Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Re: Alternative solution for NAN
Posted:
Feb 27, 2013 7:17 AM


"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=(1e10.*randi(1,size(N)))*eye(size(N)); > > > > > end > > > > > > > > > > But, the loop does not stop > > > > > > > > Your matrix N within the loop always has determinant (1e10)^(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=1e10; while(det(N) == 0) N=u.*eye(size(N)); u=u*100; end
Now, 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 ?



