Fundamental matrix rank

Is there any way one can prove that $$\text{rank}\big(F+F^{T}\big)=3$$? where F is a fundamental matrix. If you have any idea, do let me know.

I am trying to solve this problem. I am just trying to understand the first part.

• – Royi
Jul 26 '20 at 18:23

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) &\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. $$\mathrm{rank}(F+ F^\top)< 3$$), 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.
• So as I said, it's not a proof. It's barely an intuition as to why no reduction would exist to reduce $F+F^\top$ from rank-3 to rank-2. Hence, we'd $\textit{convince ourselves}$ that the resulting rank is 3. If you have a stronger argument, I would be happy to see and learn from it. But merely citing references in the literature is not sufficient. Jul 27 '20 at 4:49