Show that the only subrepresentation of $A$ is $W= \{(t,0) \mid t \in \Bbb R\}$. So $V$ is not reducible, but it is indecomposable.

Question

Let $A = \Bbb R[x]$, and $V = \Bbb R^{\oplus 2}$ be the representation with action $$\rho(x)= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}.$$

Show that the only subrepresentation of $A$ is $W= \{(t,0) \mid t \in \Bbb R\}$. So $V$ is not reducible, but it is indecomposable.

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To show that $W$ is a subrepresentation we have that for any $a = \sum_{i=0}^n c_ix^i \in A$ and $w \in W$ $$\rho(a)w = (c_n\rho(x)^n+\dots+c_0\rho(1))w$$

but taking powers of the matrix $\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ and adding constant multiples of these powers will result in a matrix of the following form $$\begin{bmatrix}u & v \\ 0 & z\end{bmatrix}$$ which means that $$\rho(a)w = \begin{bmatrix}u & v \\ 0 & z\end{bmatrix}\begin{bmatrix}t\\ 0\end{bmatrix}= \begin{bmatrix}u\\ 0\end{bmatrix} \in W$$ so $W$ is invariant under the action of $\rho$ and hence a subrepresentation.

How do I prove that $V$ is not reducible?

Answer 1

Let $V=W\oplus U$ for some sub-representation $U\subset V$. So $U$ contains some vector $v=(a,b)$, where $b\ne0$. But the $\rho(x)v-v\ne0\in U$ since $U$ is a sub-representation. But $\rho(x)v-v=(b,0)\in W$, so $W\subset U$, a contradiction.