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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.

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I am just trying to understand the first part.

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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.

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  • $\begingroup$ I guess this is what they meant when they asked to convince ourselves. I don't know what else to show. $\endgroup$ – mathuser001 Jul 26 '20 at 7:52
  • $\begingroup$ I really don't understand how this is an answer. This is the symmetric part of the Fundamental Matrix and by geometry can be shown to have rank 3 in the general case. $\endgroup$ – Royi Jul 26 '20 at 18:18
  • $\begingroup$ 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. $\endgroup$ – Tolga Birdal Jul 27 '20 at 4:49
  • $\begingroup$ @Royi The citations you gave don't have anything regarding this problem in hand $\endgroup$ – mathuser001 Jul 27 '20 at 7:24
  • $\begingroup$ Have you looked at the chapter which deals with the symmetric part of the matrix? $\endgroup$ – Royi Jul 27 '20 at 16:26

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