Consider a $2 \times 2$ matrix of smooth vector fields on a smooth manifold $M$: $$X = \begin{pmatrix} X_{11} & X_{12} \\ X_{22} & X_{21} \\ \end{pmatrix}$$ I'm thinking about $X$ as an operator on $C^\infty(M)^2$ or, if you prefer, as an operator on the sections of the trivial vector bundle $M \times \mathbb{R}^2$.
Note: It is fine with me if you want to replace $M$ by $\mathbb{R}^n$, I am mostly just trying to notationally distinguish it from the time dimension.
Suppose we want to talk about the associated time-evolution operators $e^{tX}: C^\infty(M)^2 \to C^\infty(M)^2$. What do we mean by that? Maybe it would be cleaner to talk instead about the following system of PDEs \begin{align*} \tfrac{\partial}{\partial t}u_1(t,x) &= X_{11}u_1(t,x) + X_{12} u_2(t,x) \\ \tfrac{\partial}{\partial t}u_2(t,x) &= X_{21}u_1(t,x) + X_{22} u_2(t,x) \end{align*} where $u_1,u_2 \in C^\infty(\mathbb{R}\times M)$ with the $X_{ij}$ acting in the obvious way along the fibers $\{t\}\times M$. Writing $u_t(x)=(u_1(t,x),u_2(t,x))$, the idea is then that $e^{tX} u_0 = u_t$.
Now, there is a pretty clear issue with existence. For example, if $M=\mathbb{R}$ with variable $x$ and $$ X = \begin{pmatrix} 0 & -\tfrac{\partial}{\partial x} \\ \tfrac{\partial}{\partial x} & 0 \\ \end{pmatrix} $$ then we are basically looking at the Cauchy Riemann equations and it will only make sense to apply $e^{tX}$ to $u_0 \in C^\infty(\mathbb{R})^2$ if $u_0$ is the restriction to $\mathbb{R}$ of a holomorphic function.
Well OK fine. So $e^{tX}$ may only be defined on some domain in $C^\infty(M)^2$. We can live with that. There is also, however, the issue of uniqueness to contend with. It does not make sense to talk about the operators $e^{tX}$ unless the answer to the following question is "yes":
Question: Suppose $u_1$ and $u_2$ are smooth, real-valued functions on $\mathbb{R} \times M$ which satisfy \begin{align*} \tfrac{\partial}{\partial t}u_1(t,x) &= X_{11}u_1(t,x) + X_{12} u_2(t,x) \\ \tfrac{\partial}{\partial t}u_2(t,x) &= X_{21}u_1(t,x) + X_{22} u_2(t,x). \end{align*} If $u_1$ and $u_2$ both vanish on $\{0\} \times M$, does it follow that $u_1=u_2=0$ everywhere?
Comment:
This is clearly a special case of a more general PDE uniqueness question, but I am not much versed in that subject. For example, suppose $Y$ is a $2 \times 2$ matrix of smooth vector fields on a smooth manifold $N$ (above $N=\mathbb{R}\times N$), and $w \in C^\infty(N)^2$ belongs to the kernel of $Y$. If $w$ vanishes on a some hyper surface in $N$, does it follow that $w$ vanishes on all of $N$? The answer to this question is clearly "no", because we can take $Y=0$ and then every $w \in C^\infty(N)^2$ will be a solution. So, we need some assumptions about the behaviour of $Y$ near the hypersurface. A good guess is that this should all be related to ellipticity somehow, but are such strong assumptions really necessary to get uniqueness? Or am I missing some more obvious arguments?