Re: Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian
- To: mathgroup at smc.vnet.net
- Subject: [mg86232] Re: [mg86179] Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Thu, 6 Mar 2008 03:01:22 -0500 (EST)
- References: <fqb9ai$na0$1@smc.vnet.net> <200803021856.NAA19943@smc.vnet.net> <200803050838.DAA19250@smc.vnet.net>
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}}
>
>
- References:
- Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian
- From: David Reiss <dbreiss@gmail.com>
- Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian
- From: David Reiss <dbreiss@gmail.com>
- Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian