Let $A$ be an $n \times n$ nonsingular matrix that is semisimple, and all the eigenvalues of $A$ are purely imaginary complex numbers. and let $\mathbf{x}(t)$ be the solution of the initial value problem $\dot x=Ax$ with $\mathbf{x}(0)=\mathbf{x}_{0} \not = 0$.
Show that then there are positive constants $m$ and $M$ such that for all $t \in \mathbf{R}, m \leq|\mathbf{x}(t)| \leq M$
I know that the coordinates of $\mathbf{x}(t)$ are linear combinations of $cos(b_{j}t)$ and $sin(b_{j}t)$ which $ib_{j}$ are eigenvalues of $A$, and from this the existence of $M$ easily followed.
Although $\mathbf{x}(t)$ doesn't intersect the origin, how do i get the lower bound $m$? If the imaginary part of eigenvalues are different the solution isn't necessarily periodic, so i think the image of $\mathbf{x}(t)$ isn't compact?!
There exists a basis transform so that the transformed $A$ is real and anti-symmetric. In the Euclidean norm for the coordinates of this basis, the norm of the solution is constant. Now apply the equivalence of norms in finite dimension to conclude the existence of upper and lower bounds as in the claim.