Why should the trace of a 3d rotation matrix have these properties? On the Wikipedia article about Rotation Matrices (https://en.wikipedia.org/wiki/Rotation_matrix#Determining_the_angle), 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.
 A: $3$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.
A: 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 $\theta$.
