Linear Representations: Show that no $W^0$ exists.

Question

Given the following linear representation and subrepresentation $W$, show that there exists no $W^0$ such that $\mathbb{R}^2 = W \oplus W^0$.

Let $\rho: (\mathbb{Z}, +) \to GL(\mathbb{R}^2)$ be given by $1: \to \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, and $W = \begin{bmatrix} 1 \\ 0\end{bmatrix}$ be a subrepresentation. Show that $\nexists$ a $W^0$ s.t. $\mathbb{R^2} = W \oplus W^0$.

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I have tried to show that the following sum diverges:

$\dfrac{1}{|\mathbb{Z}|}\sum_{g \in \mathbb{Z}} \rho (g)\cdot p_{W} \cdot \rho (g^{-1})$

But I have had no luck in doing so. Another option I've tried is assuming that $\mathbb{R}^2 = W \oplus W^0$, but I've had no luck in showing that it's a contradiction. Any help here would be quite lovely.

Thanks.

Answer 1

I'm assuming here that the question is to show that there does not exist a $W^{0}$ such that $$\mathbb{R}^{2} = W \oplus W^{0}$$ and $W^{0}$ is an invariant subspace of $\mathbb{R}^{2}$. If you did not meant that in the problem, I apologize.

So suppose you have a complement $W^{0}$ which is also a subrepresentation of $\mathbb{R}^{2}$. Then, $W^{0}$ is one dimensional and is hence $\mathbb{R}\cdot (x, y)$ for $(x, y)\not = 0$. Then, as $W^{0}$ is invariant under the action of $\mathbb{Z}$,

$$\left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right) \left(\begin{array}{c} x\\ y \end{array}\right) = \left(\begin{array}{c} x + y\\ y \end{array} \right) \in W^{0}.$$

Since $W^{0}$ is a complement for $W$, $y \not =0$, which gives us a contradiction. Essentially, one dimensional subrepresentations of representations of $G$ correspond to eigenspaces of every element $g \in G$, and this representation has only $W$ as an eigenspace.