why does (e'^T) E = 0 ? Is there a proof for this? Thanks!
arvinwu168
Is it because the line according to l' = Ex becomes a point when x = e? But how does a point make a line equation 0?
mpotoole
Here's my justification.
So recall from this slide that the epipole is a point that lies on every epipolar plane (or epipolar line)! As a result, all point $x'$ on the corresponding camera, there exists an epipolar plane that contains that point $x'$ and the epipole $e$. In other words, $x'^T E e = 0$ for all $x'$. And this can only be true if $E e = \mathbf{0}$.
Yet another way to prove that this is the case is to use the definition for the essential matrix here, and make use of the fact that the epipole and the baseline (translation vector) are collinear!
why does (e'^T) E = 0 ? Is there a proof for this? Thanks!
Is it because the line according to l' = Ex becomes a point when x = e? But how does a point make a line equation 0?
Here's my justification.
So recall from this slide that the epipole is a point that lies on every epipolar plane (or epipolar line)! As a result, all point $x'$ on the corresponding camera, there exists an epipolar plane that contains that point $x'$ and the epipole $e$. In other words, $x'^T E e = 0$ for all $x'$. And this can only be true if $E e = \mathbf{0}$.
Yet another way to prove that this is the case is to use the definition for the essential matrix here, and make use of the fact that the epipole and the baseline (translation vector) are collinear!