Réf: Question about Scilab-poly

[michel.marzin at ext.mpsa.com]

Sylvestre,

his matrix was :
A=
1 1 3
1 2 4
0 1 1

in short term,

the poly of this matrix is : det( I*x - A) = x^2*(x-4)
solution by hand of course, and i don't see myself do it on a 1000*1000
matrix.

under my Scilab 4.12 :   5.121D-16 + 2.902D-16x -4 x^2 + x^3

under my Scilab 5.0.3 :  - 7.019D-17 + 1.792D-15x - 4x^2 + x^3

Regarding A, i assume that all value almost at zero are zeros.
i mean, det(A)=0 so at least zero is a root of   my "poly" and i assume at
least that 5.121D-16 is "zero" or - 7.019D-17.

A famous Scilab alternative do that also but you have to take care of the
truncation of the values in this case.
>> poly(a)
ans =

    1.0000   -4.0000    0.0000    0.0000

"good" result in this case, but if i assume that in my matrix, the zero is
now 0.000001, it's now a wrong result.

regards
----------------------------------------------------------------------
Michel



                                                                           
             Sylvestre Ledru                                               
             <sylvestre.ledru                                              
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Le mardi 16 décembre 2008 à 12:46 +0200, Sotos a écrit :
> Hallo,
>
> I have been using Scilab for about 3  months,as  a new  student.
> My question...
> What kind of number is that at the bottom...(4.406D-16).
Note that Scilab 5 is results closer that what you expect:

-->a=[1 1 3
-->1 2 3
-->0 1 1 ]
 a  =

    1.    1.    3.
    1.    2.    3.
    0.    1.    1.

-->p= poly(a,"z")
 p  =

              2   3
  - 1 + z - 4z + z

Sylvestre