Getting inf in matrix?

I wrote a code where I’m implementing an algorithm. The code works, but the values in my Z matrix are all inf and I don’t know why. I added a picture so you can see it. r in R^d should a sample drawn uniformly at random from the Fourier transformation of the Gaussian kernel. Gamma is just numbers uniformly at random in (0,2pi). cos(...) is at maximum 1, so I don’t understand why I get inf. Help is appreciated!
n=2000;
d=1000;
s=50;
Gamma=10;
S=randn(n,s);
U=qr(randn(s,d));
F=randn(n,d);
D=zeros(s,s);
for i=1:s
D(i,i)=1-(i-1)/d;
end
X = S*D*U + F/Gamma;
% This is the data matrix
-----------------------------------
function [G_2] = rnca(A,m)
[n,d]=size(A);
G=zeros(n,n);
for k=1:n
for j=1:n
G(k,j)=(1/2*pi)^(d/2)*exp(-norm(A(k,:)-A(j,:))^2/2);
end
end
Z=zeros(n,m);
gamma=0+(2*pi-0)*rand(1,m);
D=fft(G);
r=D(randsample(size(D,1),m),1:d);
for i=1:n
for l=1:m
Z(i,l)=sqrt(2/m)*cos(dot(r(l,:),A(i,:))+gamma(l));
end
end
G_2=Z*Z';
end

7 Comments

We can't run pictures. Please post your code as text.
r is not in R^d because the fourier transform is complex valued. You are taking cos of a complex number and the result of that has value that is exponential growth.
Desiree
Desiree on 19 Apr 2020
So how can I fix it?
is your expectation that the fft will produce a lot of values that are exactly 0? doesn't happen much due to roundoff even for cases that theory say should be 0.
Desiree
Desiree on 19 Apr 2020
The values of fft(G) are huge... which I guess it isn't supposed to be...
what is mean(G)?
Desiree
Desiree on 19 Apr 2020
Now I discovered that I missed () when defining G matrix as I forgot to put () on 2*pi.
Now indeed the fft(G) reproduces lots of 0 values and Z ain't inf anymore. So I guess now it is fine?

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on 19 Apr 2020

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on 19 Apr 2020

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