From: unbewusst.sein@... (=?ISO-8859-1?Q?Une_B=E9v?= =?ISO-8859-1?Q?ue?=)
Date: 2007-11-23T04:40:04+09:00
Subject: Re: eigenvector discrepency was (Re: [Matrix] eigenvalues, eigenvectors in Ruby ???)

Axel Etzold <AEtzold@gmx.de> wrote:

> What is the problem that you are actually trying to solve ?

I've two molecules described with a matrix :
a = Mol.new( [  [ 0, 1, 1, 0 ], 
                [ 1, 0, 1, 0 ],
                [ 1, 1, 0, 1 ],
                [ 0, 0, 1, 0 ] ] )

b = Mol.new( [  [ 0, 1, 1, 1 ], 
                [ 1, 0, 0, 0 ], 
                [ 1, 0, 0, 1 ], 
                [ 1, 0, 1, 0 ] ] )

for example the first one in the first row of a says the carbon numbered
1 is connected to the carbon number 0.

i know, from construction a and b differs by a permuation p such that :

p * a * p^(-1) = b

only the numbered of carbon atoms is different in that case.

to state that a === b (within a permutation) it is suffciant to have :

eigenvalues of a === eigenvalues of b without respect to the order

however to find p i need the eigenvectors as given by the article :

http://en.wikipedia.org/wiki/Permutation_matrix

in the aboove case:
p = 
[ [ 0, 0, 1, 0 ], 
  [ 0, 0, 0, 1 ], 
  [ 1, 0, 0, 0 ], 
  [ 0, 1, 0, 0 ] ]

and computing :

p * a * p^( -1 ) = 
[ [ 0, 1, 1, 1 ], 
  [ 1, 0, 0, 0 ], 
  [ 1, 0, 0, 1 ], 
  [ 1, 0, 1, 0 ] ]

gives b

or computing :

p^( -1 ) * b * p = 
[ [ 0, 1, 1, 0 ], 
  [ 1, 0, 1, 0 ], 
  [ 1, 1, 0, 1 ], 
  [ 0, 0, 1, 0 ] ]

i do have a first version of a script using GSL, working as expected.
however with ExtendedMatrix i'm unable to find p from eigenvectors.

using GSL or ExtendedMatrix i get the same eigenvalues :
GSL:
a.eigval = [ 2.170e+00 3.111e-01 -1.000e+00 -1.481e+00 ]
b.eigval = [ 2.170e+00 -1.481e+00 3.111e-01 -1.000e+00 ]
ExtendedMatrix
a.eigval = Vector[-1.0, 2.17008648662603, -1.48119430409202,
0.311107817465982]
b.eigval = Vector[-1.48119430409202, 0.311107817465982, -1.0,
2.17008648662603]



with GSL, if i compare the eigenvectors :
a.eigvec = [
   5.227e-01  3.682e-01 -7.071e-01  3.020e-01 
   5.227e-01  3.682e-01  7.071e-01  3.020e-01 
   6.116e-01 -2.536e-01  0.0       -7.494e-01 
   2.818e-01 -8.152e-01  0.0        5.059e-01 ]
b.eigvec = [
   6.116e-01  7.494e-01  2.536e-01  0.0
   2.818e-01 -5.059e-01  8.152e-01  0.0 
   5.227e-01 -3.020e-01 -3.682e-01 -7.071e-01 
   5.227e-01 -3.020e-01 -3.682e-01  7.071e-01 ]

( i've rounded to 0.0 the values below e-15)

from column 0 it's easy to see what aa to be permuted
(all the values are the same but in a different order)
the first 5.227e-01 in a.eigvec column 0 is found in the third row of
b.eigvec column 0 then p[0, 2] = 1
the second 5.227e-01 in a.eigvec column 0 is found in the last row of
b.eigvec column 0 then p[1, 3] = 1
the 6.116e-01 (row 2) in a.eigvec column 0 is found in the first row of
b.eigvec column 0 then p[2, 1] = 1
the 2.818e-01 (row 3) in a.eigvec column 0 is found in the second row of
b.eigvec column 0 then p[3, 1] = 1

giving :

[ [ 0, 0, 1, 0 ], 
  [ 0, 0, 0, 1 ], 
  [ 1, 0, 0, 0 ], 
  [ 0, 1, 0, 0 ] ]

unfortunately i don't have any theoretical suggestification of this
procedure...

-- 
Une B�vue