is non-zero. We have two subcases. First assume that $\det b \ne 0$. Then consider the map $$b^{-1}a : V_1 \longrightarrow V_1.$$ We claim that this map is indecomposable. Suppose we have a decomposition $f \oplus g$ of $b^{-1}a$ as follows $$\xymatrix{ V_1 \ar[r]^{b^{-1}a} \ar[d]_{\cong} & V_1 \ar[d]^{\cong}\\
A \oplus B \ar[r]_{f \oplus g} & A \oplus B
 }$$ Set $C:= b(A)$ and $D:= b(B)$. Then it is straightforward to verify that $M = (V_1, V_2; a, b)$ decomposes as $$(A, C; a|_{A}, b|_{A}) \; \bigoplus \; (B, D; a|_{B}, b|_{B}).$$ Since $M$ is indecomposable it follows that the map $b^{-1}a$ is indecomposable. Now since $b^{-1}a$ is indecomposable we can choose a basis $\{g_0, g_1, \cdots, g_{n-1}\} $ of $V_1$ such that the rational canonical form of $b^{-1}a$ has only one block which corresponds to some power of an irreducible polynomial $f(x)^r =
\sum_{i = 0}^{n-1} \theta_i x^i$. This means we have $$\begin{aligned}
b^{-1}a \; (g_i) &= & g_{i+1}  \hspace{3 mm} \text{for} \;\; 0 \le i \le n-2, \\
b^{-1}a \; (g_{n-1}) &=& \sum_{i=0}^{n-1} \theta_i g_i. \hspace{3
mm} (*)\end{aligned}$$

Now the vectors $f_i := b(g_i)$ for $0 \le i \le n-1$ define a basis for $V_2$ because $b$ is non-singular. 