$K=D\inv_x(vP)$ and $\omega = v D\inv_x(vP) + D_x P$, and then the first equation becomes $v_t = D_x^2P +D_x(vD\inv_x(vP)) -4\lambda^2 P$. If we now put $\lambda =\tfrac{1}{2}$ then this yields $$\label{mkdv-lax-pair-flow}
v_t = (\R-1)P$$ holding for an arbitrary differential function $P(v,v_x,v_{xx},\ldots)$, where $\R$ is the mKdV recursion operator . When $P=-v_x$ is given by the $x$-translation symmetry $\X=P\partial/\partial_v$ of this recursion operator, the zero-curvature equations – yield $K= -\tfrac{1}{2} v^2 + c$ and $\omega = -v_{xx} -\tfrac{1}{2}v^3 +c v$, where $c$ is an arbitrary constant given by the freedom in $D\inv_x$. The flow equation then gives $$v_t = (c+1)v_x - \tfrac{3}{2}v^2 v_x -v_{xxx}$$ which is the mKdV equation up to a convective term $(c+1)v_x$. This term vanishes if $c=-1$, whereby $K= -\tfrac{1}{2} v^2 - 1$ and $\omega = -v_{xx} -\tfrac{1}{2}v^3 -v$, so thus the matrices – become $$U = \tfrac{1}{2} 
\begin{pmatrix}
1 & v \\ -v & -1
\end{pmatrix}, 
\quad
V = \tfrac{1}{2} 
\begin{pmatrix}
-\tfrac{1}{2} v^2 - 1 & -v_{xx} -\tfrac{1}{2}v^3 -v -v_x
\\
v_{xx} +\tfrac{1}{2}v^3 +v -v_x & \tfrac{1}{2} v^2 + 1
\end{pmatrix} .$$ These matrices satisfy the zero-curvature equation whenever $v$ is a solution of the mKdV equation . We can obtain a matrix Lax pair from them 