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Let $v_1,v_2\in \Bbb R^4$ be unit vectors with $\langle v_1,v_2\rangle =\cos \theta$. Let $H_i=I-2v_iv_i^t$ be the Householder transformation for $i=1,2$. I am asked to compute the angle of the rotation $H_2\circ H_1:\Bbb R^4\to \Bbb R^4$. We have (Are the product and sum of two Householder matrices also Householder matrices?) $H_2H_1=I-2v_2v_2^t-2v_1v_1^t+4v_2v_2^tv_1v_1^t$. But I don't know what "the angle of a rotation in $\Bbb R^4$" means. How do we compute the angle of this map? Should I compute the eigenvalues? (https://en.wikipedia.org/wiki/Rotation_matrix#Properties)

blancket
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Let's assume $u,v$ are linearly independent unit vectors and use the notation $H_u=I-2uu^\ast$. Since both reflections $H_u,H_v$ pointwise fix the orthogonal complement $\mathrm{span}\{u,v\}^\perp$, so too does their composition $R=H_v\circ H_u$. Therefore, this composition restricts to a rotation of the 2D plane $\mathrm{span}\{u,v\}$. Really, this works in any dimension, not just 4D.

Any 2D rotation is a rotation by an angle. Now it's worth thinking about Euclidean geometry. Let $A$ be any point, $B$ its reflection across $u$, and $C$ the reflection of $B$ across $v$. Thus, $R(A)=C$. The vector $u$ bisects the angle $\angle AB$ and the vector $v$ bisects the angle $\angle BC$. The angle $\theta=\angle uv$ satisfies $\cos\theta=\langle u,v\rangle$. These facts allow us to determine $R$'s angle of rotation in terms of $\theta$.

Of course, once you know $R$ is a plane rotation, you can find its angle $\alpha$ algebraically using the double angle formula. Extending any orthonormal basis for $R$'s plane of rotation to one of space, $R$ is a block matrix with one $I_{n-2}$ block and one $2\times2$ rotation matrix block. The trace is then $\mathrm{tr}(R)=(n-2)+\cos(\alpha)$. On the other hand, we can distribute

$$R \,=\, H_vH_v \,=\, (I-2vv^\ast)(I-2uu^\ast) \,=\, I-2uu^\ast-2vv^\ast+\langle u,v\rangle vu^\ast$$

and use the fact $\langle u,v\rangle=v^\ast u=\mathrm{tr}(vu^\ast)$ to get

$$\mathrm{tr}(R) ~=~ n-2-2+4\cos^2\theta ~=~ (n-2)+\cos(2\theta).$$

coiso
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