Let T be a linear transformation on $\Bbb R^n$ that leaves a subspace $E\subset \Bbb R^n$ invariant and let T(x) = Ax with respect to the standard basis for $\Bbb R^n$. Show that if x(t) is the solution of the initial value problem $$x' = Ax$$
$$x(0) = x_0$$
with $x_0 \in E$ , then $x(t) \in E$ for all $t \in \Bbb R$.
My thoughts: I understand that the subspace being invariant means that for all $x \in E$, $T(x) \in E$. The matrix A in this case is the transformation T on x. I'm struggling to see how to structure this proof though. The solution manual for this particular textbook gives a proof using sequences, but it feels like there should be a simpler route. Does anyone want to take a stab at explaining how to go about this?
Thank you :)
The solution to
$x' = Ax, \; x(0) = x_0 \tag 1$
is
$x(t) = e^{At} x_0; \tag 2$
expanding $e^{At}$ in into its power series then yields
$x(t) = \displaystyle \sum_0^\infty \dfrac{A^n t^n}{n!} x_0 = \sum_0^\infty \dfrac{t^n A^n x_0}{n!}; \tag 3$
since $E$ is $A$-invariant and
$x_0 \in A, \tag 4$
we have
$Ax_0 \in E; \tag 5$
now if
$A^k x_0 \in E \tag 6$
for some non-negative integer $k$, then the $A$-invariance of $E$ again implies
$A^{k + 1} x_0 = A(A^k x_0) \in E; \tag 7$
the conclusion of this simple induction is that
$A^k x_0 \in E, \; \forall k \in \Bbb Z, \; k \ge 0; \tag 8$
the convergence of the series for $e^{At}$ implies the rightmost series in (3) converges; since every term in this series is a vector in the closed subspace $E$ (closed since finite dimensional), the series converges to an element of $E$; thus
$x(t) \in E, \tag 9$
as was to be proved.