SIFT: why it is invariant to affine tranlsation?

Question

I am studying SIFT. I am a little confused in some of Lowe's algorithm.

In order to have a descriptor invariant to translations like scaling, we must get rid of the scaling factor in LoG (Laplacian of Gaussian). I know that the LoG is represented like this:

Then the scale invariant LoG would look like this:

But how the DoG can be like this?:

In other words, how by approximation,using DoG, we get rid of that scale factor in SIFT. I know how DoG is computed, but the thing is I cannot understand how we get rid of that factor to have a invariant-to-scale LoG approximation. Any precise explanation is really appreciated.

EDIT: some more questions, Please answer precisely. Thanks :-) 1- At the final step of SIFT, we calculate feature vector using Gradient Orientation. to do do, the keypoint's rotation is subtracted from each orientation. I cannot understand why we subtract them?

2- What about Illumination dependence?

Anyone can explain how we remove Illumination and Orientation dependency?

Answer 1

I consider that you already know the proof of how we get DoG from LoG, so I won't go further with it.

According to your question "how by approximation,using DoG, we get rid of that scale factor in SIFT". DoG works as a feature detector to detect blobs in the given image (blob like shapes) which you can say it is a bit kind of edge detectors here is a photo of some blobs detected using DoG:

These blobs detected at different scales using the scale space pyramid. Using larger values of sigma you get bigger blobs. Experimentally DoG give you the best notion of scale as you can see here :

So, experimentally DoG gives you the most robust features that are scale invariant by constructing the scale space and computing the DoG and finally extracting local maxima points.

Useful links:

1- Lecture PPT

2- Best known video lecture for SIFT