Decompositions of rotations around the origin in 3D and more. Suppose that I have a solid in 3D with its center (maybe the centroid, but any point will do) is positioned at the origin.
Suppose I choose any line passing through the origin and apply a rotation by some angle.
Is that rotation decomposable in a sequence of $3$ rotations by some $3$ angles around the $3$ cartesian space axes $x,y,z$?
In general, is that true for every solid/polytope in $n$ dimensions? Meaning that any rotation around any line passing through the origin can be decomposed in a sequence of $n$ rotations about the $n$ axes of the cartesian coordinate system?
 A: In 3 dimensions, the answer is yes. We can work backwards. Suppose that your arbitrary rotation sends the unit vectors $\mathbf i, \mathbf j, \mathbf k$ (corresponding to the positive $x$-, $y$-, $z$-axes) to unit vectors $\mathbf u$, $\mathbf v$, and $\mathbf w$. We can find:

*

*A rotation $R_1$ about the $z$-axis that puts $\mathbf u$ on the $(x,z)$-plane. If $\mathbf u$ is already pointing in the direction of the $z$-axis nothing needs to be done; otherwise, a rotation with an angle of $\theta$ turns $\mathbf u$'s projection onto the $(x,y)$-plane by $\theta$, and we just get that projection to point in the positive or negative $x$ direction.

*A rotation $R_2$ about the $y$-axis that sends $R_1(\mathbf u)$ to $\mathbf i$. This is possible, because $R_1(\mathbf u)$ and $\mathbf i$ are both unit vectors in the $(x,z)$-plane. Note that rotations preserve angles, so our three unit vectors always stay orthogonal. Since $R_2(R_1(\mathbf u)) = \mathbf i$, we know that $R_2(R_1(\mathbf v))$ and $R_2(R_1(\mathbf w))$ are both in the $(y,z)$-plane: orthogonal to $\mathbf i$.

*A rotation $R_3$ about the $x$-axis that sends $R_2(R_1(\mathbf v))$ to $\mathbf j$. This is possible, because both are unit vectors in the $(y,z)$-plane. Note that $R_3$ does not affect the $x$-axis.

Since $R_3(R_2(R_1(\mathbf u))) = \mathbf i$ and $R_3(R_2(R_1(\mathbf v))) = \mathbf j$, we know by orthogonality that $R_3(R_2(R_1(\mathbf w)))) = \pm \mathbf k$. Putting $\mathbf w$ on $-\mathbf k$, however, would be a reflection, and we know that only orientation-preserving transformations happened, throughout.
So the composition $R_3 \circ R_2 \circ R_1$ brings $\mathbf u, \mathbf v, \mathbf w$ back to $\mathbf i$, $\mathbf j$, and $\mathbf k$. This means it must be the inverse of our original rotation. Therefore the original rotation we started with must be $R_1^{-1} \circ R_2^{-1} \circ R_3^{-1}$, which is a composition of three rotations about the axes.

In $n \ge 4$ dimensions, the question does not make sense; you cannot simply choose a line through the origin and an angle, and rotate every point around that line. (Or to put it differently, there are many different rotations by that angle that fix the line you chose.)
A simple rotation, instead, rotates around an $(n-2)$-dimensional subspace. (You could say that it is a rotation in the plane orthogonal to that subspace.) There are also more complicated transformations that we also think of as rotations.
We can decompose any orientation-preserving transformation as a composition of the $\binom n2$ rotations in the $(x_i,x_j)$-plane for all pairs $\{i,j\}$. This has a recursive strategy similar to the 3D strategy above: we show how to undo a transformation using a sequence of these rotations.
First, we do rotations in the $(x_i, x_n)$-plane for all $i$ until the $n^{\text{th}}$ basis vector is in the correct position. What's left is the $\binom{n-1}{2}$ rotations in the $(x_i,x_j)$-plane for $i,j<n$, which preserve the $x_n$-axis. So they don't break what we've done already, and we can just pretend we're working in $(n-1)$ dimensions. Repeat until we get down to $n=3$ or even $n=2$, which we understand.
