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In a few places I have seen general results for rotation matrices (such as finding their eigenvalues etc.) proved by proving them for the case where the matrix is

$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{bmatrix} $$

Why is it possible to reduce all rotations in $\mathbb R^3$ to this matrix WLOG?

(for example on pg.6 here: https://www2.bc.edu/~reederma/Linalg17.pdf)

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  • $\begingroup$ Just change basis so that the $x$-axis is the axis of the rotation (so the rotation is happening only in the $yz$-plane. $\endgroup$ – Nick Feb 23 '16 at 18:32
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Let $R$ be some rotation matrix, and $R^*$ the matrix you propose. Let $f_1$ be the direction that the rotation matrix did not change. Choose $f_2$ and $f_3$ perpendicular to $f_1$ and each other. Now transform the rotation in the standard basis to the orthogonal basis orthogonal basis $\langle f_1,f_2,f_3 \rangle$.

Choose $B = (f_1 | f_2 | f_3)$, then $B^{-1} = B^T$ (due to orthogonality) and $R^* = B R B^T$.

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  • $\begingroup$ So this means that the eigenvalues of R will be precisely the eigenvalues of R*? $\endgroup$ – Jen Feb 23 '16 at 19:44
  • $\begingroup$ That is correct. You can show that this also holds for every orthogonal matrix, because of this reasoning: $p(\lambda)=\det(R-\lambda I) = \det( (R-\lambda I)^T ) = \det (R^T - \lambda I^T) = \det (R - \lambda I)$. Solving $p(\lambda)=0$ gives the eigenvalues, so you're done ;) $\endgroup$ – Jasper Feb 24 '16 at 14:53
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Rotations are a rigid transformation. If we have a rigid transformation, the eigenvalues, including possible complex values, must be |z| = 1, If they are not, then we have some kind of compression / dilation, and the transformation is not rigid.

In R^3 the characteristic polynomial to determine eigenvalues is a cubic polynomial. Which means that it must have one real root. and since |z| = 1 it must be 1 or -1.

If the product of the eigenvalues = -1, then the sphere is flipped inside-out.

and since the characteristic polynomial has real coefficients, complex eigenvalues are in conjugate pairs.

that means the eigenvalues are (1, e^it, e^-it) which is equivalent to (1, cos t + i sin t, cos t - i sin t), and could include (1,1,1) and (1,-1,-1).

What does all this mean? There is always some axis of rotation which is the eigenvector associated with the eigenvalue 1, and then there is the degree of rotation, t.

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