Consider a square matrix $A\in\mathbb{R}^{N\times N}$. We know that it's eigenspace is $A-$invariant. Suppose that $\mathcal{S}\subset\mathcal{R}^N$ is $A-$invariant and $\mathcal{S}\neq\mathcal{R}^N$. Is it necessary that $\mathcal{S}$ is a subset of the subspace spanned by all the eigenvectors of $A$? It seems correct to me, don't know how to prove it!
Clearly, if $\mathcal{S}$ is one-dimensional, it belongs to an eigenspace.
On
We may assume that the considered matrix is in the form
$A=diag(\lambda_1I_{i_1}+N_1,\cdots,\lambda_kI_{i_k}+N_k)$ where the $(\lambda_j)_j$ are distinct complex numbers and the $(N_j)_j$ are nilpotent.
$\textbf{Proposition 1}$. $E$ is $A$-invariant IFF $E=E_1\oplus\cdots \oplus E_k$ where $E_j$ is $N_j$-invariant.
The easy case is when $A$ is cyclic, that is, in the Jordan decomposition of $A$, for every $\lambda_j$, there is exactly one Jordan block associated to $\lambda_j$. According to Proposition 1, to study its invariant spaces, it suffices to obtain the invariant spaces of a Jordan block.
$\textbf{Proposition 2}$. If $J$ is a Jordan block of dimension $n$, then there is a finite number of $J$-invariant spaces. More precisely, these invariant are $[0],[e_1],[e_1,e_2],\cdots,[e_1,\cdots,e_n]$.
$\textbf{Remark.}$ i) In particular, a cyclic matrix admits a finite number of invariant spaces; moreover the converse is true.
ii) If $A$ is not cyclic, it is much more difficult to find its invariant spaces.
No. Let
$$M:=\begin{pmatrix} 0& 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}.$$
Then $M$ has two eigenvectors $(1,0,0)^T$ and $(0,1,0)^T$. Yet, take $V$ to be the subspace spanned by $(0,1,0)^T$ and $(0,1,1)^T$, which is invariant under $M$ but not spanned by any subset of its eigenvectors.