Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian
- To: mathgroup at smc.vnet.net
- Subject: [mg86115] Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian
- From: David Reiss <dbreiss at gmail.com>
- Date: Sun, 2 Mar 2008 13:56:23 -0500 (EST)
- References: <fqb9ai$na0$1@smc.vnet.net>
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=A0am, 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 - > =A0 0.88758 \[ImaginaryI], -7.37965 + 2.32729 \[ImaginaryI], -7.37965 - > =A0 2.32729 \[ImaginaryI], -4.46655 + 2.59738 \[ImaginaryI], -4.46655 - > =A0 2.59738 \[ImaginaryI], 4.36971, 3.21081, -2.32456 + > =A0 2.10914 \[ImaginaryI], -2.32456 - 2.10914 \[ImaginaryI], > =A02.04366+ 0.552265 \[ImaginaryI], > =A02.04366- 0.552265 \[ImaginaryI], -0.249588 + > =A0 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, > =A0 0, -Sqrt[3], 0, 0, Sqrt[3], 0, 0, 0, 0, 0, 0, 0}, {-Sqrt[3], 0, -4, > =A0 2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), 2 Sqrt[2/3], 0, 0, Sqrt[3], 0, > =A0 0, 0, 0, 0, 0}, {0, -Sqrt[3], > =A0 2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), -4/3, -(2 Sqrt[2])/3, 0, 0, 0, > =A0 Sqrt[3], 0, 0, 0, 0, 0}, {0, 0, 2 Sqrt[2/3], -(2 Sqrt[2])/3, 7/3, 0, > =A0 =A00, 0, 0, Sqrt[3], 0, 0, 0, 0}, {Sqrt[3], 0, 0, 0, 0, -4, 0, > =A0 2 (-1/(4 Sqrt[3]) + Sqrt[3]/4), 0, 0, 2 Sqrt[2/3], 0, 0, 0}, {0, > =A0 Sqrt[3], 0, 0, 0, 0, -4, 0, 2 (-1/(4 Sqrt[3]) + Sqrt[3]/4), 0, 0, > =A0 2 Sqrt[2/3], 0, 0}, {0, 0, Sqrt[3], 0, 0, > =A0 2 (-1/(4 Sqrt[3]) + Sqrt[3]/4), 0, -14/3, > =A0 2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), 2 Sqrt[2/3], (2 Sqrt[2])/3, 0, > =A0 0, 0}, {0, 0, 0, Sqrt[3], 0, 0, 2 (-1/(4 Sqrt[3]) + Sqrt[3]/4), > =A0 2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), -2, -(2 Sqrt[2])/3, 0, ( > =A0 2 Sqrt[2])/3, 0, 0}, {0, 0, 0, 0, Sqrt[3], 0, 0, > =A0 2 Sqrt[2/3], -(2 Sqrt[2])/3, -7/3, 0, 0, > =A0 2 (1/(3 Sqrt[2]) + (2 Sqrt[2])/3), Sqrt[10/3]}, {0, 0, 0, 0, 0, > =A0 2 Sqrt[2/3], 0, (2 Sqrt[2])/3, 0, 0, -16/3, > =A0 2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), 2 Sqrt[2/3], 0}, {0, 0, 0, 0, 0, > =A0 =A00, 2 Sqrt[2/3], 0, (2 Sqrt[2])/3, 0, > =A0 2 (-1/(4 Sqrt[3]) + (3 Sqrt[3])/4), -8/3, -(2 Sqrt[2])/3, 0}, {0, 0, > =A0 =A00, 0, 0, 0, 0, 0, 0, 2 (1/(3 Sqrt[2]) + (2 Sqrt[2])/3), > =A0 2 Sqrt[2/3], -(2 Sqrt[2])/3, 1/2, > =A0 2 (-Sqrt[5/3]/16 - Sqrt[15]/16)}, {0, 0, 0, 0, 0, 0, 0, 0, 0, Sqrt[ > =A0 10/3], 0, 0, 2 (-Sqrt[5/3]/16 - Sqrt[15]/16), 7/2}}
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- From: David Reiss <dbreiss@gmail.com>
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- Re: Mathematica 6 obtains imaginary eigenvalues for a Hermitian