[scilab-Users] Problem with Floating point and determinant of a matrix

[Calixte Denizet]

Le lundi 27 décembre 2010 à 14:43 -0800, Prof. Dr. Reinaldo Golmia Dante
a écrit :
> Hi Calixte,
> 
> Thank you for your comment.
> 
> However, I do not still understand your suggestion: "--> 
> n=10;T=floor(rand(n,n-1)*200-100);A=[T,sum(T,2)];det(A)"
> 
> A does not show as a singular matrix for Scilab. I executed it on Scilab, as 
> follows:
> 
> -->n=10;T=floor(rand(n,n-1)*200-100);A=[T,sum(T,2)];det(A)
>  ans  =
>  
>   - 12228.029  
> 
> 
> What does it mean ? The determinant of A is different to zero, so A is 
> nonsingular matrix. Could you be clear in your
> example ?
> 

The last column of A is the sum of the previous columns, so the
determinant should be equal to 0 !! So A is not invertible... If you
calculate rcond(A), you'll see that A is very ill-conditioned. Take
care, there is big difference between numerical computations and
symbolic ones.

Cheers,

Calixte

> In regards to the paper, I am going to read it.
> 
> Thank you for your attention.
> 
> All best,
> Reinaldo.
> 
> 
> 
> ----- Original Message ----
> From: Calixte Denizet <calixte at contrib.scilab.org>
> To: users at lists.scilab.org
> Sent: Mon, December 27, 2010 7:52:05 PM
> Subject: Re: [scilab-Users] Problem with Floating point and determinant of a 
> matrix
> 
> Hi Reinaldo,
> 
> Try that:
> --> n=10;T=floor(rand(n,n-1)*200-100);A=[T,sum(T,2)];det(A)
> 
> A is built to be singular...
> 
> Have a look at this article, it could interested you:
> http://www.eng.nus.edu.sg/civil/REC2010/documents/papers/038.pdf
> 
> Calixte
> 
> Le lundi 27 décembre 2010 à 13:15 -0800, Prof. Dr. Reinaldo Golmia Dante
> a écrit :
> > Hi all,
> > 
> > I understand that Scilab stores the real numbers with foating point numbers, 
> > that is, with limited precision, and the computed value (answer) is not exactly 
> >
> > equal to 0 (page 23 - manual "Introduction to Scilab").
> >  
> > Take at look for those examples:
> >  
> > A = [1 2 3; 4 5 6; 7 8 9]
> > A  =
> >  
> >    1.    2.    3.  
> >    4.    5.    6.  
> >    7.    8.    9.  
> >  
> > -->det(A)
> >  ans  =
> >  
> >    6.661D-16  <---------- it should be 0
> >  
> > -->inv(A)
> >  ans  =
> >  
> >  10^15 *
> >  
> >  - 4.5035996    9.0071993  - 4.5035996  
> >    9.0071993  - 18.014399    9.0071993  
> >  - 4.5035996    9.0071993  - 4.5035996            <---------- it should appear 
> 
> > an error message because the matrix A is not invertible (or singular).
> > 
> >  
> > -->det(inv(A))
> >  ans  =
> >  
> >    9.007D+15    <-------------- The determinant of invertible matrix A^(-1) 
> > does not exist.
> >  
> > Other example:
> > 
> > -->B = [1 1; 1 1]
> > B  =
> >  
> >    1.    1.  
> >    1.    1.  
> >  
> > -->det(B)
> >  ans  =
> >  
> >    0.  <-------- it is correct !!
> >  
> > -->inv(B)
> >        !--error 19    <-------- it is correct !!
> > 
> >  
> > The previously examples show two integer matrices A and B. The determinant of 
> > matrix A is quite zero, but not,
> > and this can propagate an error in case the Scilab developer uses that result 
> > into other future calculations or algorithms.
> > The determinant of matrix B is equal to 0 and the answer is correct. In case 
> >the 
> >
> > Scilab developer uses that value,
> > he or she can use the simple statement for testing like to:
> >  if ( det(matrix) <> 0 ) then
> > <action 1>                    // The Scilab developer knows that the matrix is 
> 
> > invertible (or nonsingular)
> > else
> > <action 2>                    // The Scilab developer knows that the matrix is 
> 
> > not invertible (or singular)
> > end
> >  
> > My doubt: "How can I proceed to design any algorithm, which uses matrix, if the 
> >
> > determinant of 
> > 
> > the matrix could not be zero and, as the same time, that matrix is not 
> > invertible ?".
> > How can I manage this uncertainty ?
> >  
> > I appreciate to hearing from you some hints to solve this uncertainty.
> >  
> > Thank you in advance.
> >  
> > All best,
> > Reinaldo.
> >  
> > PS: I am not MATLAB user, but has MATLAB got this limited precision for 
> > determinants ?
> > 
> > 
> >      
> 
> 
>