I'm struggling with this proof:
Let $\phi(t)$ a square matrix of size $n$ with $C^1$ functions such that $$\phi(0)=I_n \qquad \text{and} \qquad \phi(s+t)=\phi(s)\phi(t)$$ $\forall s,t\in\mathbb{R}$. Prove that there exists a square matrix $A\in M_n(\mathbb{R})$ such that $\phi(t)=e^{tA}$, for all $t$.
I'm almost sure that the matrix $\phi$ is kind of solution of an linear ODE. But I don't realize how i'm supposed to construct the matrix A
We differentiate $\phi(t)$ by means of the given formula
$\phi(s + t) = \phi(s)\phi(t), \tag 1$
viz:
$\phi(s + t) - \phi(t) = \phi(s) \phi(t) - \phi(t) = (\phi(s) - I)\phi(t); \tag 2$
$\dfrac{\phi(s + t) - \phi(t)}{s} = \dfrac{\phi(s) - I}{s} \phi(t); \tag 3$
$\phi'(t) = \displaystyle \lim_{s \to 0}\dfrac{\phi(s + t) - \phi(t)}{s} = \lim_{s \to 0} \dfrac{\phi(s) - I}{s} \phi(t) = \phi'(0) \phi(t) = A\phi(t), \tag 4$
where
$A = \lim_{s \to 0} \dfrac{\phi(s) - I}{s}; \tag 5$
we know these limits as well as the derivative $\phi'(t)$ exist because we are given that the entries of $\phi(t)$ are $C^1$ functions; the unique solution to (4), which exists by virtue of the fact that $A\phi$ is a Lipschitz continuous function of $\phi$, is
$\phi(t) = \phi(0) e^{At}; \tag 6$
since
$\phi(0) = I_n, \tag 7$
we find
$\phi(t) = I_n e^{At} = e^{At}. \tag 8$