Date: Feb 27, 2013 6:08 AM
Author: Tony Kittler
Subject: Re: Alternative solution for NAN
"Torsten" wrote in message <kgko7b$e4d$1@newscl01ah.mathworks.com>...

> "Carl S." wrote in message <kgkmon$aeo$1@newscl01ah.mathworks.com>...

> > "Torsten" wrote in message <kgklbh$6ph$1@newscl01ah.mathworks.com>...

> > > "Carl S." wrote in message <kgki2g$rie$1@newscl01ah.mathworks.com>...

> > > > The following code gives NAN (Not a Number) values

> > > > [U,D]=eig(N);

> > > >

> > > > To solve this problem, I wrote that

> > > > while(det(N) == 0)

> > > > N=(1e-10.*randi(1,size(N)))*eye(size(N));

> > > > end

> > > >

> > > > But, the loop does not stop

> > >

> > > Your matrix N within the loop always has determinant (1e-10)^(size(N))

> > > which may become very small if N is large.

> > >

> > > :( Are there any alternative solution instead of this loop to solve the NAN problem ?

> > >

> > > Depends on the original matrix N.

> > >

> > > Best wishes

> > > Torsten.

> >

> > Dear Torsten,

> > The matrix N has standard deviation values of grayscale images. So, it changes for each image. How to solve the NAN problem in this case ?

>

> Did you check whether the matrix N already contains NaN values ?

>

> Best wishes

> Torsten.

Yes, I have checked with this codes

if isnan(N)==1

h=1;

else

h=0;

end

This code return that h=0