Linear rotation spanning 4d Let $M$ be a real $4 \times 4$ matrix  Let $x$ be the time-dependent 4-dimensional vector evolving according to the linear ODE 

$$ \frac{ d x }{dt} = M x $$ 

Can you give an example of a matrix $M$ such that the trajectory will span all 4 dimensions for any initial condition?

In other words, find an $M$ such that starting from any vector $x_0$, the subsequent trajectory will not be confined in a plane, or even in a certain 3d space, but will evolve periodically in the 4d space. 

I tried to answer this by starting from the fact that any two eigenvectors can be non-orthogonal. But how the pairs of eigenvectors corresponding to imaginary eigenvalues? If the 4d matrix has two pairs of imaginary eigenvalues, won't the two cyclic behavior of the ODE be uncoupled? And if so, we can always choose an initial condition orthogonal to one plane, so the system will rotate exclusively in the other plane?

But then finding an M as described above is impossible. Please enlighten me!
Let $Mu=\lambda u$. So, $x(t) = e^{\lambda t} u$ is a solution to $\dot{x} = Mx$. This means if you start on $u$, you will stay on $u$ forever. If $\lambda$ is real, then $x(0) = u$ selection can only span a 1d space. If it is complex then $x(0) = \alpha \operatorname{Re}u + \beta \operatorname{Im}u$ selection can only span a 2d space for any real $\alpha, \beta$.

In summary, there exists an initial condition where the solutions can span at most 2d space.
Let $Mu=\lambda u$. So, $x(t) = e^{\lambda t} u$ is a solution to $\dot{x} = Mx$. This means if you start on $u$, you will stay on $u$ forever. If $\lambda$ is real, then $x(0) = u$ selection can only span a 1d space. If it is complex then $x(0) = \alpha \operatorname{Re}u + \beta \operatorname{Im}u$ selection can only span a 2d space for any real $\alpha, \beta$.

In summary, there exists an initial condition where the solutions can span at most 2d space.