How can we construct a differential equation from a system of differential equation? Suppose we have a linear differential equation of order $n$. All of us know how to write it down as a system of linear differential equation as $X' = A_{n \times n} X_{n \times 1}$.
My question is about the converse. I am not getting a satisfactory answer anywhere.
Suppose an arbitrary $n \times n$ matrix $A$ is given. How can we construct a linear differential equation of order $n$ whose matrix representation is $X' = AX ? Is it always possible?
My intuition is NO. But I can not explain it properly. If possible please give an example explaining the process. If not possible give a proof or counterexample. 
Thank you for your help.
 A: Look up the "cyclic vector" method (CVM) and the Danilevski-Barkatou-Zürcher algorithm
(DBZ), e.g. see Bostan,Chyzak, and Panafieu, Complexity Estimates for Two Uncoupling Algorithms, 2013. and Kovacic, Cyclic Vectors And Picard-Vessiot Extensions (1996).
A: You can do it, if you are allowed to do coordinate transformations. In this case you can do it if


*

*Either $A$ has distinct eigenvalues, or

*If $A$ has a repeated eigenvalue, then that eigenvalue has only one eigenvector.


Under these conditions, you can transform A to the form
$$
\pmatrix{0&1&0& \cdots &0 \\0&0&1&\cdots &0\\ 0&0&0&\ddots&0\\0&0&0&\cdots &1\\
-d_1&-d_2&-d_3&\cdots & -d_n}$$
One way to find the transformation, is to start with a vector $b$ and calculating $A b$, $A^2b$ through $A^{n-1}b$. Then the transformation matrix is given by
$$
T=\pmatrix{b &A b &A^2 b& \cdots & A^{n-1}b } \tag 1$$
One can show that under the stated assumptions, there exists a $b$ so that the above matrix is invertible.
Added in response to comments
The following statements are equivalent:


*

*A system of $n$ first order linear differential equation of the form $dX/dt = A X$ can be converted to $$
\frac{d^n y}{d t^n} + a_1 \frac{d^{n-1} y}{d t^{n-1}} + a_2 \frac{d^{n-2} y}{d t^{n-2}}\cdots + a_0 = 0$$

*There exists a non-zero vector, $c$, so that the matrix $Q$ defined below has full rank
$$
Q = \pmatrix{c \\c A \\\vdots\\ c A^{n-1} }$$ 

*There exists a vector $b$ so that the matrix $T$ defined in (1) is full rank.

*The minimal polynomial of $A$ is equal to its characteristic polynomial

*The two conditions mentioned at the start of his answer hold


Refer to any standard Modern Control Theory book for proof of the above. THe requirement for $T$ to be full rank is called controllability condition and the requirement that $Q$ be full rank is called observability condition.
In particular, if $A=\pm I$, the identity matrix and $n>1$ then $A$ does not satisfy the last condition and hence it is not possible to transform $X'= \pm X$.
