Show that the representation $\mathbb Z\ni a\mapsto\begin{pmatrix}1& a\\0&1\end{pmatrix}$ is not completely reducible

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.

1. 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.

2. 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).

   Depending on the answer to question (1), 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).

This CW answer intends to remove the question from the unanswered queue.

As you already noted yourself in the comments you can take $V=\mathbb{C}^2$ and then a Jordan normal form argument is perfectly fine.