prove that there is a group representation of $\Bbb Z$ which is not totally decomposable over $\Bbb C$ prove that there is a group representation of $\Bbb Z$ which is not totally decomposable over $\Bbb C$
what I tried - 
let $\mu: \Bbb Z \to GL_{2*2} \Bbb (C)$
$$\mu(x)(v) =  \begin{pmatrix}
        1 & x \\
        0 & 1 \\
        \end{pmatrix} \begin{pmatrix}
        a  \\
        b  \\
        \end{pmatrix}$$
and then showed invariant field 
$L =  \begin{pmatrix}
        a  \\
        0  \\
        \end{pmatrix}$
that does not have an invariant complement.
I have used this way to prove that there is a group representation of $\Bbb Z/p\Bbb Z$ that is not decomposable over a field $\Bbb F_p$
So I think i might have a mistake somewhere and that it is not the solution in this case. 
Any help will be appreciated.
 A: Let us show that the $1d$ subspace $L$ of $\mathbb{C}\cong\mathbb{R}^2$, generated by $\begin{pmatrix}
        a  \\
        0  \\
        \end{pmatrix}$, $a\in\mathbb{R}$, does not have an invariant complement: 
Let $\begin{pmatrix}
        a  \\
        b  \\
        \end{pmatrix}$, with $\frac{b}{a}=\textrm{const}\neq 0$ be a vector,  generating an $1d$ invariant subspace of $\mathbb{C}\cong\mathbb{R}^2$ which is a complement of $L$ (note that a necessary condition for being a complement is $b\neq 0$). Thus for any $x\in\mathbb{Z}$:
$$\mu(x)\begin{pmatrix}
        a  \\
        b  \\
        \end{pmatrix}=
        \begin{pmatrix}
        1 & x \\
        0 & 1 \\
        \end{pmatrix} 
       \begin{pmatrix}
        a  \\
        b  \\
        \end{pmatrix}=
        \begin{pmatrix}
        a+bx \\
        b  \\
        \end{pmatrix}$$
But $\frac{b}{a}=\frac{b}{a+bx}$ for any $x\in\mathbb{Z}$, implies $b=0$, leading to the conclusion that no complement of $L$ is invariant under $\mu$. 
Thus, $\mu:\mathbb{Z} \to GL_{2\times 2}(\mathbb{R})$ is an indecomposable representation of the additive group $\mathbb{Z}$ on the $2d$ real vector space $\mathbb{C}$.
