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.
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Sign up to join this communityIs 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.
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.