While considering the zero curvature equation $U\_t - V\_x + [U, V] = 0$, I developed a similar problem, albeit one that discards time dependence entirely. For a given $U(x)$ and $C(x)$, find $V(x)$ such that $[U, V] - V\_x = C(x)$ holds. After considering this [question](https://math.stackexchange.com/questions/1307098/is-there-any-inverse-commutator-for-matrices), I believe that the system would either have a single solution or an infinite class of solutions depending on the choice of $C$ and $U$. In the [question](https://math.stackexchange.com/questions/1307098/is-there-any-inverse-commutator-for-matrices), I found the method involving the Kronecker product to be more suited for matrix functions as I ultimately hope to prove that the compatibility condition is isospectral. (I do *not* want to assume that the eigenvalues are constant across space, as I believe that it would produce a circular argument.)


I had also wondered if the system could be transformed by finding $\gamma(x)$ and $\eta(x)$ such that $[U + \gamma, V + \eta] = [U,V] - V\_x$. Ultimately, I realized such an attempt would reduce to solving $[\eta, \gamma] + [V, \gamma] + [\eta, U] = V\_x$, which is no simpler than the original equation. Most of the difficulty arises from the lack of commutativity of the matrix product.


**Question:** For a given $U(x)$ and $C(x)$, can one solve for $V(x)$ such that $[U, V] - V\_x = C(x)$ holds?



On the equation $[U, V] - V_x = C(x)$
Yes, you can always do this, as follows:


First, consider the equation $M\_x = -M U$ with the initial condition $M(0) = I\_n$. This linear equation with initial condition has a unique solution and $M(x)$ will be invertible for all $x$. Now, set $V = M^{-1}AM$ for some matrix $A$ and consider the equation
$$
V\_x - UV + VU + C = 0.
$$
This expands to
$$
M^{-1}A\_xM + C = 0,
$$
or
$$
A\_x = -MCM^{-1}
$$
which clearly has a unique solution for any given value of $A(0)$. Thus, there is always an $n^2$-dimensional family of solutions for any given $U(x)$ and $C(x)$.


