- #1
mef51
- 23
- 0
Homework Statement
Suppose that ##s \to A(s) \subset \mathbb{M}_{33}(\mathbb{R})## is smooth and that ##A(s)## is antisymmetric for all ##s##. If ##Q_0 \in SO(3)##, show that the unique solution (which you may assume exists) to
$$\dot{Q}(s) = A(s)Q(s), \quad Q(0) = Q_0$$
satisfies ##Q(s) \in SO(3)## for all ##s##.
Homework Equations
For ##Q \in \mathbb{M}_{33}(\mathbb{R})##, if:
- ##Q^TQ = I, \quad Q \in O(3)##
- additionally, if ##det(Q) = 1, \quad Q \in SO(3)##
- Note: ##Q^TQ = QQ^T## for ##Q \in SO(3)##
- ##A## is antisymmetric if ##A^T = -A##
The Attempt at a Solution
Since I only know that ##Q_0## is in ##SO(3)##, I suspect that ##\dot{Q}(s)## must be 0. In that case, since the derivative is zero, ##Q(s)## must be constant and ##Q(s)=Q_0## for all ##s##. So I try to prove that ##A(s)Q(s)## is zero.
Based on a hint someone gave me I'll start by taking the derivative of ##{Q_0}^TQ_0 = I##:
$$
({Q_0}^TQ_0)' = I' \\
\dot{Q_0}^TQ_0 + {Q_0}^T\dot{Q_0} = 0_{33} \\
{Q_0}^T\dot{Q_0} = - \dot{Q_0}^TQ_0 \\
Q_0{Q_0}^T\dot{Q_0} = - Q_0\dot{Q_0}^TQ_0 \\
Q_0{Q_0}^T\dot{Q_0} = - Q_0\dot{Q_0}^TQ_0 \\
\dot{Q_0} = - Q_0\dot{Q_0}^TQ_0 \\
$$
Now I plug in that expression for ##\dot{Q_0}## in the differential equation..
$$
- Q_0\dot{Q_0}^T Q_0 = A_0Q_0 = -{A_0}^T Q_0
$$
I'm not really sure what this gives me though. I'm also not confident that you can even use the product rule like that. It certainly isn't true for matrices in general.
Any help or pokes in the right direction are appreciated!
mef