Services & ResourcesWolfram ForumsMathGroup Archive

[Date Index] [Thread Index] [Author Index]

Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian

• To: mathgroup at smc.vnet.net
• Subject: [mg86290] Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian
• From: David Reiss <dbreiss at gmail.com>
• Date: Fri, 7 Mar 2008 02:32:47 -0500 (EST)
• References: <fqb9ai$na0$1@smc.vnet.net> <200803021856.NAA19943@smc.vnet.net>

At the expense of stating the obvious, I suspect that most readers of
this newsgroup understand this point...

-_David


On Mar 6, 3:09 am, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote:
> Yes, since it uses quite different algorithm. Generally algorithms  
> used for numerical solution of equations or various matrix  
> computations are quite different from (and much faster than) those  
> used in the symbolic case.
>
> Andrzej Kozlowksi
>
> On 5 Mar 2008, at 09:38, David Reiss wrote:
>
> > Thanks....  I stand  (or sit) corrected.
>
> > I note that although N[Eigenvalues[matrix]] does not give correct
> > results (with the bug that Daniel Lichtblau explains),
>
> > Eigenvalues[N[matrix]]
>
> > does indeed work as expected...
>
> > --David
>
> > On Mar 3, 4:48 am, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote:
> >> Eigenvalues or characteristic values of a  are defined (or rather, 
> >> can be defined) as the roots of the characteristic polynomial - and it
> >> does not matter is the matrix is invertible or not. Indeed, for a
> >> nilpotent matrix, such as
>
> >> M = {{-1, I}, {I, 1}}
>
> >> we have
>
> >> In[39]:= Eigenvalues[M]
> >> Out[39]= {0, 0}
>
> >> and
>
> >> In[40]:= CharacteristicPolynomial[M, x]
> >> Out[40]= x^2
>
> >> Moreover, the problem has nothing to do with numerical precision
> >> because in this case the exact eigenvalues do not satisfy the
> >> characteristic polynomial and in fact are the exact roots of a
> >> different polynomial of the same degree (as shown in my first post in
> >> this thread). Very weird.
>
> >> Andrzej Kozlowski
>
> >> On 2 Mar 2008, at 19:56, David Reiss wrote:
>
> >>> Note that your matrix is not invertible (its determinant is  
> >>> zero).  So
> >>> this is the source of your problem...
>
> >>> Hope that this helps...
>
> >>> -David
> >>> A WorkLife FrameWork
> >>> E x t e n d i n g MATHEMATICA's Reach...
> >>>http://scientificarts.com/worklife/
>
> >>> On Mar 1, 4:57 am, Sebastian Meznaric <mezna... at gmail.com> wrote:
> >>>> I have a 14x14 Hermitian matrix, posted at the bottom of this
> >>>> message.
> >>>> The eigenvalues that Mathematica obtains using the
> >>>> N[Eigenvalues[matrix]] include non-real numbers:
> >>>> {-9.41358 + 0.88758 \[ImaginaryI], -9.41358 -
> >>>>  0.88758 \[ImaginaryI], -7.37965 + 2.32729 \[ImaginaryI],
> >>>> -7.37965 -
> >>>>  2.32729 \[ImaginaryI], -4.46655 + 2.59738 \[ImaginaryI],
> >>>> -4.46655 -
> >>>>  2.59738 \[ImaginaryI], 4.36971, 3.21081, -2.32456 +
> >>>>  2.10914 \[ImaginaryI], -2.32456 - 2.10914 \[ImaginaryI],
> >>>> 2.04366+ 0.552265 \[ImaginaryI],
> >>>> 2.04366- 0.552265 \[ImaginaryI], -0.249588 +
> >>>>  1.29034 \[ImaginaryI], -0.249588 - 1.29034 \[ImaginaryI]}.
> >>>> However, if you do Eigenvalues[N[matrix]] it obtains different
> >>>> results
> >>>> {-9.09122, -7.41855, -7.41855, -7.2915, 4.33734, -4., -4.,  
> >>>> 3.2915, \
> >>>> -3.24612, -2.38787, -2.38787, 1.80642, 1.80642, 0}.
>
> >>>> These results agree with  
> >>>> Solve[CharacteristicPolynomial[matrix,x],x].
> >>>> Therefore I assume that the latter are correct. Has anyone seen  
> >>>> this?
> >>>> I am using 6.0.0.
>
> >>>> Here is the matrix:
> >>>> {{-6, 0, -Sqrt[3], 0, 0, Sqrt[3], 0, 0, 0, 0, 0, 0, 0, 0}, {0, -6,
> >>>>  0, -Sqrt[3], 0, 0, Sqrt[3], 0, 0, 0, 0, 0, 0, 0}, {-Sqrt[3], 0, 
>
> >>>> -4,
> >>>>  2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), 2 Sqrt[2/3], 0, 0, Sqrt[3], 
>
> >>>> 0,
> >>>>  0, 0, 0, 0, 0}, {0, -Sqrt[3],
> >>>>  2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), -4/3, -(2 Sqrt[2])/3, 0, 0, 
> >>>> 0,
> >>>>  Sqrt[3], 0, 0, 0, 0, 0}, {0, 0, 2 Sqrt[2/3], -(2 Sqrt[2])/3,
> >>>> 7/3, 0,
> >>>>   0, 0, 0, Sqrt[3], 0, 0, 0, 0}, {Sqrt[3], 0, 0, 0, 0, -4, 0,
> >>>>  2 (-1/(4 Sqrt[3]) + Sqrt[3]/4), 0, 0, 2 Sqrt[2/3], 0, 0, 0}, {0,
> >>>>  Sqrt[3], 0, 0, 0, 0, -4, 0, 2 (-1/(4 Sqrt[3]) + Sqrt[3]/4), 0, 0, 
> >>>>  2 Sqrt[2/3], 0, 0}, {0, 0, Sqrt[3], 0, 0,
> >>>>  2 (-1/(4 Sqrt[3]) + Sqrt[3]/4), 0, -14/3,
> >>>>  2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), 2 Sqrt[2/3], (2 Sqrt[2])/3, 
>
> >>>> 0,
> >>>>  0, 0}, {0, 0, 0, Sqrt[3], 0, 0, 2 (-1/(4 Sqrt[3]) + Sqrt[3]/4),
> >>>>  2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), -2, -(2 Sqrt[2])/3, 0, (
> >>>>  2 Sqrt[2])/3, 0, 0}, {0, 0, 0, 0, Sqrt[3], 0, 0,
> >>>>  2 Sqrt[2/3], -(2 Sqrt[2])/3, -7/3, 0, 0,
> >>>>  2 (1/(3 Sqrt[2]) + (2 Sqrt[2])/3), Sqrt[10/3]}, {0, 0, 0, 0, 0,
> >>>>  2 Sqrt[2/3], 0, (2 Sqrt[2])/3, 0, 0, -16/3,
> >>>>  2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), 2 Sqrt[2/3], 0}, {0, 0, 0, 
>
> >>>> 0, 0,
> >>>>   0, 2 Sqrt[2/3], 0, (2 Sqrt[2])/3, 0,
> >>>>  2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), -8/3, -(2 Sqrt[2])/3, 0},  
>
> >>>> {0, 0,
> >>>>   0, 0, 0, 0, 0, 0, 0, 2 (1/(3 Sqrt[2]) + (2 Sqrt[2])/3),
> >>>>  2 Sqrt[2/3], -(2 Sqrt[2])/3, 1/2,
> >>>>  2 (-Sqrt[5/3]/16 - Sqrt[15]/16)}, {0, 0, 0, 0, 0, 0, 0, 0, 0,  
>
> >>>> Sqrt[
> >>>>  10/3], 0, 0, 2 (-Sqrt[5/3]/16 - Sqrt[15]/16), 7/2}}

• Follow-Ups:
  • Re: Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>

• References:
  • Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian
    • From: David Reiss <dbreiss@gmail.com>