Why is the exterior power $\bigwedge^kV$ an irreducible representation of $GL(V)$? $\newcommand{\GL}{\operatorname{GL}}$
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?
 A: Pick a basis $e_1, \dots e_n$ of $V$ so that we can identify $GL(V)$ with $GL_n(F)$ (we'll start out working with an arbitrary base field $F$ and then restrict $F$ later). Write $T$ for the subgroup of $GL_n(F)$ consisting of diagonal matrices. An element of $T$ consists of some diagonal elements $(t_1, \dots t_n)$ and acts on $\Lambda^k(V)$ by sending $e_i$ to $t_i e_i$, then extending multiplicatively. 
What this means is that each pure tensor $e_{i_1} \wedge e_{i_2} \wedge \dots \wedge e_{i_k} \in \Lambda^k(V)$ is a simultaneous eigenvector for every element of $T$; said another way, it spans a $1$-dimensional (hence simple) subrepresentation of $\Lambda^k(V)$, considered as a representation of $T$. (These are the "weight spaces" of this representation.) Since $\Lambda^k(V)$ is the direct sum of these $1$-dimensional subspaces, it follows that $\Lambda^k(V)$ is semisimple as a representation of $T$. 
The significance of semisimplicity is that any $GL(V)$-subrepresentation of $\Lambda^k(V)$ is also a $T$-subrepresentation, and subrepresentations of semisimple representations are semisimple; they must also have the same simple components, in the same or smaller multiplicities. Moreover, if $F$ is any field except $\mathbb{F}_2$ (over $\mathbb{F}_2$, unfortunately, $T$ is the trivial group), the different $1$-dimensional representations above are all nonisomorphic. The conclusion from here is that any $GL(V)$-subrepresentation of $\Lambda^k(V)$ must be a direct sum of weight spaces. 
But now we're done (again, for any field $F$ except $\mathbb{F}_2$), for example because $GL(V)$ acts transitively on these weight spaces. 
