How to prove that the composition of 2 rotations about an axis $l_1$ and $l_2$ is the rotation?
I know that we can represent rotation about the axis $l$ at angle $\phi$ as the composition of 2 symmetries relative to the axis passing through the planes the angle between them is equal $ \frac{\phi}{2} $. But how using it prove that the composition of 2 rotations about an axis $l_1$ and $l_2$ is the rotation?
Thanks for the help!
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$\begingroup$Well, if you agree that a rotation $R$ can be represented as a matrix so that $R R^T=I$, then the same is true for a composition $R_1 R_2$. So, $R_1 R_2$ is an orthogonal matrix and if $R_1, R_2$ have positive determinant (they are rotations, not reflections), so has $R_1 R_2$. But we are in dimension $3$, so the characteristic polynomial of $R_1 R_2$ is of order three and must have a real root. Therefore, there exists a real eigenvector; the eigenvalue can only be $1$ or $-1$. If it is $1$, you have an axis and are done. If it were $-1$, then the other two eigenvalues can be either complex numbers $z$ and $z^*$, but that would contradict $\det=1$, or the pair $\{1, -1\}$ in which case you again have found an eigenvector corresponding to $1$.
Note that in $\mathbb{R}^4$, it might happen that the composition of two rotation has no "axis" (no line needs to be fixed).
$\endgroup$ $\begingroup$Two ways to see this in a visual mode:
First way.
First of all, two reflections give rise to a rotation, the angle of the rotation being the double of the angle formed by the planes defining the reflections. Reciprocally, a rotation can be seen as the product of two reflection, and we have the freedom to choose one of the reflection, except for the fact that its plane must contain the axis of the rotation.
Once you are convinced of this, if you have two rotations in 3D, $R_1=\{e_1,\phi_1\}$ and $R_2=\{e_2,\phi_2\}$, you can fix a reflection along the plane defined by the axis of $e_1$ and $e_2$, call it $T$. So we have that$$ R_1=TU $$and$$ R_2=VT $$and therefore$$ R_2 R_1=VTTU=VU $$is the product of two reflection and hence a rotation.
In the picture, we have expressed reflections by means of the intersection of their defining planes with a unitary sphere.
Second way.
We can express rotations by giving two points in a unitary sphere. This way, two points $A$ and $B$ determine the rotation $R_{AB}$ whose axis is the orthogonal line to the plane $OAB$ and whose angle is twice the distance from $A$ to $B$ (the arc of the great circle resulting from the intersection of plane $OAB$ with the sphere). Observe that giving a rotation, we have some freedom to choose one of the points.
With this convention to define rotations, suppose you have two rotations $R_1$ and $R_2$. We can choose the point $B$ of intersection of the two great circles defined by both rotations in the unitary sphere. Then we have two points $A$ and $C$ such that$$ R_1=R_{AB} $$and$$ R_2=R_{BC} $$
This way, we get a spherical triangle $ABC$. Now, we are going to construct three more triangles congruent with the original one. We take symmetric point along the great circles and give them names the way you see in the figure:
Triangles $T_1=AB'C'$, $T_2=A''BC''$ and $T_3=A'''B'''C$ are all congruent with $ABC$ since they have in common with it an angle and two sides. But observe that $R_{BC}R_{AB}$ take $T_1$ to $T_3$, the same that it does $R_{AC}$. And since three points determine any symmetry of the sphere, we are done:$$
R_{BC}R_{AB}=R_{AC}
$$
These two geometric ideas are of great importance in Clifford algebras and spinors.
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