form [itex]cos(\theta)\pm i sin(\theta)[/itex] where [itex]\theta[/itex] is the angle of rotation. Two pairs of complex rotation give two simultaneous rotations in four space but are again of the form [itex]cos(\theta)+ i sin(\theta)[/itex]. What those mean depends upon how you are writing vectors in four space. If you are talking about a matrix representing a rotation matrix projectively, then you can renormalize to make the last row [0 0 0 1] and the last column [tex]\begin{bmatrix}0 \\ 0 \\ 0\\ 1\end{bmatrix}[/tex]. The 3 by 3 matrix made up of the first three rows and columns will have one eigenvalue of 1 (the corresponding eigenvector gives the axis of rotation) and two complex conjugate eigenvalues of modulus 1. They will be of the form [itex]cos(\theta)+ i sin(\theta)[/itex] where [itex]\theta[/itex] is the angle of rotation. 



#3
Dec1909, 12:54 PM

P: 5

Hello!
Thank you very much for your detailed reply. I must admit that I'm pretty new to transformation matrices and have not yet entirely understood the mathematical meaning of eigenvalues and eigenvectors although I try hard to understand everything I can read about it, but with some help I surely learn a lot faster. I'm rotating in a projective threedimensional space, that's why I use a 4x4 matrix. To give a more specific example, I have a transformation matrix that is the following: [tex]\begin{bmatrix}0.893 & 0.060 & 0.447 & 20 \\ 0.157 & 0.97 & 0.184 & 15 \\ 0.423 & 0.235 & 0.875 & 45 \\ 0 & 0 & 0 & 1 \end{bmatrix}[/tex] This transformation matrix should transform the object with a translation of 20 15 45 and a rotation of 15 25 10 (xyz). Now the eigenvalues. I don't know if I've understood the meaning of them correctly, but if yes the eigenvalues for this matrix should be in the identity matrix which is: [tex]\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}[/tex] So if I'm still on the right track, the eigenvectors, which are the axes of rotation, are simply [tex]\left(1, 0, 0, 0)[/tex] for x [tex]\left(0, 1, 0, 0)[/tex] for y [tex]\left(0, 0, 1, 0)[/tex] for z Am I still on the right track or am I totally and fatally wrong? 


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