Let $\rho : \mathbb Z \to \mathrm{GL}_2(\mathbb C)$ be the representation defined by $\rho(1) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. I'd like to show that $\rho$ is not completely reducible.
i) I have one preliminary question (which is probably a silly one) - for what vector space $V$ is $\mathrm{GL}(V) \cong \mathrm{GL}_2(\mathbb C)$?
Firstly, I noted that $\rho(1)$ has an eigenvector, so the representation is not irreducible. So if it were completely reducible, it would have to break up as a direct sum of two 1-dimensional sub representations. But a 1-dimensional subrep is given by an eigenvector - but $\rho$ only has one eigenvalue, which has a 1-dimensional eigenspace. So this can't happen.
ii) Is this reasoning OK?
Once I've shown that the representation isn't irreducible, the problem is equivalent to showing that $\rho(1)$ cannot be diagonalised (which I've done by showing that the sum of the dimensions of the eigenspaces is 1, not 2).
Dependant on the answer to question i), I could have reduced (excuse the pun) the amount of work by considering Jordan Normal Form ($\rho(1)$ is in JNF but isn't diagonal, so isn't diagonalisable).
Thanks.