On the Wikipedia article about Rotation Matrices (), the article states that the trace of the matrix will be equal to 1 + 2 cos(theta), where the theta represents the angle of the rotation in axis/angle form.
How is this property found? There doesn't appear to be any derivation on the site, and I can't see any reason why it might be the case.
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$\begingroup$$3$D rotation is defined as fixing a pole and rotating the orthogonal sub space to that pole (a unit vector). For instance, if our pole is the vector $(0,0,1)$, we rotate the orthogonal subspace given by the $x-y$ plane.
The sub space is roared according the the rotational matrix.
Defined by:
$$\begin{bmatrix}\cos (\theta) &-\sin (\theta) \\\sin (\theta) & \cos (\theta) \end{bmatrix}$$ .
Choosing basis suitably, we can make $v_1$ our first basis vector and this is fixed by the rotation. While the other bases will be transformed according to our rotation angle. Therefore, all rotation matrices are similar to:
$$\begin{pmatrix} 1&0&0\\ 0&\cos(\theta)&-\sin(\theta)\\ 0&\sin(\theta)&\cos(\theta)\\ \end{pmatrix} $$
Similar matrices have same trace so it follows.
Edit: I should have a book somewhere explaining this in detail, if you want, let me know so that I can find the book and post an image.
$\endgroup$ 3 $\begingroup$Note that when we say that $R$ is a rotation matrix, we're really saying that there exists an orthonormal basis $\{v_1,v_2,v_3\}$ such that
- $R v_3 = v_3$ (i.e. $v_3$ is the "axis of rotation")
- $R$ rotates vectors in the plane spanned by $v_1,v_2$.
So for instance, in the case that $v_1,v_2,v_3$ are the standard basis (that is, $\hat i, \hat j, \hat k$), $R$ is the matrix of the rotation in the $xy$-plane, namely$$ R_{\theta} = \pmatrix{\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1}. $$When $v_1,v_2,v_3$ is taken to be an arbitrary orthonormal basis, the resulting matrix is$$ R = VR_{\theta}V^{-1} $$where $R_{\theta}$ is the matrix above and $V$ is the matrix whose columns are $v_1,v_2,v_3$ (in other words, the change-of-basis matrix). Note: because $V$ has orthonormal columns, $V^{-1} = V^T$.
Now, it turns out that for any matrix $A$ and any invertible $S$, we have$$ \operatorname{trace}(SAS^{-1}) = \operatorname{trace}(A). $$In more abstract terms, the trace of the matrix associate with a linear transformation will always be the same, regardless of the basis chosen. So, our new rotation $VR_{\theta}V^{-1}$ has the same trace as $R_{\theta}$, the "standard" rotation by $\theta$.
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