Can a rotation matrix be written as the convex combination of two rotation matrices?

Question

I want to say that for all $R,S \in SO(3)$ and $0 < t < 1$ (note the strict inequalities) with $R \not = S$, we must have $tR + (1-t)S \not \in SO(3)$. I've been trying to think up a counterexample to this but can't imagine any. Is this true? I know that if some $t$ does satisfy this, then we must have $(1-t^2-(1-t)^2)I=t(1-t)(R^TS + S^TR)$, but can't seem to go beyond this. Any ideas?

Answer 1

A more succinct variation on previous answers: since each column (and row) of a "rotation matrix" (more generally any orthogonal matrix) is a vector on the unit sphere, any nontrivial convex combination of two rotation matrices induces the same convex relation on their first columns. But all points on the unit sphere are extreme points: it is impossible that a nontrivial convex relation can be satisfied by three distinct points on the unit sphere. Thus the convex combination must be trivial; i.e., all three rotation matrices must agree in their first column. Ditto for all other such columns.Since all columns agree, the matrices agree.

Answer 2

From Jyrki's answer: Additive rotation matrices

"Assume contrariwise that for certain rotations $R_1,R_2,R_3$ the equation" $$ tR_1\vec{x}+(1-t)R_2\vec{x}=R_3\vec{x}$$ (*the formula changed by me wrt Jyrki's answer to meet conditions of the problem)

"holds for all $\vec{x}\in\Bbb{R}^3$. If this works for the triple $(R_1,R_2,R_3)$ then multiplying $(*)$ from the left by $R_3^{-1}$ we see that it also works for the triple $(R_3^{-1}R_1,R_3^{-1}R_2,I_3)$. So without loss of generality we can assume that $R_3$ is the identity mapping."

Below my continuation:

therefore for some rotation matrices $Q_1=R_3^{-1}R_1$, $Q_2=R_3^{-1}R_2$

$tQ_1\vec{x}+(1-t)Q_2\vec{x}=I\vec{x}$

$tQ_1\vec{x} =(t-1)Q_2\vec{x}+ I\vec{x}$

Assume now $\vec{u}$ is the axis vector of $Q_2$ i.e. $Q_2\vec{u}=\vec{u}$

Then
$tQ_1\vec{u} =(t-1)Q_2\vec{u}+ I\vec{u}=t\vec{u}$

From this

$Q_1\vec{u}=\vec{u}$ so $Q_1$ has the same axis as $Q_2$.

So from initial assumption we have that there are two rotation matrices $Q_1,Q_2$ about the same axis that

$tQ_1 +(1-t)Q_2 =I$

But taking now any unit vector $\vec{v}$ lying in the plane of rotation, vector which being rotated with all possible angles defines on this plane an unit circle, we have that point defined by vector

$tQ_1\vec{v} +(1-t)Q_2\vec{v}$ should lie on the internal part of the segment with endpoints $Q_1\vec{v}$ and $Q_2\vec{v}$ (these endpoints are lying on the unit circle), and consequently this point doesn't lie on the the unit circle as point $I\vec{v}$ does,

which leads to
contradiction.

Answer 3

Let $C = (1 - t) A + t B$ where $A$ and $B$ are the two given rotation matrices, such that

$ A = [a_1, a_2, a_3]$ , $B = [b_1, b_2, b_3] $, and $C = [c_1, c_2, c_3] $

Then, it follows that

$c_1 = (1 - t) a_1 + t b_1$

We want $c_1$ to be a unit vector, so

$1 = (1 - t)^2 (a_1^T a_1) + t^2 (b_1^T b_1) + 2 t (1 - t) (a_1^T b_1 )$

Since $a_1 $ and $b_1$ and unit vectors then

$ 1 = (1 - t)^2 + t^2 + 2 t (1 - t) (a_1^T b_1)$

$ 0 = - 2 t + 2 t^2 + 2 t (1 - t) (a_1^T b_1 ) $

Since $t \ne 0$, then

$ 0 = -2 + 2 t + 2 (1 - t) (a_1^T b_1) $

Thus

$ 0 = -2 (1 - t) + 2 (1 - t) (a_1^T b_1) $

Since $ t \ne 1$ then

$ a_1^T b_1 = 1 $

which means the angle between $a_1$ and $b_1$ is zero. Similarly,

$a_2^T b_2 = 1 $ and $ a_3^T b_3 = 1 $

And this means, that for $C$ to be a rotation matrix, matrices $A$ and $B$ must be identical.