# Rank of image intensity matrix

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?

## 1 Answer

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.

• 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. – BRabbit27 Apr 16 '14 at 9:16
• By "lighting the scene from 3 orthogonal directions" you don't mean 3 images with different incoming light-directions, do you? – BRabbit27 Apr 16 '14 at 9:17
• 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. – BRabbit27 Apr 16 '14 at 9:20
• 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. – LKlevin Apr 16 '14 at 9:37
• 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. – BRabbit27 Apr 16 '14 at 10:37