$Q^T\dot{Q}=\text{constant}$ if $Q$ is an orthogonal matrix?

Question

I was told that if $Q\in O(n)$ (the orthogonal group, i.e. $Q^TQ = I$), then $Q^T\dot{Q}=\text{constant}$. I wanted to show this so I took the derivative $\dot{Q}^T\dot{Q}+Q^T\ddot{Q}$. But why is this zero?

Answer 1

I presume $\dot{Q}$ means that $Q$ is a differentiable function of some real variable $t$ with values in $O(n)$, and you're taking the derivative with respect to $t$. Then it's not true. For example, consider

$$ Q = \pmatrix{\cos(t^2) & \sin(t^2)\cr -\sin(t^2) & \cos(t)}$$ where $$ Q^T \dot{Q} = \pmatrix{0 & 2t\cr -2t & 0}$$

What is true is that $Q^T \dot{Q}$ is skew-symmetric, i.e. $$0 = \dfrac{d}{dt}(Q^T Q) = Q^T \dot{Q} + \dot{Q}^T Q = Q^T \dot{Q} + (Q^T \dot{Q})^T$$