Is it possible to write a discretized differentiation operator as a Kronecker sum? The following is a question that was posted and then deleted. Since I found the question interesting, I am posting it anew.

I have this differential operator 
$$L=\begin{bmatrix}
0 & -\partial_x  \\
-\partial_x & 0 
\end{bmatrix}$$
and I have to discretize (say with 2nd order finite differences). Let's call $A$ the matrix that discretize $\partial_x$. Then, the discretization of $L$ will result in a block diagonal  matrix like 
$$\begin{bmatrix}
\mathcal{O} & -A \\
-A & \mathcal{O} 
\end{bmatrix}$$
Of course, I could diagonalize, but my question is: "is it possible to write this as a Kronecker sum?"  Something like $I \otimes A + A \otimes I$?
The zeros along the diagonal in some sense don't look so promising, but I really don't know how to even disprove this. 
 A: In short: the answer is no except in the case that $A$ is a multiple of the identity.
In long: the matrix we are considering is given by
$$
M = \pmatrix{0&-1\\-1&0} \otimes A.
$$
The question is whether there exists a matrix can be expressed in the form
$$
M = B \otimes I_n + I_2 \otimes C
$$
for matrices $B,C$. 
As we deduce from the work below: if such matrices $B,C$ exist, then they must have the form
$$
B = p_1 \pmatrix{0&-1\\-1&0} + p_2 I_2, \quad C = q_1 I_n + q_2 A.
$$
Thus, we must have
$$
M = (p_2 + q_1)I_{2n} + p_1 \pmatrix{0&-1\\-1&0} \otimes I_n + q_2 A \otimes I_2.
$$
Looking at the top-right $n \times n$ block of $M$ allows us to deduce that if $A$ is not a multiple of the identity, then no solution exists.
If $A$ is a multiple of the identity, that is $A = \lambda I$, we see that we have a solution with $B = \lambda \pmatrix{0&-1\\-1&0}$ and $C = 0$.

Consider the linear map $\Phi:\Bbb R^{2n \times 2n}$ to $\Bbb R^{n^2 \times 4}$ defined so that for $v_1,v_2 \in \Bbb R^n$ and $w_1,w_2 \in \Bbb R^2$, we have
$$
\Phi:(v_1v_2^T) \otimes (w_1w_2^T) \mapsto (v_1 \otimes v_2)(w_1 \otimes w_2)^T.
$$
That is, 
$$
\Phi: P \otimes Q \mapsto \operatorname{vec}_{n^2 \times 4}^{-1}[\operatorname{vec}_{n \times n}(P) \otimes \operatorname{vec}_{2 \times 2}(Q)]
$$
where $\operatorname{vec}_{m \times n}: \Bbb R^{m \times n} \to \Bbb R^{mn}$ denotes the row-stacking vectorization operator.  Denote 
$$
a = \operatorname{vec}(A),\quad 
b = \operatorname{vec}(B),\quad 
c = \operatorname{vec}(C),\\
v_1 = \operatorname{vec}\pmatrix{0&-1\\-1&0}, \quad 
v_2 = \operatorname{vec}(I_n), \quad 
v_3 = \operatorname{vec}(I_2).
$$
After applying $\Phi$ to both sides of the equations involving $M$, we end up with the following equivalent formulation of your problem. Given that $\tilde M = \Phi(M)$ satisfies $\tilde M = v_1a^T$, we want to find vectors $b,c$ such that
$$
\tilde M = bv_2^T + v_3c^T = \pmatrix{b & v_3}\pmatrix{v_2 & c}^T.
$$
From $\tilde M = v_1a^T$, we know that $\tilde M$ has rank $1$ with column space spanned by $v_1$ and row space spanned by $a$. From the column and row space conditions, we must have:


*

*$v_1 \in \operatorname{span}(v_3,b) \implies b = p_1 v_1 + p_2 v_3$ with $p_2 \neq 0$ 

*$a \in \operatorname{span}(v_2,c) \implies c = q_1 v_2 + q_2 a$ with $q_2 \neq 0$.

