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.


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 $1$.

  • 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

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