determinant of hessian matrix of a greyscale image is too small

I am trying to find determinant of hessian matrix of a 50x50 grayscale image. Determinant of matrix I am getting is a very small value i.e 4.7612e-134. I think I am missing something. My code is below. Thanks

 % computing second derivatives in each direction first
 [gx, gy] = gradient(double(sliceOfImageK2));
 [gxx, gxy] = gradient(gx);
 [gyx, gyy] = gradient(gy);

 hessianMatrix = [gxx gxy; gxy gyy];
 determinantHessianMatrix = det(hessianMatrix);

tayyab

[Walter Roberson]

Could you attach the image?

[Tayyab Ateeq]

here is the image. Actually it is a small patch extracted from a larger image. Thanks

[Walter Roberson]

Most of that image is practically constant white, which would have a gradient of 0. The variation from pure white would be caused by trying to analyze a JPEG image that lossy compression has been applied to.

But aside from that, there is something in your procedure which seems to lead to a determinant of 0 when evaluated symbolically. I note that the rank() of the matrix appears to be 99 instead of 100, so the determinant is going to come out 0 in that case. I am not sure why the rank comes out 99, but perhaps there are good algebraic reasons.

[Tayyab Ateeq]

The white portion is not part of the image, only central part is which is grayscale. I have extracted this patch from a large image which is in nii format. I saved it in jpg in order to upload it here. Now I have uploaded png file of same patch to give clear understanding. Second thing is rank. The dimensions of hessianMatrix are 50 x 50 but I just checked that its rank is 49 so you are right thats why determinant is zero. Is there any workaround to make it work? Thanks

[Walter Roberson]

You put together two 50 x 50 gradient matrices in each direction, forming a 100 x 100 matrix. The hessianMatrix cannot be 50 x 50 afterwards.

[John D'Errico]

Are you applying the gradient to the image itself? Why would you do that? You have posted jpg images, but not sliceOfImageK2.

[Tayyab Ateeq]

"Why would you do that?" How do you mean?

Answer 1

[John D'Errico]

Nope. You are not missing a thing, well, except in your understanding of why you should not be using a determinant.

Determinants will almost always have numerical problems. So, when doing virtually any scientific computing task, you should almost never compute a determinant. There are always better ways to achieve what you need to do.

[Walter Roberson]

If you have the symbolic toolbox you can reduce the numeric error by using

   determinantHessianMatrix = det(sym(hessianMatrix));

The question persists as to whether using det() is something you should be doing.

[Tayyab Ateeq]

The above command gives -391057855557969437726928269430339828061829017700937539811671619117157870174814093035735068965980807382182295......Is it ok to have such a large value for determinant. Also how can I achive this without using det(). Thanks

[John D'Errico]

You seem to have this understanding that if the determinant is small (or big), that it means something. It does not.

What is the determinant of an identity matrix? It is clearly 1.

det(eye(1000))
ans =
     1

Yep. Wow. That was easy. It must be non-singular.

So then is the diagonal matrix with 1/3 all down the main diagonal also of full rank? Is it singular? That seems a pretty simple change. And clearly the matrix is no more singular than the first one.

But det thinks it is singular. Exactly so.

det(eye(1000)/3)
ans =
     0

And that result is not just a tiny number, but flat out zero.

det(eye(1000)/3) == 0
ans =
  logical
   1

But then why does this happen?

det(eye(1000)*3)
ans =
   Inf

Is that matrix any more or less singular than the others?

Has det told us ANYTHING of value here? NO. Only that using det was a waste of time.

Now, we don't really know why you want to use det there. Only that you have the idea that it tells you something, and that you are worried about the size of the numbers it generates.

[Walter Roberson]

"Also how can I achive this without using det()."

You have not told us what you are trying to achieve, except by defining the problem as requiring det(). If you define the problem as being the calculation of det() then it is not clear that there is any benefit from finding a different method of calculating what det() would calculate.

[Tayyab Ateeq]

Actually my objective is to detect particular black spots from brain mri. I have managed to extract all black spots in an image(which is nii file). Now I intend to apply SVM to classify right black dots from false positives. I have extracted all these black spots as patches and extracting features from them. I want to use determinant of hessian as a feature. Afterwards I also want to use eigen values and trace of hessian as features but I am stuck while computing determinant of hessian. I did not explain this before to avoid complexity.

[Walter Roberson]

"I want to use determinant of hessian as a feature"

Either "Don't DO that", or use a high precision calculation system.

Are you trying to determine critical points? https://en.wikipedia.org/wiki/Second_partial_derivative_test ? If so then notice that they are taking the hessian of a function, not the hessian of a matrix.

Answer 2

[FOROUGH ASKARISIAHOOIE]

Use the function "imgradientxyz" instead of "gradient", and use the latest version of Matlab.