Exactly Half of a 3D Transformation matrix

Suppose that your transformation consists of a rotation by non-zero angle $theta$ (about the origin), whose $3 times 3$ matrix we denote $R$, followed by the translation $x mapsto x + d$. That is, the full transformation (in standard coordinates) is given by $T(x) = Rx + d$.

Let $J$ denote the $90^circ$ rotation about the same axis as $R$. Note that $R$ can be written in the form $R = cos theta I + sin theta J$, where $I$ denotes the $3 times 3$ identity matrix. Its "square root" can be written in the form $S = cos (theta/2)I + sin (theta/2)J$. It is easy to see that $S$ should be the rotational component of our "halfway" transformation.

Note that if we are given the transformation matrix $R$, then the angle satisfies $$ operatorname{trace}(R) = 1 + 2 cos theta, $$ which allows us to compute $cos theta, sin theta,$ and $J$ from the matrix $R$.

We will describe the halfway transformation as $H(x) = Sx + e$, and we wish to solve for the vector $e$. We must select $e$ such that $T(x) = H(H(x))$. That is, $$ Rx + d = S(Sx + e) + e = S^2x + Se + e = Rx + (I + S)e. $$ In other words, we must select $e$ such that $(I+S)e = d$. That is, it suffices to take $e = (I + S)^{-1}d$, where we note that $(I + S)$ must be invertible because it is a rotation by $theta/2$, which is not a multiple of $180^circ$ (so there is no $x$ for which $Sx = -x implies (I + S)x = 0$).

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We can go a step further an compute this inverse explicitly. We have $$ (I + S)^{-1} = ((1 + cos (theta/2))I + sin(theta/2)J)^{-1}. $$ It turns out that because we have $J^2 = -I$, we can treat these coefficients as the real and complex parts of a complex number. So, following the process by which one would compute $(a + bi)^{-1}$, we find that $$ (I + S)^{-1} = frac{1}{(1 + cos(theta/2))^2 + sin^2(theta/2)}((1 + cos (theta/2))I - sin(theta/2)J) \ = frac{1}{2 + 2 cos (theta/2)}((1 + cos (theta/2))I - sin(theta/2)J), $$ which is to say that we have $$ e = frac{1}{2 + 2 cos (theta/2)}((1 + cos (theta/2))I - sin(theta/2)J)d. $$