How to show the solution of $\dot{x} = Ax $ is an invariant subspace? Consider the linear dynamical system $\dot{x} = Ax $ in $V$ a finite dimensional vector space.
The definition of an invariant subspace $U$ is as follows:
For all $x_0 \in U$, (initial condition), the solution $x(t)$ with initial condition $x(0)=x_0$ lies in $U$, i.e. $x(t) \in U$ for all $t \in \mathbb{R}$.
Intuitively, it means if we start in a subspace we never leave that subspace.
Prove that a subspace $U \subset V$ is an invariant subspace if and only if $AU \subset U$.
 A: Given the linear, constant coefficient differential equation
$\dot x = Ax, \tag 1$
it is well-known that the solution with
$x(0) = x_0 \tag 2$
is
$x(t) = e^{At}x_0; \tag 3$
now if
$AU \subset U, \tag 4$
and 
$x_0 \in U, \tag 5$
then
$Ax_0 \in U, \tag 6$
and indeed, for any non-negative $k \in \Bbb Z$,
$A^kx_0 \in U \tag 7$
as well; thus
$e^{At}x_0 = \left (\displaystyle \sum_0^\infty \dfrac{A^k t^k}{k!} \right ) x_0 = \displaystyle \sum_0^\infty \dfrac{t^k A^k x_0}{k!} \in U, \; \forall t \in \Bbb R, \tag 8$
since the series on the right converges to an element in $U$, a closed sub-space, being finite dimensional.
Now suppose every solution initialized at $x_0 \in U$ remains in $U$ for all $t$; then
$x(t) = e^{At}x_0 \in U, \; \forall t \in \Bbb R; \tag 9$
since $U$ is a linear subspace of $V$, the tangent vector to any curve in $U$ is also in $U$, whence
$\dot x(t) = Ae^{At}x_0 \in U, \forall t \in \Bbb R; \tag{10}$
setting $t = 0$ yields
$Ax_0 = Ae^{A(0)}x_0 = \dot x(0) \in U; \tag{11}$
but since $x_0 \in U$ is arbitrary, we have
$AU \subset U, \tag{12}$
and we are done.  $OE\Delta$.
