Fundamental matrix rank
This is not a proof, but maybe an intuition why this conjecture can be true for the points in general position. From the properties of rank we know that:
$$
\mathrm{rank}(F) = \mathrm{rank}(F^\top) = 2.
$$
Hence:
$$
\begin{align}
\mathrm{rank}(F+ F^\top) &amp;\leq \mathrm{rank}(F)+\mathrm{rank}(F^\top)\leq 4.
\end{align}
$$
Because $(F+F^\top)\in \mathbb{R}^{3\times 3}$, $\mathrm{rank}(F+ F^\top)\leq 3$. To have any rank deficiency here (e.g. $\mathrm{rank}(F+ F^\top)&lt; 3$), we should have an additional rank constraint on $F$, which we donot seem to have. The only additional constraint is that $F$ has 7 degrees of freedom and that $\det(F)=0$. These non-linear constraints are not sufficient to reduce the rank.