Worked-out solution

$\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{0 & 2\\-2 & 0\\}$

1. $\quad$The characteristic polynomial of $A$ is equal to $$\det(A-\lambda\cdot I)=\left|\begin{array}{cc} -\lambda & 2 \\ -2 & -\lambda \end{array}\right|= \lambda^2 +4$$
   $\quad$The eigenvalues of $A$ are the zeros of the characteristic polynomial: $-2\ii$ and $2\ii$.
   
   $\quad$Let $\cv{v_1\\ v_2}$ be an eigenvector corresponding to the eigenvalue $2\ii$: $\matrix{0 & 2\\-2 & 0\\}\cdot \cv{v_1\\ v_2}=2\ii\cv{v_1\\ v_2}$.
   
   $\quad$Then: $\left\{\begin{aligned} 2v_2 &= 2\ii\cdot v_1 \\[0.2cm] -2v_1 &= 2\ii\cdot v_2\end{aligned}\right.$
   
   $\quad$Thus: $v_2=\ii\cdot v_1$ and $v_1$ can be freely chosen.
   $\quad$An eigenvector corresponding to the eigenvalue $2\ii$ (with integral coefficients) is $\vec{v}=\cv{1\\\ii}$.
   $\quad$Similarly, $\vec{v}=\cv{1\\ -\ii}$ is an eigenvector corresponding to the eigenvalue $-2\ii$.
   
   
2. $\quad$The general solution is $$\vec{x}(t) =\alpha \cdot\cv{\cos(2t)\\ -\sin(2t)}+ \beta\cdot \cv{\sin(2t)\\ \cos(2t)}$$ $\quad$with constants $\alpha$ and $\beta$.
   
   
3. $\quad\vec{x}(0)=\cv{2\\0}$ means that$x_1(0)=2$ and $x_2(0)=0$.
   $\quad$Substitution in the general solution (b) gives: $\alpha=2$ and $\beta=2$.
   $\quad$Thus: $\vec{x}(t)=2\cv{\cos(2t)\\ -\sin(2t)}$.
   
   $\quad$The trajectory is a circle with the origin as centre and radius 2: $$x_1^2+x_2^2=4$$ $\quad$This is also illustrated in the phase portrait below.
   $\quad$