dimension of a subspace of matrix space

Question

Let $V$ be the linear space of $n\times n$ matrix over some field $F$. Let $W$ be a subspace. $\forall A,B\in W$ we have $AB\in W$. And $\dim W>1$. Every non-zero matrix in $W$ is invertible. Prove that $ \dim W$ is even.

Answer 1

I don't know the answer, but this seems relevant: $W$ must be a skew-field.

Fix non-zero $B \in W$. Then the map $\phi_B : A \mapsto BA$ is linear and invertible on $V$. Restricting this map to $W$, certainly $\phi_B|_W$ is injective, and hence it must be surjective as it is a linear operator on the finite-dimensional space $W$.

Note also that $\phi_B^{-1}(A) = B^{-1}A$. Since $\phi_B^{-1}|_W = (\phi_B|_W)^{-1}$, it follows that $B^{-1} B = I \in W$. Moreover, $\phi_B^{-1}(I) = B^{-1} \in W$, hence $W \setminus \lbrace 0 \rbrace$ is a group. Already $W$ is an additive group, and matrix multiplication satisfies the distributive laws. Thus, $W$ is a skew-field.

In a previous edit, I said it must be a field, but of course, I didn't verify commutativity of the multiplication.