4D Left Isoclinic Rotations as double rotations

For this matrix, $a$ is the cosine of the rotation angle.

Proof sketch: First show that $R$ is an orthogonal matrix, so that $|Ru| = |u|$ for all $u$. Now recall that $u^T v =|u| |v|\cos\theta$ where $\theta$ angle between two vectors $u$ and $v$. So then, to compute the angle between $u$ and $Ru$, show that $u^T R u = a u^T u$.

There are always many ways to split $\Bbb{R}^4$ into a direct sum of two planes, stable under $R$ (and hence both rotated by the same angle).

The eigenvalues of $R$ are complex, $\lambda_{\pm}=a\pm i\sqrt{b^2+c^2+d^2}$, both with multiplicity two. Let $\vec{v}_1, \vec{v}_2\in\Bbb{C}^4$ be two linearly independent eigenvectors belonging to $\lambda_+$. Then their componentwise complex conjugates $\vec{v}_1^*$ and $\vec{v}_2^*$ belong to $\lambda_-$. Consider the spaces $$ T_j=\{z\vec{v}_j+z^*\vec{v}_j^*\mid z\in\Bbb{C}\}, $$ $j=1,2$. Clearly $T_1$ and $T_2$ are both subsets of $\Bbb{R}^4$ and $2$-dimensional vector spaces over $\Bbb{R}$. Furthermore $$ R(z\vec{v}_j+z^*\vec{v}_j^*)=\lambda_+z\vec{v}_j+\lambda_-z^*\vec{v}_j^*\in T_j, $$ so the two planes are stable under $R$.

Replacing $\vec{v}_1,\vec{v}_2$ with a different pair will produce different pairs of planes.

The common angle of rotation (assuming normalized to a unit quaternion) is $\arccos a$.