Rigorous way to show a continuous function of matrices is orthogonal

Question

Suppose $Q(t)$ is a continuous set of real matrices with $t \in \mathbb R$ a parameter. Suppose $Q(0)$ is orthogonal and suppose $$Q'(t) = S(t) Q(t)$$ where $S(t)$ is a continuous set of real skew-symmetric matrices. I conclude that $Q(t)$ is orthogonal everywhere.

My poor informal way is this pseudo-inductive method:

Suppose $Q(t')$ is orthogonal:

$Q(t'+\delta t) = Q(t') + \delta t\times S(t')Q(t')$

$Q(t'+\delta t) \times Q^*(t'+\delta t)$

$=(Q(t') + \delta t\times S(t')Q(t')) \times (Q(t') + \delta t\times S(t')Q(t'))^*$

$=(Q(t') + \delta t\times S(t')Q(t')) \times (Q^*(t') - \delta t\times Q^*(t')S(t'))$

letting the $\delta t ^2$ go to zero, this becomes:

$=I$

proving $Q(t'+\delta t)$ is orthogonal.

So with $Q(0)$ orthogonal, and with the inductive step, I conclude $Q(t)$ is orthogonal everywhere.

What is the rigorous/formal way to do this?

Answer 1

Look at $\Phi : t \mapsto (Q^tQ)(t)$. If you differentiate it, you get $0$, so it's constant, equal to $\Phi(0)=I$.

$$\Phi^{'}(t)=(Q^{t})^{'}(t)Q(t)+Q^t(t)Q^{'}(t)=0$$