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I want to add one thing to @Dima's answer, just so things are absolutely clear: $E = (K')^TFK$ (note the transpose) You can see how this also captures that the fundamental matrix takes image coordinates as inputs and not normalized 3D points. Let $x$, $x'$ be 2 normalized 3D points: $(x')^T*E*x = (x')^T(K')^TFKx = (K'x')^TF(Kx)$ I can't comment yet because I ...


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I tried it on square grayscale images myself (just hoping they would be invertible, which was always the case (not much luck involved here, since invertible matrices lie dense)). I added a few examples so you can judge the inverses yourself. Their precise look of course depends on the particular normalization (I used cv2.normalize and also normalized ...


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