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For matrices that are elements of $SO(3)$ is there a formula for the eigenvectors corresponding to the eigenvalue $1$ in terms of the entries of the matrix?

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Let $A \in SO(3)$. The matrix $A-A^T$ is skew-symmetric, hence of the form $$ \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} $$ for some $a$, $b$, $c$. Then the vector $(-c,b,-a)^T$ is an eigenvector of $A$ with eigenvalue $1$.

This works since if $F$ is a rotation by the angle $\phi$ around the vector $n$, then the transformation $F-F^{-1}$ geometrically equals $\sin\phi$ times the operation of taking the cross product with $n$.

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In computer graphics quaternion spherical linear interpolation is common. Since finding the eigenvector is straightforward would a pure matrix approach to spherical linear interpolation be competitive to quaternions? – user782220 Jun 7 '12 at 22:01
I have no idea about that! I should add that this method fails if $\phi=\pi$ (since then $a=b=c=0$), but in that case any nonzero column of $I+A$ will give you the eigenvector instead. – Hans Lundmark Jun 8 '12 at 4:24

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