Re: Eigenvalues and eigenvectors of a matrix with nonpolynomial elements.

*To*: mathgroup at smc.vnet.net*Subject*: [mg51313] Re: [mg51284] Eigenvalues and eigenvectors of a matrix with nonpolynomial elements.*From*: DrBob <drbob at bigfoot.com>*Date*: Thu, 14 Oct 2004 06:35:50 -0400 (EDT)*References*: <200410120557.BAA19223@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

Why not let Mathematica do the work? 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]}}; mat // MatrixForm Eigensystem@mat (voluminous output omitted) Your use of FindRoot below didn't find the eigenvalues of a single matrix; it found the values of k that make the matrix singular. At each value of k (whether it's one of those roots or not), the matrix has a different Eigensystem. Bobby On Tue, 12 Oct 2004 01:57:47 -0400 (EDT), Goyder Dr HGD <h.g.d.goyder at cranfield.ac.uk> wrote: > 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 > > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Eigenvalues and eigenvectors of a matrix with nonpolynomial elements.***From:*"Goyder Dr HGD" <h.g.d.goyder@cranfield.ac.uk>