Diagonalization of a matrix of differential operators Dear community, 
i have a question regarding differential operators acting on vector valued functions and how to "diagonalize" them.
To explain my question i will use an example:
Let $V^k$ be the space of twice differentiable functions $U:[0,2\pi] \to \mathbb{R}^k$
with periodic boundary conditions.
Consider the differential operator $L:V^2\to V^2$ defined via
$L :=\begin{pmatrix} -\partial^2 & 0 \\ 0 & -\partial^2 \end{pmatrix}$
That means it acts on $U\in V$, $U(t)=(u(t),v(t))^T$, by mapping it to $LU=(-u''(t),-v''(t))^T$.
$L$  is self adjoined with respect to the scalar product
$(U,V) := \int_0^{2\pi} U^T V \ dt$. So it has real eigenvalues.
It commutes with the operator $J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.
In fact, the spectrum of $L$ consists of all $\lambda=k^2$ with $k \in \mathbb{Z}$ with multiplicity 4. The spectrum of $-\partial^2:V^1 \to V^1$ is the same, except that the multiplicities of the eigenvalues are halved. This is evident from the diagonal form of $L$.
Now, what happens for other self-adjoined operators on $V^2$ that commute with $J$?
For example consider the operator $M:V^2 \to V^2$ defined by
$M :=\begin{pmatrix} -\partial^2 & -\partial \\ \partial & -\partial^2 \end{pmatrix}$
It is also self-adjoined with respect to the above mentioned inner product and it also commutes with $J$. Is it possible to "diagonalize" this operator into a form 
$\begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix}$
with a scalar differential operator $m: V^1 \to V^1$ having the same spectrum as $M$ except for the multiplicities?
Any references would be appreciated. Thanks.
 A: Your matrix is a matrix with coefficients in the ring $\mathbb C[\partial]$ of differential operators with constant coefficients, which happens to be a principal ideal domain. Therefore we can take your matrix — let's call it $A$ — to its Smith normal form. In this case, the normal form is $$S=\left(
\begin{array}{cc}
 \partial  & 0 \\\\
 0 & \partial ^3+\partial 
\end{array}
\right)$$
Indeed, if we let $$P=\left(
\begin{array}{cc}
 -1 & 0 \\\\
 -\partial  & 1
\end{array}
\right)\qquad\text{and}\qquad Q=\left(
\begin{array}{cc}
 0 & 1 \\\\
 1 & -\partial 
\end{array}
\right),$$ which are invertible, we have $$PAQ=S.$$ This means that you can change coordinates to get a diagonal matrix.
(Of course, in your situation you want to consider $A$ as defining an endomorphism, so you probably want to restrict changes of coordinates which coincide in the domain and the codomain of the map...)
A: Another approach, for this particular example is to try to solve the equation $AMA^{-1} - m I = 0$, where $A$ is an invertible $2$-by-$2$ matrix of functions and $m$ is a scalar differential operator.  There are a number of solutions to this.  For example,
$$
A = \begin{pmatrix}\cos(\tfrac12x) & \sin(\tfrac12x)\cr 
                   -\sin(\tfrac12x) & \cos(\tfrac12x) \end{pmatrix}
$$
and $m = -\partial^2 - \tfrac14$.  Unfortunately, $A$ is $4\pi$-periodic, not $2\pi$-periodic.  In fact, there are no $2\pi$-periodic solutions.  
You can interpret this in two ways.  One is that you shouldn't have imposed the $2\pi$-periodicity in the first place, since the conjugacy question for differential operators is really a local one.  The other is that $A$ actually represents an isomorphism between two rank $2$-vector bundles over the circle, one with trivial transition, and the other with a twist (so that the sections of the second bundle are represented by functions $U:\mathbb{R}\to \mathbb{R}^2$ such that $U(t+2\pi) = -U(t)$).   
