Pathological cases for subspaces invariant under the action of skew-symmetric matrices. Let $V$ be a dimension $n$ vector space over an arbitrary field $k$. I was asked to show that if $S$ is the vector space of matrices $M$ such that $M+M^t=0$ then there does not exist subspace $W \leq V$ that is proper and nonzero and such that $S \cdot W \subseteq W$ (i.e. invariant under the action by $S$). I think that it requires some additional hypothesis on the field and on the dimension. But I am not completely sure of my arguments.
I believe I have a proof with the set of hypothesis $\{n\geq 3,\text{char}(k)\neq 2\}$ and another with the set of hypothesis $\{n = 2, \text{ there is no }\alpha\in k \text{ such that }\alpha^2 = -1\}$. Then I made this list of pathological cases:

*

*If $n=1$ the result holds trivially.

*If $n \geq 2$ and $\text{char }k = 2$ the span of the vector $(1 ,\dots, 1)$ is invariant under the action by $S$ so the claim does not hold.

*If $n = 2$ and $\text{char }k \neq 2$ then there is a $W \leq V$ that is nonzero, proper and $S$-invariant if and only if there exists some $\alpha \in k$ such that $\alpha^2 = -1$.

Proof of item 3:

Given such an $\alpha$, the span of the vector $(1,\alpha)$ is
invariant under the action of $S$.
Conversely, suppose that $W \leq V$ is nonzero and proper and $S$-invariant. Then we have that $\text{dim }W = 1$ and so there is
some nonzero vector $w=(w_1,w_2)$ and $\lambda\in k$ such that $$(w_1,w_2) = \lambda(-w_2,w_1).$$ It follows that $w_1 = - \lambda^2 w_1$. Neither $w_1$
or $\lambda$ can be zero because we took $w$ to be nonzero (and $w_1 = 0$ gives $w_2=0$), but then $\lambda^2=-1$. Which concludes the proof.

I'm just a bit unsure because the exercise was given to me with no extra assumptions, so I wanted to check with somebody here. Thanks in advance.
BTW is there a more elegant way to summarise all of this? Am I overcomplicating things?
 A: People often get careless about this sort of thing (I mean the people who assigned you this exercise)!
Here's how I'd organize it. I'll write $F$ for the underlying field. Note that transpose isn't a well-defined operation on endomorphisms on a bare vector space; I'll work with $V = F^n$. This vector space of matrices is more typically called the orthogonal Lie algebra $\mathfrak{o}(n, F)$ (with the possible exception of characteristic $2$ where it is arguably the wrong object to consider). It is the Lie algebra of matrices $X$ preserving the standard "inner product" $\langle -, - \rangle : F^n \times F^n \to F$ in the sense that
$$\langle Xv, w \rangle = \langle v, Xw \rangle$$
(it is not entirely obvious that this is the correct definition of "preserving" but it is), and in particular it is closed under the commutator bracket $[X, Y] = XY - YX$, although we won't use this; this is at least context. $F^n$ is the defining representation of this Lie algebra, and what you're proving is that it's (usually) irreducible.
When $\text{char}(F) \neq 2$, $\mathfrak{o}(n, F)$ is ${n \choose 2}$-dimensional with basis given by $\omega_{ij} = E_{ij} - E_{ji}$, where $i < j$ and $E_{ij}$ is the matrix with $ij$-entry equal to $1$ and all others equal to zero. A subspace $W \subseteq F^n$ is invariant iff $\omega_{ij} W \subseteq W$. If $n \le 1$ there's nothing to prove so assume that $n \ge 2$.
Write $e_i$ for the standard basis of $F^n$, so that $E_{ij} e_k = e_i \delta_{jk}$ (where $\delta_{jk}$ is the Kronecker delta). We compute that
$$\omega_{ij} \left( \sum_{\ell=1}^n c_{\ell} e_{\ell} \right) = c_j e_i - c_i e_j$$
which gives
$$\omega_{ij} \omega_{jk} \left( \sum_{\ell=1}^n c_{\ell} e_{\ell} \right) = \omega_{ij} \left( c_k e_j - c_j e_k \right) = c_k e_i.$$
It follows that if $n \ge 3$ and $w = \sum c_{\ell} e_{\ell} \in W$ is a nonzero vector in a nonzero invariant subspace $W$ then $W$ contains every basis vector (by setting $k$ to be such that $c_k \neq 0$ then letting $i$ take every possible value) and hence must be $F^n$.
If $n = 2$ then $\mathfrak{o}(n, F)$ is $1$-dimensional spanned by $\omega_{12} = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right]$ and then we can argue as you did for item 3. This matrix $J$ is a very special matrix: it squares to $-1$, and over $\mathbb{R}$ it's the matrix generating the one-parameter subgroup $\theta \mapsto \exp(J \theta) = \left[ \begin{array}{cc} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{array} \right]$ of rotation matrices. Over any field it generates a subalgebra of $M_2(F)$ isomorphic to $F[x]/(x^2 + 1)$. So it makes sense that its behavior would depend on the presence of square roots of $-1$ in the underlying field. In fact you can show that the action of this algebra on $F^2$ is its regular representation, so it's irreducible iff $F[x]/(x^2 + 1)$ is a field iff $x^2 + 1$ is irreducible over $F$ iff it doesn't have a root.
When $\text{char}(F) = 2$, skew-symmetry becomes equivalent to symmetry and so $\mathfrak{o}(n, F)$ is ${n+1 \choose 2}$-dimensional, and we need new basis elements $\omega_{ii} = E_{ii}$ for the diagonal matrices. (Again, arguably this is the wrong object to study and no longer deserves the name "orthogonal Lie algebra," although explaining why is a bit of a digression.) Here I don't reproduce your conclusion in item 2; we have
$$\omega_{ii} \left( \sum c_{\ell} e_{\ell} \right) = c_i e_i$$
hence
$$\omega_{ij} \omega_{jj} \left( \sum c_{\ell} e_{\ell} \right) = c_j e_i$$
so we conclude as before that if $w = \sum c_{\ell} e_{\ell} \in W$ is a nonzero vector in a nonzero invariant subspace $W$ then $W$ contains every basis vector and hence $W = F^n$, as above and for all $n \ge 2$.
