Say we have a matrix $M$, an invertible matrix $P$ and a Jordan matrix $J$ such that $$ P^{-1} M P = J = \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{pmatrix}. $$
Say I want to find the eigenspace for $M$. Intuitively, I would say that $$ \left\langle \begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 1\\ 0 \end{pmatrix} \right\rangle, $$ but I have no idea if this is correct.
Intuitively I would also say that the list of invariant subspaces of $M$ would be: $$ \left\langle \begin{pmatrix} 2\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 2\\ 1\\ 0\\ 0 \end{pmatrix} \right\rangle, \quad \left\langle \begin{pmatrix} 0\\ 0\\ 2\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 2\\ 1 \end{pmatrix} \right\rangle, $$ and then any combination of the "eigenspaces" $$ \begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 0\\ 1\\ 0 \end{pmatrix}. $$ And then of course $\mathbb{R}^{4}$ and $0$.
Am I correct? If not, is there any way of finding the eigenspaces and invariant subspaces of $M$ from the Jordan normal form $J$.
You're right: the eigenvectors are indeed ${}^{\mathrm t\mkern-1mu}(0,0, 1,0)$ and ${}^{\mathrm t\mkern-1mu}(1,0, 0,0)$, BUT when their coordinates are calculated in the new (Jordan) basis. To have their coordinates in the standard basis, just apply the change of basis matrix $P$. This means the eigenvectors are simply the 1st and 3rd columns of $P$.
Similarly for the generators of the invariant subspaces.