Matrix Exponent - equivalent of a rotation matrix

Question

Every Rotation Matrixcan be represented as a power of e with exponent a skew symmetric matrix.

In particular, if we have a rotation matrix ${R}\in\mathbb R^{3 \times 3,}$ then there will be a skew symmetric $B$ such that $R=\mathrm{e}^B.\tag 1$ How do we prove that and also for a given $R$ how do we calculate $B$?

Thanks.

Answer 1

A rotation matrix $R$ is orthogonally diagonalizable with eigenvalues of absolute value one, i.e., $$ R=U^* D U, $$ where $D=\mathrm{diag}(d_1,\ldots,d_n)$, with $\lvert d_j\rvert=1$, for all $j=1,\ldots,n$, and $U^*U=I$. Clearly, as $\lvert d_j\rvert=1$, there exists a $\vartheta_j\in\mathbb R$, such that $$ d_j=\mathrm{e}^{i\vartheta_j}, \quad j=1,\ldots,n. $$ Now it is clear that if $B=U^*\varTheta U$, where $\varTheta=\mathrm{diag}(i\vartheta_1,\ldots,i\vartheta_n)$, then $$ R=\mathrm{e}^B. $$ We have used here the fact that, if $A=V^*DV$, then $\mathrm{e}^A=V^*\mathrm{e}^DV$.

Answer 2

What we have given is a rotation matrix R. We know there exits a form $e^B$ for every R. Now as per the paper $B=log R \tag 1$

$ \theta = \arccos\left( \frac{\mathrm{trace}(R) - 1}{2} \right) \tag 2$ $ \log R = \{ \begin{matrix} 0 & \mathrm{if} \; \theta = 0 \\ \frac{\theta}{2 \sin(\theta)} (R - R^\top) & \mathrm{if} \; \theta \ne 0 \; \mathrm{and} \; \theta \in (-\pi, \pi) \end{matrix} \tag 3$