Given the matrix $A$:
$$A= \begin{pmatrix} 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 0 \\ \end{pmatrix} $$
$a)$ Find the general solution for the system of first order linear differential equations $y' = Ay$. (HINT: you should get 4 independent eigenvectors).
After solving $a)$ you should be able to answer $b)$.
$b)$ Find out for which positive integer $n$ the following statement holds:
Let C be an $n \times n$ invertible matrix with pure imaginary eigenvalues. If $y : \Re \rightarrow \Re^n$ is a solution of $y' = Cy$ then $y$ is limited (EDIT: what is meant is that $y$ is bounded).
What I have tried:
$a)$ The characteristic polynomial of $A$ (let me skip showing the determinant; the result is correct) is:
$$\Delta (\lambda) = \lambda^4 + 2\lambda^2 + 1$$
Which roots are:
$$\lambda_1 = i$$ $$\lambda_2 = -i$$
Where each has multiplicity of 2.
The general solution is (let me skip the algebraic steps; I will provide more details if you want to):
$y(x) = c_1\left[ \begin{pmatrix} 1 \\ -2i \\ -1 \\ 0 \\ \end{pmatrix}xe^{ix} + \begin{pmatrix} -1 \\ i \\ -1 \\ i \\ \end{pmatrix} e^{ix}\right] + c_2\left[\begin{pmatrix} 1 \\ 2i \\ -1 \\ 0 \\ \end{pmatrix}xe^{-ix} + \begin{pmatrix} 1 \\ i \\ 1 \\ i \\ \end{pmatrix} e^{-ix}\right]$
EDIT: The above equation may be wrong. The correct answer may be:
$y(x) = c_1 \begin{pmatrix} 1 \\ -2i \\ -1 \\ 0 \\ \end{pmatrix}xe^{ix} + c_2\begin{pmatrix} -1 \\ i \\ -1 \\ i \\ \end{pmatrix} e^{ix} + c_3\begin{pmatrix} 1 \\ 2i \\ -1 \\ 0 \\ \end{pmatrix}xe^{-ix} + c_4\begin{pmatrix} 1 \\ i \\ 1 \\ i \\ \end{pmatrix} e^{-ix}$
$b)$ I don't see what is meant here by saying 'then $y$ is limited'. My guess is that what is limited is the number of independent eigenvectors; Thus for our case $n=4$ and the statement holds. Am I right?
An explanation on $b)$ would be appreciated.