Why should the trace of a 3d rotation matrix have these properties?

Question

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.

Answer 1

$3$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.

Answer 2

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

1. $R v_3 = v_3$ (i.e. $v_3$ is the "axis of rotation")
2. $R$ 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 associated 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 $\theta$.