Indecomposable but not irreducible representation and direct sums I have two questions concerning representation theory of groups:
1) Consider the group $G=(\mathbb{R},+)$ and the representation $$\rho: G\to\text{GL}(\mathbb{R}^2), r\mapsto\begin{pmatrix}1&r\\0&1\end{pmatrix}.$$
Why is this representation indecomposable but not irreducible?
It is not irreducible as the space $\{(r,0), r\in \mathbb{R}\}$ is invariant under this action.
But why is this indecomposable? 
2) If I have an exercise like 'Show that $V=V_1\oplus V_2$ as $G$-representations: What do I have to show? The first thing is that this is a direct sum of vector spaces, i.e. $V=V_1 + V_2$ and trivial intersection. But what else?
 A: 1) Suppose on the contrary that $\rho$ is decomposable, i.e. exist $G$-invariant and linearly independent vectors $v = (v_{1}, v_{2} )^{t}, w = (w_{1}, w_{2} )^{t} \in \mathbb{R}^{2}$. Then, the following must be true:
\begin{align}
\rho_{r} v = \lambda v \tag{1} \\
\rho_{r} w = \gamma w \tag{2}
\end{align}
$\forall r \in \mathbb{R}$ y and some $\lambda, \gamma \in \mathbb{R}$. Equation (1) implies
\begin{align}
v_{1} + v_{2} r = \lambda v_{1} \\
v_{2} = \lambda v_{2}
\end{align}
so $\lambda = 1$ and $v_{2} r = 0$ for all $r \in \mathbb{R}$. Then $v_{2} = 0$, and $\mathbb{R}v = \mathbb{R} e_{1}$. From equation (2) we reach to the same conclusion, i.e. $\mathbb{R}w = \mathbb{R} e_{1}$. But this is absurd because we supposed $v$ and $w$ are linearly independent. So $\rho$ is indecomposable.
2) For your second question, you need to show that $V_{i}$ is $G$-invariant and that $\left. \rho\right|_{V_{i}} : G \to GL(V_{i})$ is a representation.
A: For the first question, indecomposable means that you have to show that the representation is not equivalent to a direct sum of non-zero representations of $G$. This is clear in the first example, so it is indecomposable. In the second part, you have to verify the definition of direct sum of representations, see
Direct sum and tensor product of two representations of a group
