Invariant Subspaces Dynamical Interpretation

Question

Consider the linear system $\dot{x}=A_{nx n}x$ ; $x(0)=x_0$. Recall that a subspace $\mathcal{U}$ s (dynamically) invariant to the flow if for all initial conditions $x_0 \in \mathcal{U}$ the solutions $x(t) \in \mathcal{U}$ $\forall t \geq 0$. Show that a subspace $\mathcal{U}$ is dynamically invariant if and only if $A\mathcal{U}\subseteq U$

Solution so far

Assume $A\mathcal{U}\subseteq U$. That is , $v \in \mathcal{U} \implies Av \in \mathcal{U}$. Let $x(0) \in\mathcal{U}$.

Answer 1

A linear, constant coefficient system $\dot{y} = A y$ can be solved using the matrix exponential, see https://en.wikipedia.org/wiki/Matrix_exponential#Linear_differential_equation_systems . This can be rewritten in terms of the eigenvalues and eigenvectors of $A$, but that's not the point here.

I would suggest to combine the above with the observation that, if $A \mathcal{U} \subset \mathcal{U} $, then $A^2 \mathcal{U} = A (A \mathcal{U}) \subset A \mathcal{U} \subset \mathcal{U}$.

Answer 2

If $U$ is invariant, then $e^{At}v\in U$ whenever $v\in U$ and $t\ge0$ (recall that the solutions of the equation $x'=Ax$ are precisely the functions $e^{At}v$ for some $v$). Then $$ \frac{e^{At}v-v}t\in U $$ for $t>0$ (since $U$ is a subspace). Taking the limit when $t\to0^+$ we get $Av\in U$ (since $(e^{At})'=Ae^{At}$ and since $U$ is closed; it is finite-dimensional). Therefore, $AU$ is contained in $U$.

On the other hand, if $AU$ is contained in $U$ and $v\in U$, then $A^nv\in U$ for all $n\in\mathbb N$ (by induction) and so $$ e^{At}v=\sum_{n=0}^\infty\frac1{n!}t^nA^nv=\lim_{m\to\infty}\sum_{n=0}^m\frac1{n!}t^nA^nv\in U $$ for $t>0$, again since $U$ is closed (the finite sum is in $U$ because this is a subspace, and since the limit exists it is in $U$, because $U$ is closed).