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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.


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 $\vec{x}$.


[1] Multiple View Geometry, p.255, Second Edition 2004, Hartley & Zisserman, CUP (Section canonical cameras given F)

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up vote 10 down vote accepted

Are you asking for a proof that $x^T A x = 0$ for all $x \in \mathbb{F}^n$? Here $A \in \mathbb{F}^{n,n}$ is skew-symmetric and $\mathbb{F}$ is a field of characteristic $\ne 2$. If so, let $x \in \mathbb{F}^n$ and set $a := x^T A x$. Then $a = (x^T A x)^T = x^T A^T x = -x^T A x = -a$, whence $2a = 0$.

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