0
$\begingroup$

I've been reading a paper about using Matrix Completion for Photometric Stereo but I am having some troubles in section 2.2 trying to understand why irrespective of the number of pixels and the number of images $$rank(\mathbf{O}) = 3$$ where $\mathbf{O}$ is the matrix with the images linearized and stacked as column vectors.

Can someone give me a hint to understand this?

$\endgroup$
2
$\begingroup$

The images described in the paper are not independent, but are uniquely defined by a lighting direction which is 3 dimensional.

You could thus construct any conceivable image (in this model) by lighting the scene from 3 orthogonal directions (x, y and z for instance). This corresponds to the rank being 3.

Hope this helps.

$\endgroup$
  • $\begingroup$ Is there some other way you could explain it. I kinda understand what you are saying and based on what I've read intuitively makes sense but it will be haunting me until I really understand it. $\endgroup$ – BRabbit27 Apr 16 '14 at 9:16
  • $\begingroup$ By "lighting the scene from 3 orthogonal directions" you don't mean 3 images with different incoming light-directions, do you? $\endgroup$ – BRabbit27 Apr 16 '14 at 9:17
  • $\begingroup$ The other way, maybe not correct, I was thinking of it was that, any pixel has a uniquely surface-normal vector which is of dimension 3, therefore independently of the number of pixels or images we will have a unique normal, but still I am not 100% convinced of it. $\endgroup$ – BRabbit27 Apr 16 '14 at 9:20
  • $\begingroup$ That's actually equivalent to what I'm saying. From 3 images with orthogonal lighting directions we could recover the surface normal in each pixel of the image. So yes, it is because the normal has dimension 3. $\endgroup$ – LKlevin Apr 16 '14 at 9:37
  • $\begingroup$ Could you give an example, maybe more math-based, in which something similar to the case I'm dealing with happens? Like I said, intuitively I see why the rank-3 but I would like a more convincing statement. $\endgroup$ – BRabbit27 Apr 16 '14 at 10:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.