2
$\begingroup$

I have a function $F(\vec x)$ of many variables (let's say in the order of hundreds of thousands). I need to compute the determinant of the Hessian matrix at the point $x_0$.

Is there a way to compute some approximation of it without explicitly computing the determinant?

I want to compute this value because it appears in the formula I want to use to compute a rate in a transition. There is no way to avoid to compute this quantity without renounce to use this theory of transitions and developing another one by myself.

This function is the discretized version of a functional, this is the reason for the great number of variables I am dealing with.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ Are you trying to determine if the function is convex/concave or some other measure (eg curvature)? $\endgroup$
    – Richard
    May 6 '19 at 0:52
  • 5
    $\begingroup$ What makes you think that knowing the determinant of the Hessian will tell you whether or not the function is convex or concave? $\endgroup$
    – Brian Borchers
    May 6 '19 at 2:18
  • $\begingroup$ You are right, I remembered badly from some calculus lectures. I have edited the question. $\endgroup$
    – skdys
    May 6 '19 at 7:14
  • 4
    $\begingroup$ It is still very questionable that you have to have the determinant value no matter what since the determinant calculations are very numerically inefficient. Thus, usually, determinants are used in pure theoretical derivations rather than provide the recipe for scientific computations. A proper reference would go a long way here. $\endgroup$
    – Anton Menshov ♦
    May 6 '19 at 10:52
  • 3
    $\begingroup$ Isn't the determinant of such a large matrix going to be almost $0$ or almost $\infty$? If you use the determinant of a random matrix as a guide (mathoverflow.net/questions/13008; you say $n\sim 10^5$), it's going to be either infinitesimal or humongous, and therefore not useful for calculations. $\endgroup$
    – Kirill
    May 6 '19 at 15:10
3
$\begingroup$

I would place a wild guess here as an answer.

The derivations in the omitted reference used determinants for some theoretical derivations. While determinants are a very useful tool to prove certain things, they are a very lousy tool from the computational science perspective.

Now, the theory of transitions was developed (which I am not familiar with) to enable numerical computations (possibly) at a large scale. This theory would provide a method to compute the required quantities without the determinant of the Hessian; therefore, leading to a potentially accurate and scalable scheme.

You might be able to use determinants to verify your implementation at a very small scale (say 3-4 variables, and even then you might end up having trouble due to floating-point arithmetic), but going for large-scale with determinants is a no-go.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I didn't add details about the problem I am studying to keep the question as abstract as possible and thus be useful to other people maybe. However, here is some additional info. I want to apply Kramers formula for transition rates in the case of a multidimensional system, you can find more information in this paper link.springer.com/article/10.1007/BF01770354 The formula derived there contains the determinant of the Hessian evaluated in the minimum of the well and in the saddle point. $\endgroup$
    – skdys
    May 6 '19 at 13:11
  • 2
    $\begingroup$ @skdys that paper is behind a paywall (most people won’t have access to it). mind adding the relevant context and equations into your original question? $\endgroup$
    – GoHokies
    May 6 '19 at 14:50
  • $\begingroup$ You'll have to work with non-normalized probability distributions. The normalization constant ("partition function") I suspect shows up in your formula can not be computed in practice. $\endgroup$
    – Wolfgang Bangerth
    May 9 '19 at 16:28

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.