Solution
= Solution
Take $t\mapsto A$ and $r\mapsto B$. Then $A^2=I$ and
$$
ABA^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}=B^{-1},
$$
so these <matrices> define a representation. The line $\mathbb Ce_1$ is fixed by both <matrices> and is a one-dimensional subrepresentation.
Any complementary line has a generator $e_2+ce_1$. But
$$
B(e_2+ce_1)=e_2+(c+1)e_1,
$$
which cannot lie on the same line. Thus no complement is $B$-invariant, and therefore no $G$-invariant complement exists.