2
$\begingroup$

My syllabus states the following procedure to determine quickly the rotational axis of a 3x3 matrix $A$ that is orthogonal with determinant 1, but it's not completely clear:

"If $A$ $\in$ SO(3) and $A$ is not symmetric and $u$ is an eigenvector with eigenvalue 1 (so $u$ determines the axis of rotation), then construct the skew-symmetric part of $A$: $$S = \frac{1}{2}(A-A^T) = \begin{bmatrix} 0 & \omega_3 &-\omega_2 \\ -\omega_3 & 0 & \omega_1 \\ \omega_2 & -\omega_1 & 0 \end{bmatrix}$$

From $Au = u$, it follows that $Su = 0$, so the axis of rotation is determined by the vector $[\omega_1,\omega_2,\omega_3]^T$.

If $A$ is symmetric then $S = 0$. $A$ is in that case the identity matrix (in which case this theorem is ofcourse true), or $A$ is orthogonal equivalent with $diag(1,-1,-1)$ (in which case the theorem doesn't hold)."

I understand that if $u$ determines the axis of rotation, then $u$ is in the kernel of $S$, but I don't understand why it follows that $[\omega_1,\omega_2,\omega_3]^T$ (which is in the kernel of $S$) has to determine the axis of rotation, which seems to imply that the kernel of $S$ is 1-dimensional (if this is true, then the problem is solved, because the 1-dimensional space of vectors that determine the axis of rotation are already in the kernel, so there can be no other vectors in the kernel of $S$).

The second thing I don't understand is why $A$, if symmetric and not $I$, has to be orthogonal equivalent with $diag(1,-1,-1)$.

I hope someone can help me out.

$\endgroup$
3
$\begingroup$

For your first question, $\ker S$ is indeed one-dimensional if $S$ is nonzero. You may verify that if $\omega_i\neq0$, then the two rows of $S$ containing $\omega_i$ are linearly independent.

For the second question, every real symmetric matrix $A$ is orthogonally equivalent to a real diagonal matrix that contains the real eigenvalues of $A$ as diagonal entries. Now $A$ is also real orthogonal, so these eigenvalues have to be $\pm1$. Since $\det A=1$ and $A\neq I$, the only possibility is that two of these eigenvalues are $-1$ and the remaining one is $1$.

$\endgroup$
  • 1
    $\begingroup$ Thank you. That was exactly what I was looking for. It was way less difficult than I thought :D. $\endgroup$ – yarnamc Mar 24 '13 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.