Simple proof for uniqueness of solutions of linear ODEs? Consider the system of linear ODEs $\dot{x}(t)=Ax(t)$, $x(0)=x_0\in\mathbb{R}^n$. Does anyone know a simple proof showing that the solutions are unique that does not require resorting to more general existence/uniqueness results (e.g., those relating to the Picard iteration) nor solving for the solutions explicitly?
 A: If you just need uniqueness, use the Bellman-Gronwall lemma. Suppose $x,y$ are two solutions, then $x(0) = y(0) = x_0$, and both $x,y$ satisfy the ODE. In particular, if we let $\delta = x-y$, we have $\delta(t) = \int_0^t A \delta(\tau) d\tau$, and hence $\|\delta(t)\| \leq \int_0^t \|A\|  |\delta(\tau)\| d\tau$, from which it follows (using the integral form of the Bellman-Gronwall lemma) that $\|\delta(t)\| = 0$ for all $t \geq 0$.
A: Since the students are engineers, why don't you want to show them explicit solutions, which surely they'd need to see anyway?  If we knew about a matrix exponential $e^{At}$, then to show $x(t) = e^{At}x_0$ let's look at the $t$-derivative of $e^{-At}x(t)$, which is 
$$
e^{-At}x'(t) + (-Ae^{-At})x(t) = e^{-At}Ax(t) - Ae^{-At}x(t).
$$
From the series definition of the matrix exponential, $A$ and $e^{Bt}$ commute if $A$ and $B$ commute, so $A$ and $e^{-At}$ commute. Thus
$$
(e^{-At}x(t))' = e^{-At}Ax(t) - Ae^{-At}x(t) = Ae^{-At}x(t) - Ae^{-At}x(t) = 0.
$$
Therefore $e^{-At}x(t)$ is a constant vector, and setting $t = 0$ tells us this constant vector has to be $x(0) = x_0$. Thus $e^{-At}x(t) = x_0$, so $x(t) = e^{At}x_0$ if we know that $e^{At}$ and $e^{-At}$ are inverses of each other.
Note that this solution can be thought of as a higher-dimensional version of the integration-free proof that the only solution of the 1-dim. ODE $y'(t) = ay(t)$ with $y(0) = y_0$ is $y_0e^{at}$: if $y(t)$ is a solution then the derivative of $e^{-at}y(t)$ is
$$
e^{-at}y'(t) - ae^{-at}y(t) = e^{-at}(ay(t)) - ae^{-at}y(t) = 0.
$$
Thus $e^{-at}y(t)$ is a constant function, and at $t = 0$ we see the value is $y(0) = y_0$, so $e^{-at}y(t) = y_0$. Thus $y(t) = y_0e^{at}$. In higher dimensions we just need to be more careful about the order of multiplication (e.g., the way the product rule is formulated for matrix-valued functions). 
A: The general solution of the initial value problem $X'=AX$ with $X(0)=X_0$ is    $$X(t)=e^{At}X_0$$ Let $Y(t)$ be an another solution satisfying  $Y(0)=X_0$. I want to show that $X(t)=Y(t)$. Define $$Z(t)=exp(-tA)\cdot Y(t)$$ $$\qquad\frac{dZ}  {dt}=-\exp(-tA)\cdot Y(t)+\exp(-tA)\cdot Y(t) =0$$ therefore $Z(t)$ is a constant. Setting $t=0$ we have $Z(t)=X_0$ so  $Y(t)=\exp(At)X_0$ with  $Y(0)=X_0$ this completes the proof.
