Is it possible to convert a linear system in an ODE of higher order The opposite direction is quit simple. But under which circumstances is it possible to convert a system of first order ODE of dimension n into a single ODE of order n?
 A: This is not a complete answer (yet), and I encourage other folks to build on it:

A set of autonomous coupled linear first-order ODEs can always be written in the form
$$
y' = A y
$$
where $y$ stands for a vector of $n$ functions and $A$ is an $n \times n$ matrix.  By differentiating both sides further, we obtain
$$
y'' = A y' = A^2 y \\
y''' = A y'' = A^3 y \\
\vdots \\
y^{(n)} = A^n y.
$$
Now, we can take an arbitrary linear combination of these $n$ derivatives to obtain an expression of the form
$$
(c_n y^{(n)} + c_{n-1} y^{(n-1)} + \dots + c_1 y' + c_0 y) = (c_n A^n + c_{n-1} A^{n-1} + \dots + c_1 A + c_0 I)y.
$$ 
But if we pick the $c_i$ coefficients as the coefficients of the characteristic polynomial of $A$ (i.e., $c_n$ is the coefficient of $\lambda^n$ in the polynomial $\det(\lambda I - A)$), then by the Cayley-Hamilton theorem the right-hand side of this equation will vanish.  This implies that for any autonomous system, each component of $y$ will individually satisfy the $n$th-order equation 
$$
c_n y_i^{(n)} + c_{n-1} y_i^{(n-1)} + \dots + c_1 y_i' + c_0 y_i = 0.
$$
where the $c_n$ coefficients are those of the characteristic polynomial of $A$.
This is, of course, only a sufficient condition;  certainly there are non-autonomous systems that can be converted into a single $n$th-order equation as well.  It is also possible for the $n$th-order equation to possess solutions that are not solutions of the original system;  for example, blindly applying this procedure to the system
$$
y_1' = y_1 \qquad y_2' = y_2
$$
i.e., $A = I$, we obtain the equation
$$
y_1'' - 2 y_1' + y_1 = 0,
$$
and similarly for $y_2$.  However, there are solutions to the latter equations that are not solutions to the original system.
