Let $V$ be a real $n$-dimensional vector space. For $1<k<n$ we have a natural representation of $\GL(V)$ via the $k$ exterior power:
$\rho:\GL(V) \to \GL(\bigwedge^kV)$, given by $\rho(A)=\bigwedge^k A$. I am trying to show $\rho$ is an irreducible representation. Let $0\neq W \le \bigwedge^kV$ be a subrepresentation. If we can show $W$ contains a non-zero decomposable element, we are done.
Indeed, suppose $W \subsetneq \bigwedge^kV$. Then, there exist a decomposable element $\sigma=v_1 \wedge \dots \wedge v_k \neq 0$, such that $\sigma \notin W$. We assumed $W$ contains a non-zero decomposable element $\sigma'=u_1 \wedge \dots \wedge u_k \neq 0$. Define a map $A \in \GL(V)$ by extending $u_i \to v_i$. Then
$$\rho(A) (\sigma')=\bigwedge^k A(u_1 \wedge \dots \wedge u_k)=\sigma \notin W,$$
while $\sigma' \in W$, con
So, the question reduces to the following: Why does every non-zero subrepresentation contain a non-zero decomposable element?
I asked an even more naive question here-whether or not every subspace of dimension greater than $1$ contains a non-zero decomposable element?