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$newcommand{GL}{operatorname{GL}}$

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 $0neq W le bigwedge^kV$ 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?

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.