# Why commutativity gives an invariant subspace?

In A Generic Approach to Invariant Subspace Attacks the authors defined an invariant subspace as: $$F(u + A) = v + A$$

Where $$F:\mathbb{F}_2^n \to \mathbb{F}_2^n$$,$$u, v \in \mathbb{F}_2^n$$, and $$A$$ is a linear subspace.

Then they claimed that if $$M\circ F = F\circ M$$ for a linear subspace $$M$$ then $$A:=ker\left(M+Id\right)$$ is a linear subspace(Lemma 2).

I couldn't figure out why this is true for any bijective $$F$$ function. Any idea how to prove it?

• This question would probably be a better fit for Mathematics, as it seems to be pure linear algebra. – Ilmari Karonen Jan 26 '19 at 19:08