2
$\begingroup$
sample = RandomVariate[NormalDistribution[0, 1.2], {10, 4, 4}]; 
Tranp= Transpose /@ sample; 
GOEe = sample + Tranp; 
Eigs = Map[Eigenvalues, GOEe]; 
Histogram[Eigs] 

Findingfit = FindFit[Eigs,A*1/(s*Sqrt[2*Pi])* Exp[-(1/2)*((x - u)/s)^2], {s, u, A}, x]
functionalform = Normal[Findingfit] 

Plot[functionalform, {x, Min[Eigs], Max[Eigs]}]

Hello, I need help to show that these eigenvalues of ten matrices 4 by 4 fall within a Gaussian distribution. I use Findfit function, but I get error messages such as "-7.92278 is not a valid variable", etc.

I really appreciate your help!

$\endgroup$
4
$\begingroup$

I wonder if your task is to describe the distribution of individual eigenvalues after being sorted by their absolute values (which is what Mathematica does when it returns the 4 eigenvalues). If so the following might be considered:

sample = RandomVariate[NormalDistribution[0, 1.2], {10000, 4, 4}];
Tranp = Transpose /@ sample;
GOEe = sample + Tranp;
Eigs = Map[Eigenvalues, GOEe];
pr = {{-12, 12}, {0, 0.4}};
Grid[{{SmoothHistogram[Transpose[Eigs][[1]], Automatic, "PDF", 
    PlotRange -> pr, PlotLabel -> "1st eigenvalue"],
   SmoothHistogram[Transpose[Eigs][[2]], Automatic, "PDF", 
    PlotRange -> pr, PlotLabel -> "2nd eigenvalue"]},
  {SmoothHistogram[Transpose[Eigs][[3]], Automatic, "PDF", 
    PlotRange -> pr, PlotLabel -> "3rd eigenvalue"],
   SmoothHistogram[Transpose[Eigs][[4]], Automatic, "PDF", 
    PlotRange -> pr, PlotLabel -> "4th eigenvalue"]}}]

Smooth histograms of eigenvalues

The sign for an individual eigenvalue is arbitrary which is why you see the bimodal distributions.

If the task is to consider the distribution of a randomly selected eigenvalue, then obtaining a smooth histogram is what you want (and not fit it to a normal because it clearly isn't distributed as a normal distribution):

SmoothHistogram[Flatten[Eigs], Automatic, "PDF"]

Smooth histogram for a randomly selected eigenvalue

P.S. (If you learned in a class to fit random samples from probability distributions with FindFit or NonlinearModelFit, demand your money back.)

$\endgroup$
3
$\begingroup$

FindFit returns a list of replacements, not a fitted function, so Normal doesn't do anything to it. I think you meant to use NonlinearModelFit there. Nevertheless, it would not have been the right tool for the job, as you are not doing a regression, but you are fitting a distribution to your data.

(I've increased the number of matrices in your sample to $10^6$ for significance.)

First of all, you will want to Flatten your lists of eigenvalues, so they are presented as a single long list:

sample = RandomVariate[NormalDistribution[0, 1.2], {1*^6, 4, 4}];
tranp = Transpose /@ sample;
GOEe = sample + tranp;
eigs = Flatten@Map[Eigenvalues, GOEe];

Histogram[eigs]

histogram of eigenvalues by counts

Then you can fit a (normal) distribution to the data to find its descriptor parameters (but not the expression of a PDF! Your data does not represent a frequency, so it cannot be fit to a PDF directly. You would have had to bin it and count the contents of each bin):

distpars = 
 FindDistributionParameters[eigs, NormalDistribution[mu, sigma]]

(* Out: {mu -> 0.00278758, sigma -> 3.79351} *)

Here are your histogram and the corresponding PDF of the best-fit normal distribution. I want to caution you though: they don't seem like a particularly good fit to me.

Show[
  Histogram[eigs, Automatic, "PDF"],
  Plot[PDF[NormalDistribution[mu, sigma] /. distpars, x], {x, -20, 20}]
]

superimposed calculated PDF of the best fit normal and histogram of data as a PDF

$\endgroup$
1
  • 1
    $\begingroup$ Note, selecting only the smallest Eigenvalue eigs=Flatten[Eigenvalues[#,-1]&/@GOEe] seems to fit a normal distribution really well instead. $\endgroup$ – Julien Kluge Jan 24 at 23:24

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.