Let $\mathcal{E}_n$ be the set of idempotents in the semigroup $\mathcal{B}_n$ of $n \times n$ Boolean relation matrices. The relation $E \leq F$ iff $EF=FE=E$ is called the natural partial order on $\mathcal{E}_n$.
Let $R(A), C(A)$ denote the row space and column space of the matrix $A \in \mathcal{B}_n$. Then $E \leq F$ iff $R(E) \subseteq R(F)$ and $C(E) \subseteq C(F)$ is a partial order on $\mathcal{E}_n$.
Is this the same relation as the natural partial order described above?. In other words, is it true that:
$R(E) \subseteq R(F)$ and $C(E) \subseteq C(F)$ iff $EF=FE=E$.
I was able to show $EF=FE=E$ implies $R(E) \subseteq R(F)$ and $C(E) \subseteq C(F)$.
I also checked all the cases for $n \leq 4$.
Can someone give a counter example or a proof of the forward direction in my conjecture above.
If $R(E)\subseteq R(F)$, then for each $i$, the $i$-th row of $E$ is equal to $x_i^TF$ for some row vector $x_i^T$. It follows that $E=XF$ for some matrix $X$ and $EF=XF^2=XF=E$. Similarly, $C(E)\subseteq C(F)$ implies that $FE=E$.