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MathGroup Archive 2004

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Eigenvalues and eigenvectors of a matrix with nonpolynomial elements.

  • To: mathgroup at smc.vnet.net
  • Subject: [mg51284] Eigenvalues and eigenvectors of a matrix with nonpolynomial elements.
  • From: "Goyder Dr HGD" <h.g.d.goyder at cranfield.ac.uk>
  • Date: Tue, 12 Oct 2004 01:57:47 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

I need to find the eigenvalues and eigenvectors of matrices where the elements depend on a variable, k, in a nonpolynomial manner. Thus, according to my (limited) knowledge of eigensystems, the eigenvalues are the values of k that make the determinant zero and there should be an eigenvector associated with each eigenvalue. A simple warm-up example is given below; my actual cases will be more complicated. I wish to know if the method I have put together below, using engineering rather than maths, is suitable and what accuracy I can expect. 

First, here is the warm-up problem; the matrix is called mat and depends on the variable k. I start by working out the determinant.

In[39]:=
mat = {{-1, 0, -1, 0}, {-Cos[k], -Sin[k], -Cosh[k], -Sinh[k]}, {1, 0, -1, 0}, 
    {Sin[k], -Cos[k], -Sinh[k], -Cosh[k]}}; 

In[40]:=
d = Det[mat]

Out[40]=
2*Cosh[k]*Sin[k] - 2*Cos[k]*Sinh[k]

Second, I find one of the roots. There are an infinite number of roots and Ted Ersek's method for finding all the roots in an interval might help. I assume that each root is an eigenvalue: is this always true?

In[41]:=
rt = FindRoot[d, {k, 6, 8}]

Out[41]=
{k -> 7.068582745628732}

Third, I attempt to get the eigenvector. It has been necessary to "go numeric" and so I need a method that will cope with numerical issues. I try singular value decomposition because I feel that the information is in there somewhere.

In[43]:=
{u, w, v} = SingularValueDecomposition[mat /. rt]; 

Examination of the singular values indicates that the last one is zero. I assume that this means that this singular value is associated with a zero determinant. I select the column of v associated with the zero singular value and hope that this is the eigenvector.

In[46]:=
vec = Transpose[v][[-1]]; 

A check shows that the selected vector  when multiplied by the matrix evaluated at the eigenvalue gives approximately zero: this looks hopeful. Am I on the correct track? 

In[47]:=
(mat /. rt) . vec

Out[47]=
{3.883820346708995*^-17, 1.2588519636425044*^-13, -1.1055045534287126*^-16, 1.255173265671261*^-13}

Are there better methods? In general how do I set up the number of figures to give me a target accuracy for the eigenvalue and vector?

Thanks
Hugh Goyder


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