Show that there is a unique matrix $ A $ such that $ \varphi (t) = e ^ {tA} $. Let $ \varphi:\mathbb R\to\mathcal M_{n\times n}(\mathbb R)$ be a function from $\mathbb R$ to the space of $n\times n$ real-valued matrices, and suppose that each component of $\varphi$ is a $C^1$ function. If $\varphi(0)=I$ (identity), and $\varphi(t + s) = \varphi (t)  \varphi (s)$ for all $ t, s \in \mathbb R $, show that there is a unique matrix $ A $ such that $ \varphi (t) = e ^ {tA} $. 
Hint: Consider $ A = \varphi (0)'$. 
 A: Here's a more elementary approach which doesn't rely on Lie groups:
Since each component of $\varphi(t) \in \mathcal M_{n \times n}(\Bbb R)$ is of class $C^1$, $\varphi(t)$ is a differentiable matrix function of $t$.  Note that $\varphi(s)$ and $\varphi(t)$ commute for any $s, t \in \Bbb R$, since
$\varphi(s)\varphi(t) = \varphi(s + t) = \varphi(t + s) = \varphi(t)\varphi(s).  \tag{1}$
We compute $\varphi'(t)$ directly from first principles, using the formula $\varphi(s + t) = \varphi(s) \varphi(t)$.  We have:
$\varphi'(t) = \lim_{h \to 0}\dfrac{\varphi(t + h) - \varphi(t)}{h} = \lim_{h \to 0}\dfrac{\varphi(t)\varphi(h) - \varphi(t)}{h} = \lim_{h \to 0}\dfrac{\varphi(h)\varphi(t) - \varphi(t)}{h}$
$= \lim_{h \to 0}\dfrac{(\varphi(h) - I)\varphi(t)}{h}= \lim_{h \to 0}\dfrac{(\varphi(h) - I)}{h}\varphi(t) = \varphi'(0)\varphi(t), \tag{2}$
where we have used (1) in the derivation (4); 
$\lim_{h \to 0}\dfrac{(\varphi(h) - I)}{h} = \varphi'(0) \tag{3}$
follows directly from the hypotheses that $\varphi(t)$ is of class $C^1$ and that $\varphi(0) = I$.  We thus see that $\varphi(t)$ satisfies the linear differential equation
$\varphi'(t) = \varphi'(0)\varphi(t) \tag{4}$
with the initial condition $\varphi(0) = I$.  The unique solution to (4) with $\varphi(0) = I$ is
$\varphi(t) = e^{\varphi'(0) t}; \tag{5}$
this shows that $\varphi(t)$ has the requisite form.  To see that $\varphi'(0)$ is the only matrix $A$ such that $\varphi(t) = e^{At}$, note that if
$\varphi(t) = e^{At}, \tag{6}$
then
$\varphi'(0)\varphi(t) = \varphi'(t) = Ae^{At}; \tag{7}$
setting $t = 0$ in (7) yields $\varphi'(0) = A$; $\varphi'(0)$ is the only matrix such that (6) holds.  And we are done!  QED!!!
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: I'll explain using some properties of Lie groups but the key idea is that $\phi$ satisfies the ODE corresponding to the exponential function.
Note that since the determinant and $\phi$ are both continuous, there has to be some $\epsilon$ such that for all $t \in (-\epsilon, \epsilon)$, $\phi(t) \in GL_{n}(\mathbb{R})$. But now for any $t \in \mathbb{R}$, we can find some integer $n$ such that $\frac{t}{n} \in (-\epsilon, \epsilon)$. Hence, we have 
$$\phi(t) = \phi\left(n \frac{t}{n}\right) = \left(\phi\left(\frac{t}{n}\right)\right)^{n} \in GL_{n}(\mathbb{R}).$$
So $\phi$ is actually a $C^{1}$ homomorphism from $\mathbb{R}$ to $GL_{n}(\mathbb{R})$ and is hence a $C^{1}$ one parameter subgroup. Now, the exponential map $\exp: M_{n}(\mathbb{R}) \rightarrow GL_{n}(\mathbb{R})$ is a local diffeomorphism at $0$, and hence in some neighborhood of the identity in $GL_{n}$, a $\text{Log}$ function exists. 
Now, at this point, we could invoke general properties of Lie groups to say that such a homomorphism must be $e^{t\phi'(0)}$ but we can actually get this result by using the uniqueness of an integral curve corresponding to a vector field with given initial conditions.
So, we let $\psi(t) = e^{t A}$ where $A = \phi'(0)$. Then, 
$$\psi'(t) = A e^{tA} = A \psi(t).$$
On the other hand,since $\phi$ is a homomorphism 
$$\phi'(t) = \frac{d}{ds}\phi(t + s)|_{s = 0} = \phi(t) \phi'(0) = A \phi(t).$$
So, $\psi(t)$ and $\phi(t)$ are both integral curves of the left invariant vector field corresponding to the matrix $A$, and since $\phi(0) = \psi(0)$, $\phi$ and $\psi$ must be the same functions by the uniqueness of integral curves.
This gives existence of the matrix $A$, but the matrix $A$ must be unique because it must be $\phi'(0)$. 
