4D Left Isoclinic Rotations as double rotations According to this page, 4D left-isoclinic rotation matrices are double rotations of the same angle (and same sign). But in the same page, in the context of quaternions, it is mentioned that left-isoclinic matrices have the form:
$$R=\begin{bmatrix}a&-b&-c&-d \\ b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{bmatrix}$$ where $a,b,c,d\in\mathbb{R}$ such that $$a^2+b^2+c^2+d^2=1.$$ My question is, how can we view $R$ as a double rotation? I mean, how can we find the common angle? If two of $b,c,d$ are $0$ then it is easy to see. But what if this is not the case?
 A: 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 $u^T R u = a u^T u$.
A: 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 $\arccos a$.
A: Notice the first column! This is the matrix of left-multiplication-by-$(a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k})$. We may write this quaternion in polar form as $ a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}=\exp(\theta\mathbf{u}) $.
Solving for the unknowns $\theta$ and $\mathbf{u}$ with Euler's formula yields
$$ \begin{cases} \cos\theta=a \\ \sin\theta=\sqrt{b^2+c^2+d^2} \end{cases}, 
\qquad \mathbf{u}=\frac{~~~~b\mathbf{i}+c\mathbf{j}+d\mathbf{k}}{\sqrt{b^2+c^2+d^2}}. $$
The angle $\theta$ is convex and $\mathbf{u}$ is a unit vector. If we extend $\{\mathbf{u}\}$ to an (appropriately oriented) orthonormal basis $\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ for $\mathbb{R}^3$, then $\{1,\mathbf{u},\mathbf{v},\mathbf{w}\}$ is an orthonormal basis for $\mathbb{H}$ with the same orientation and multiplication table as $\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$.
With respect to this basis, the transformation acts as a rotation by $\theta$ in the oriented $1\mathbf{u}$-plane and its orthogonal complement, the oriented $\mathbf{vw}$-plane. (Exercise.)
There is more than one way to choose a pair of oriented, orthogonal, stable planes. The stable planes for $\exp(\theta\mathbf{u})$ are the same as for those of multiplying-by-$\mathbf{u}$. Multiplying by $\mathbf{u}$ is a complex structure on $\mathbb{H}$ making it a left complex vector space, so the stable planes are precisely the 1D complex subspaces! This forms the complex projective line $\mathbb{CP}^1\simeq S^2$, diffeomorphic to the Riemann sphere by stereographic projection.
(The space of all oriented 2D planes in 4D real space is the real Grassmanian $\mathrm{Gr}_2\mathbb{R}^4\simeq S^2\times S^2$, which can be shown in a couple ways using quaternions too.)
