I'm studying systems of linear differential equations and have the following problem:
Show that real eigenspaces corresponding to the matrix $A\in\mathcal{M}_n(\mathbb{R})$ are invariant subspaces of the phase plane, i.e., for $\lambda$ eigenvalue of $A$ and $E(\lambda,A)$ it's eigenspace, if for some $t\in\mathbb{R}$, $y(t)\in E(\lambda,A)$, then $y(s)\in E(\lambda,A)$ for all $s>t$.
My attempt is the following one:
Let $A$ be a $n\times n$ matriz with real entries, $\lambda$ a real eigenvalue of $A$ and $E(A,\lambda)$ the corresponding eigenspace. Since $\dim(E(\lambda,A))<\dim(A)=n<\infty$, there exists $\{v_1,\ldots,v_k\}$ such that it is a basis for $E(A,\lambda)$, for some $k\leq n$.
Now fix $t\in\mathbb{R}$, if $y(t)\in E(\lambda,A)$, then there exist $\alpha_1,\ldots,\alpha_k$ such that $y(t)=\sum_{j=1}^k\alpha_jv_j$.
Let's consider $s>t$. Now we have the IVP: $$ \begin{cases} y'=Ay, \\ y(t)=\sum_{j=1}^k\alpha_jv_j. \end{cases} $$ The homogeneous solution would depend on the multiplicity of the eigenvalue. However, if we consider the exponential of a matrix, $e^{Ax}=\sum_{k=0}^\infty A^kx^k/k!$, it is a fundamental matrix for the system and satisfies $\left(e^{Ax}\right)=Ae^{Ax}$. I also know, due to Lagrange Method (also known as variation of parameters) that the homogeneous solution is some multiple of $e^{(s-t)A}y_0$ (this exponential matrix satisfies common properties of the real valued function $e^x$) where $y_0\in\mathbb{R}^n$ is the initial condition (in our case the sum of $\alpha_jv_j$). Doing some algebraic work, I got that $$y(s)=e^{-tA}\bar{v}$$ where $\bar{v}$ is a vector in the eigenspace, but since the matrix $e^{-tA}$ is the inverse of the fundamental matrix, it has no linearly independent solutions to the differential equation, so I cannot asure $y(s)$ is parallel to $\bar{v}$ and conclude the exercise.
I also thought about the solution using generalized eigenvectors to fully complete de space and have a more general form but that led me to nothing enlighting. Sorry if this attempt seems pretty confusing, I'm kinda lost in here.
Just in case, I know something realted has been asked here: Question about phase portrait and invariant subspaces but I use different definitions and don't understand the answer too much.
There is an easier way. Note that everything is smooth, etc, so there are no issues about existence or uniqueness. Suppose $y$ is the solution to $y'=Ay$ such that passes through $y(t)$ at time $t$.
Let $x(s) = y(t) e^{\lambda (s-t)}$ and check that $x'(s) = Ax(s)$ and $x(t) = y(t)$. Since the solution is unique we have $y=x$.