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I tried to prove this theorem but having hard time to start with. What I have gotten hint is that starting with rigid motion f = La(left multiplication by A) where A is in O(3) (set of all orthogonal matrices) with det(A)=-1 then using eigenvalue.

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  • $\begingroup$ Cubics always have roots, so you always have an eigenvalue. One has eigenspace dimension 1, which is your axis of rotation. $\endgroup$
    – Randall
    Jan 26, 2018 at 19:37
  • $\begingroup$ @Randall How do you know there is an eigenspace of dimension $1$? $\endgroup$ Jan 26, 2018 at 19:52
  • $\begingroup$ Use a Gram-Schmidt style argument. You have a real eigenvalue, so choose a non-zero eigenvector (might as well make it unit). Expand this to a basis. The resulting matrix comes in a block which makes it clear. (I am also assuming the original determinant was 1.) $\endgroup$
    – Randall
    Jan 26, 2018 at 20:01
  • $\begingroup$ Oh, I see now. I've been assuming that the OP wants to show that every element of $SO(3)$ (besides the identity) has an axis. True, but not what was asked. I didn't read all that well. $\endgroup$
    – Randall
    Jan 26, 2018 at 20:02
  • $\begingroup$ Suggestion to the post (v2): Change -1 to +1. $\endgroup$
    – Qmechanic
    Jan 27, 2018 at 13:00

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What is a "proper" rigid motion? The map taking $x$ to $-x$ is a rigid motion that fixes the origin, but is not a rotation.

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