Let's consider a $2\times 2$ rotation matrix $R_\theta \in SO(2,\mathbb{R})$, and the following matrix obtained by repeatedly applying $n-1$ times the Kronecker product of $R_\theta$ with itself: $$Q_\theta = R_\theta{\,}^{\otimes n} = \underbrace{R_\theta\otimes\ldots \otimes R_\theta}_{R_\theta\,\mathrm{appears}\, n \, \mathrm{times}}$$
From the mixed-product property of the Kronecker product, it is straightforward to verify that the matrix $Q_\theta$ is orthogonal, i.e. $Q_\theta Q_\theta^T=I$ (where $I$ is the $2n\times2n$ identity matrix). In fact, it is equally straightforward to verify that: $$\mathcal{Q}=\{Q_\theta : \theta \in [0,2\pi) \} \subset SO(2n,\mathbb{R})$$
In other words, the set of matrices $Q_\theta$ equipped with the ordinary matrix product, form a subgroup of $SO(2n,\mathbb{R})$.
Does anyone have suggestions on how I could proceed to determine the eigenspace of $Q_\theta$?
My attempts have been so far limited to studying the case $n=2$: $$Q_\theta = (I \cos\theta + E\sin\theta) \otimes (I \cos\theta + E\sin\theta) \in SO(4,\mathbb{R})$$ where $I=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}$ and $E=\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}$.
Expaning the product in the above formula yields: $$(I\otimes I)\cos^2\theta + (E\otimes E) \sin^2\theta +(I\otimes E + E\otimes I)\sin\theta\cos\theta$$ which can be re-expressed (using the double-angle formulas) as: $$(I\otimes I)\cos(2\theta) +\frac{1}{2}(I\otimes E + E\otimes I)\sin(2\theta) + (I\otimes I + E\otimes E)\sin^2\theta$$ which I didn't find particularly insightful, as I can't easily recognize the eigenspace of rotations in that formula.
Best Answer
This is just about the $n=2$ case.
If $e_1,e_2$ is the standard basis of $V=\mathbb{R}^2$, then the matrix of $R_\theta\otimes R_\theta$ on $V\otimes V$ with respect to the ordered basis $e_1\otimes e_1,e_1\otimes e_2,e_2\otimes e_1,e_2\otimes e_2$ is readily computed (see the answer by @Jean Marie) as $$R_\theta\otimes R_\theta=\begin{pmatrix} c^2& -cs& -cs& s^2\\ cs & c^2& -s^2& -cs\\ cs & -s^2& c^2& -cs\\ s^2 & cs & cs& c^2\end{pmatrix}$$ where $c:=\cos\theta$, $s:=\sin\theta$.
The $1$-eigenspace of dimension $2$ has basis $e_1\otimes e_1 +e_2\otimes e_2, e_1\otimes e_2 -e_2\otimes e_1$: to verify this just add/subtract the relevant rows/columns of the matrix.
The $2$ dimensional space ('plane') on which $R_\theta\otimes R_\theta$ acts as rotation by $2\theta$ must be the orthogonal complement of the $1$-eigenspace, that is the space with basis $e_1\otimes e_1 -e_2\otimes e_2, e_1\otimes e_2 +e_2\otimes e_1$. Again this is easily verified by adding/subtracting the relevant rows/columns of the matrix.
Comment 1 In fact the same argument (with the same bases) would let us identify the two 'planes' for $R_\theta\otimes R_\phi$: the one on which it acts as rotation by $\theta+\phi$, and the one on which it acts as rotation by $\theta-\phi$. This in principle gives a recursive way of building up appropriate bases/planes for $(R_\theta)^{(n+1)}=(R_\theta)^{(n)}\otimes R_\theta$.
Comment 2 If one attempts to deal with $R_\theta^{(n)}$ directly it seems to me that the Chebychev polynomials have something to do with it.