Given a smooth group homomorphism $t\mapsto \psi(t)\in\mathrm{GL}(n,\mathbb{R})$, where $t\in(\mathbb{R}$,+), find an $n\times n$ matrix $A$ such that $\psi(t)=e^{tA}$ for every $t$.
A bit of context:
I have shown that if $A$ is a given $n\times n$ matrix, with a vector field $v$ on $\mathbb{R}^n$ defined by $v(x)=Ax$, then $v$ is a complete vector field. In fact, the integral curve passing through a point $x_0\in\mathbb{R}^n$ is given by $t\mapsto e^{tA}x_0$.
Furthermore, $v$ generates a one-parameter subgroup of $\mathrm {Diff}(\mathbb{R}^n)$ by the flows of $v$. The flows are exactly $e^{tA}$. I therefore also proved that $e^{(t+s)A}=e^{tA}e^{sA}$.
My Attempt:
I first notice that if it is indeed the case, then we would have $\psi(0)=I$ and $\psi^{(n)}(0)=A^n$. So my thought is to take $A=\psi'(0)$. Moreover, both $\psi(t)$ and $e^{tA}$ are smooth curves passing through $I$ at $t=0$, and $\mathbb{R}$ is their common domain. I want to find a vector field $v$ defined on $\mathrm{GL}(n,\mathbb{R})$ so that both $\psi(t)$ and $e^{tA}$ are integral curves of $v$, and therefore by the uniqueness of integral curves they are equal.
However, I have difficulty in finding such $v$ and I feel like I could have make use of what I have already proved. Any hint would be appreciated.
You don't need to know anything about vector fields to do this. Instead of $GL_n(\mathbb{R})$ it's much simpler to work in $M_n(\mathbb{R})$ because it's a vector space. Take $A = \psi'(0)$. We have that
$$\begin{align*} \frac{d}{dt} \psi(t) &= \lim_{h \to 0} \frac{\psi(t+h) - \psi(t)}{h} \\ &= \lim_{h \to 0} \frac{\psi(t) (\psi(h) - I)}{h} = \psi(t) A. \\ \end{align*}$$
Next we compute that
$$\frac{d}{dt} \psi(t) e^{-At} = \psi(t) A e^{-At} + \psi(t) (-A e^{-At}) = 0.$$
So $\psi(t) e^{-At}$ is a constant, and it has value $I$ at $t = 0$, so it is equal to $I$ everywhere.
Alternatively and more conceptually we can also use the fact that $e^{At}$ and $\psi(t)$ are both solutions to the differential equation $M'(t) = M(t) A$ (or $A M(t)$, either way) with initial condition $M(0) = I$, then appeal to the uniqueness part of the Picard-Lindelof theorem, but that's overkill here since the above computation works too and uses only the product rule.
Implicitly I am using some facts about derivatives of maps from $\mathbb{R}$ to matrices (e.g. the product rule, which requires a bit of care due to noncommutativity) but they are all straightforward facts to prove once it occurs to you to write them down at all. Everything generalizes without much additional difficulty to Banach algebras.