The condition that P'^TFP is skew-symmetric is equivalent to \vec{X}^TP'^TFP\vec{X} = 0 for all $\vec{X}$. [1]
The authors say that if A is skew-symmetric than $\vec{x}^T A \vec{x} = 0$. Sadly, there is no further proof of that.
Question:
Other than expanding the equation (which I did) is there a different proof that holds for any sized skew-symmetric matrix? If so, what is the proof?
My thoughts:
I can only assume that because (as Wikipedia tells me) a skew-symmetric matrix of odd-size has one eigenvalue equal to zero, there always exists a kernel such that $A\vec{x} = \vec{0}$ but that would be specific to odd-sized matrices and would not hold for all $\vec{x}$.
Source:
[1] Multiple View Geometry, p.255, Second Edition 2004, Hartley & Zisserman, CUP (Section canonical cameras given F)