I have an initial value problem $\mathbf x'=A\mathbf x$ $$A = \begin{pmatrix} 1 &1 &0 &0 \\ 3& -1 &0 &0 \\ 0 &0 &0 &-2 \\ 0 &0 &2 &0 \end{pmatrix}$$
My initial condition is $\mathbf x (0) = \mathbf x_0$. I need to use the matrix exponential to "determine all initial vectors $\mathbf x_0$ for which the solutions are periodic".
I've got my Eigenvalues as $2, -2, 2i, -2i$.
My understanding is that eigenvectors that span the center subspace (who have corresponding eigenvalues with real part 0) are periodic - correct? Do I only need to solve for the eigenvectors whose eigenvalues have real part being 0? If so, what happens after I get those? I'm unsure how to be approaching this.
Observe that the matrix $A$ has block form $$\begin{pmatrix} B & 0 \\ 0 & C \end{pmatrix}$$ This implies that $(x_1,x_2)$ is driven by $2\times 2$ matrix $B$, while $(x_3,x_4)$ is driven by matrix $C$; there is no interaction between these processes.
Since the eigenvalues of $C$ are $\pm 2i$, all its orbits are periodic. That is, $(x_3,x_4)$ are periodic no matter what the initial condition.
Since the eigenvalues of $A$ have nonzero real part, none of its orbits are periodic... except the trivial, zero solution, which is periodic to every period.
Hence, the conclusion: solution is periodic if and only if the starting point has the first two coordinates zero.
Using the matrix exponential isn't really necessary. But if you want to: $$ e^{tA} = \begin{pmatrix} e^{tB} & 0 \\ 0 & e^{tC} \end{pmatrix}$$ where $e^{tC}$ is sometimes the identity, but $e^{tB}$ never is, except at $t=0$.