```Date: Feb 27, 2013 7:17 AM
Author: Tony Kittler
Subject: Re: Alternative solution for NAN

"Carl  S." wrote in message <kgkpep\$h9u\$1@newscl01ah.mathworks.com>...> "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 ?> > I have tried this,u=1e-10;while(det(N) == 0)     N=u.*eye(size(N));     u=u*100;endNow, it works without giving NAN value. But, I am not sure that this algorithm correct results. Do you think that this is meaningful or I can get unexpected results ? Any suggestions ?
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