I have to prove that each orbit of
\begin{align*}
\dot{x}=
\begin{pmatrix}
\phantom{-}0 & 4 \\
-9 & 0
\end{pmatrix}x
\end{align*}
is an ellipse.
I know the eigenvalues are complex numbers, being $\lambda_1=6i ,\lambda_2=-6i.$ But I don't know how to proceed afterwards. Any insight is appreciated.
The equivalent form $$x_1'= \phantom{-}4x_2\\x_2'=-9x_1 $$ of the system will be helpful for the idea of the shape of the orbits, which are determined by $\big(x_1(t),x_2(t)\big).$ As the eigenvalues are $\pm 6i,$ we have $$x_1(t)=a\cos 6t+b\sin 6t, \;\;\;\; a,b \in \mathbb{R}.$$ From $\;x_1'=4x_2\;$ we deduce $$x_2(t)=-\frac{3}{2}a\sin 6t +\frac{3}{2}b\cos 6t.$$ With a bit of algebra we get $$\frac{x_1^2}{4}+\frac{x_2^2}{9}=a^2+b^2,$$ which is equation of an ellipse (except if $a=b=0,$ when the solution is trivial).
N.B. The choice of the constants is deduced from the eigenvectors $(\pm 2,3i).$