How to decompose a $3\times 3$ orthoginal matrix back into Reflection and Rotation components.

Question

I am writing some CAD software that keeps track of the orientation of the modeled components. The oriention is stored as a single $3\times 3$ orientation matrix that is a product of a reflection $R$ and a rotation $Q$, thus $T = RQ$. The question is, given $T$, can I reconstruct the $R$ and $Q$ components? And if so, is this deconstuction unique?

Notes:

$R$ is a Householder reflection matrix:

$R = I_3 - 2 n n^\text{T}.$ For some unit 3-vector $n$.

$\text{det}(R)$ = -1.

$Q$ is an Euler rotation matrix, $\text{det}(Q)$ = 1.

Answer 1

Rewrite $$RT = Q$$

Now $$Q = {S_1}^{-1}\underset{C_1}{\underbrace{\left(\begin{bmatrix}a&b\\-b&a\end{bmatrix} \otimes 1\right)}}S_1$$ $$R={S_2}^{-1}\underset{C_2}{\underbrace{(1\otimes 1 \otimes -1)}}{S_2}$$

Here $\otimes$ denotes block diagonalization operator. In other words:

$$(a_1\otimes a_2 \otimes ... \otimes a_N) = \begin{bmatrix}a_1&&&\\&a_2&&\\&&\ddots&\\&&&a_N\end{bmatrix}$$

For two arbitrary transformations $S_1,S_2$ and $a=\cos(\phi),b=\sin(\phi)$

We can conclude that if $RT=Q$ holds, then $RT-Q =0$ and furthermore $\|RT-Q\|$ will be minimal.

Now we can create an alternating regularized linear least squares

$$\min\left\{\|RT-Q\|_2 + \|S_2R-C_2S_2\|_2 + \|S_1Q-(C_1)S_1\|_2\right\}$$

Answer 2

No, you cannot. You have $T = (I-2nn^T)Q$. Choose an arbitrary unit vector $e$ and set $U := (I-2ee^T)T$. Then $\det U = 1$, so $U$ is a rotation, and $T = (I-2ee^T)U$.