With respect to the bases $(g_i)$ of $V_1$ and $(f_i)$ of $V_2$, it is now clear that $M$ has the description $$\xymatrix@=1.9em{
       \overset{g_{n-1}}{\bullet} \ar[dr]_b & & \overset{g_{n-2}}{\bullet}\ar[dl]^a \ar[dr] &  &
      \overset{g_{n-2}}{\bullet} \ar[dl]  & \bullet  \hspace{4 mm}
      \bullet \hspace{4 mm} \bullet
      & & \overset{g_0}{\bullet} \ar[dl] \ar[dr] \\
       & \underset{f_{n-1}}{\bullet} & & \underset{f_{n-2}}{\bullet} &  & &  \underset{f_1}{\bullet} & &
      \underset{f_0}{\bullet}
    }$$ The action of $a$ on $g_{n-1}$ can be seen by applying $b$ on both sides of the equation (\*) above: $a(g_{n-1}) = \sum_{i=0}^{n-2}
b(g_i) = \sum_{i=0}^{n-1} f_i$. These representations are the exactly ones in item 3(a).

Now if $\det(b)  = 0$, we do a change of coordinate trick. We assume that $k$ is an infinite field. If $k$ is finite, we can pass to an extension field and do a descent argument; see [@ben-1] for details. Then there exists some $\lambda_0$ in $k$ such that $\det(a + \lambda_0 b) \ne 0$. Now consider the tuple $(V_1, V_2; b, a+\lambda_0 b)$. By case (i), we know that there exist bases for $V_1$ and $V_2$ such that $a +
\lambda_0 b = I$ and $b = J_0$ (the rational canonical form [@LinearAlgebra] [^1] corresponding to any indecomposable singular transformation). This gives the representation 