Let $A\in M_3(\mathbb R)$ be a skew-symmetric matrix and $x:[0,\infty)\to\mathbb R^3$ be a solution of $$x’(t)=Ax(t),\forall t\in (0,\infty)$$ Which of the following statements are true?
$1.$ $\|x(t)\|=\|x(0)\|,\forall t\in(0,\infty).$
$2.$ for some $a\in\mathbb R^3-\{0\}, \|x(t)-a\|=\|x(0)-s\|,\forall t\in(0,\infty).$
$3.$ $x(t)-x(0)\in img(A),\forall t\in(0,\infty)$.
$4.$ $\lim_{t\to\infty}x(t)$ exists.
I am trying to solve this problem . I can see last option is wrong as because of term $cos(at)$ and $ sin(at)$ in solution, so limit does not exist for non-zero matrix as $t\to\infty$. I am unable to handle first three options. If eigen values of skew symmetric matrix $A$ are $0,\pm ai$ and corresponding Eigne vectors are $V_1,V_2$ and $V_3$, then solution is given by $x(t)=V_1+e^{ait}V_2+e^{-ait}V_3$. $x(0)=V_1+V_2+V_3$, how $\|x(t)\|=\|x(0)\|?$ How to solve first three options. Please help. Thank you.