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Trying to find the axis of rotation.

$$ M(x)= \frac{1}{2}\left[ {\begin{array}{cc} -1+\cos\ x & \sqrt{2}\sin\ x & 1+\cos\ x\\ \sqrt{2}\sin\ x & -2\cos\ x & \sqrt{2}\sin\ x\\ 1 + \cos\ x & \sqrt{2}\sin\ x & -1 + \cos\ x \end{array} } \right] $$

I know the trace $= -1$ and that this cooresponds to a rotation of order 2 but I'm not sure how to find the axis of rotation.

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  • $\begingroup$ See math.stackexchange.com/a/1788804/265466. $\endgroup$ – amd Dec 6 '17 at 0:29
  • $\begingroup$ This method seems very hard to follow... $\endgroup$ – Tommy Dec 6 '17 at 4:10
  • $\begingroup$ Seems pretty straightforward to me: plug your matrix into a relatively simple formula and read the exits from any row of the result. Using the asymmetric part of the rotation matrix is even simpler, but it doesn’t work when the rotation angle is $\pi$, as is the case here. $\endgroup$ – amd Dec 6 '17 at 7:05
  • $\begingroup$ @Tommy If you are ok, you can set as solved. Thanks! $\endgroup$ – user Dec 6 '17 at 13:51
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One way to compute the rotation axis is to find an eigenvector of $1$, i.e., to compute a basis for the kernel of $M-I$. This involves a slightly messy, but not too difficult row-reduction computation. However, there are a couple of ways to find the rotation axis without referring to eigenvalues or eigenvectors.

As explained here, the simplest way is to take the skew-symmetric part of $M$, which is a scalar multiple of the “cross-product matrix” of a vector on the rotation axis and read off the axis vector coordinates. That is, for a 3-D rotation matrix $R$, $R-R^T=\sin\theta\,K$, where $\theta$ is the rotation angle and $K$ is the axis cross-product matrix. Unfortunately, for the matrix in this question $\operatorname{tr}M=1+2\cos\theta=-1$, which means that $\theta=\pi$ and $M-M^T=0$, which gives us no information about the axis.

Fortunately, with a little more work the axis can also be extracted from the symmetric part of $R$: each of the rows/columns of $T=R+R^T-(\operatorname{tr}R-1)\,I$ consists of the coordinates of a vector on the rotation axis. This method has the advantage over using the skew-symmetric part of working for any rotation angle.

In this case, $$T = \begin{bmatrix} 1+\cos x & \sqrt2 \sin x & 1+\cos x \\ \sqrt2 \sin x & 2-2\cos x & \sqrt2 \sin x \\ 1+\cos x & \sqrt2 \sin x & 1+\cos x \end{bmatrix},$$ and so the rotation axis is $[1+\cos x, \sqrt2 \sin x, 1+\cos x]^T$, which you can verify satisfies $Mv=v$.

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  • $\begingroup$ Where is the “here” that you mention? $\endgroup$ – José Carlos Santos Dec 6 '17 at 7:31
  • $\begingroup$ @JoséCarlosSantos Oops. Forgot to include the link—the same one as in my comment to the question. $\endgroup$ – amd Dec 6 '17 at 7:46
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HINT

you have to find a vector s.t. $Av=v$ thus the eigenvectors corresponding to eigenvalue = 1.

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  • $\begingroup$ So $Av - Iv = 0$ / $(A-I)v = 0$ where v is the vector cooresponding to the axis of rotation? Or is there a better way to compute this given how messy the matrix is. $\endgroup$ – Tommy Dec 5 '17 at 22:14
  • $\begingroup$ Computing it, I only get (0, 0, 0) which must be wrong... $\endgroup$ – Tommy Dec 6 '17 at 4:19
  • $\begingroup$ it means that 1 is an eigenvalue! thus a fixed axes exists $\endgroup$ – user Dec 6 '17 at 4:20
  • $\begingroup$ Sorry if I seem confused but my question lies more with determining what the axis is? Not whether it exists. Not sure how to do that. $\endgroup$ – Tommy Dec 6 '17 at 4:23
  • $\begingroup$ you have to find the eigenvector associated to v, that is the solution to $(A-I)v=0$ $\endgroup$ – user Dec 6 '17 at 4:30

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