Magnitude of rotation in R^3 from rotation matrix

Given a rotation matrix $Q \in R^3$, how to find a magnitude of rotation. I guess it makes sense to say that identity matrix $I$ is considered 0 magnitude and the rotation of $-I$ is considered maximum magnitude. It would also makes sense that $mag(Q) = mag(Q^T)$.

Is there a way of computing this without going through the trouble of finding Euler angles and then deriving metric using identity $cos^2(\alpha) + cos^2(\beta) + cos^2(\gamma) = 1$?

• hmm, I have doubts whether $-I$ we can treat as a rotation matrix in 3d ... its determinant $=-1$ – Widawensen Nov 30 '16 at 20:18
• I am pretty sure it is. Any orthogonal matrix can be considered a rotation matrix since the rotation has to follow property $Q Q^T = Q Q^{-1} = I$, it follows that determinant be +- 1. Also, you can verify that vector under this transformation remains constant length, since no scaling is done. Therefore it is rotation of some sort. But to make the matter concrete, consider vector $v = [1, 1, 1]$ – Joonatan Samuel Dec 1 '16 at 10:20
• I was taught that determinant must be +1. But maybe it is just more general approach.., – Widawensen Dec 1 '16 at 13:08
• What's the connection between the Euler angles, the direction cosines identity and the rotation magnitude ??? – Yves Daoust Dec 1 '16 at 13:10
• On negative determinant. en.wikipedia.org/wiki/… – Joonatan Samuel Dec 1 '16 at 13:28

You don't really define what is meant by magnitude and I don't think that the Euler angles representation is appropriate for that purpose.

I suggest to use the axis/angle representation, and this method: https://en.wikipedia.org/wiki/Rotation_matrix#Determining_the_angle.

• ..with axis/angle rotation he rather would not be able to go from $I$ to $-I$. He needs something additional.... – Widawensen Dec 1 '16 at 13:12
• @Widawensen: $-I$ is not a rotation. – Yves Daoust Dec 1 '16 at 13:18
• How is -I not a rotation? It takes vector v and rotates it to negative v. Any matrix transformation that doesn't change length of v is considered rotation, and negative I definitely doesn't change the length of v. – Joonatan Samuel Dec 1 '16 at 13:28
• @JoonatanSamuel: no, it's a reflection, $|-I|=-1$. – Yves Daoust Dec 1 '16 at 13:30
• en.wikipedia.org/wiki/… – Joonatan Samuel Dec 1 '16 at 13:37

I found it useful to think about the transformation rotation matrix $Q \in R^{3 \times 3}$ affects given vector $v$. That means what does $$\vec u = \vec v \times Q^T$$ look like. After that I can compare the initial vector $\vec v$ and compare it to $\vec u$ by dot product between them since $$cos\ \phi = \frac{\vec u \cdotp \vec v}{||\vec u|| \ ||\vec v||}$$.

$\vec v \in R^3$; $\vec v$ should be unit vector, let us choose $[1, 0, 0]$. giving the metric from $[-1; 1]$ of $$metric(Q) = (\vec v Q^T) \cdot \vec v;\ v = [1, 0, 0]$$

This simplifies to $metric(Q) = Q_{11}$, which agrees with maximum of 1, when $Q = I$, minimum of -1 when $Q = -I$ and $metric(Q) = metric(Q^T)$