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

*To*: mathgroup at smc.vnet.net*Subject*: [mg51389] Re: [mg51315] Re: [mg51284] Eigenvalues and eigenvectors of a matrix with nonpolynomial elements.*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Fri, 15 Oct 2004 02:48:26 -0400 (EDT)*References*: <200410141035.GAA14868@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Goyder Dr HGD wrote: > Goyder Dr HGD 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. > > > Daniel Lichtblau wrote: > > >>No, eigenvalues for a matrix mat are values lambda for which > > mat - lambda*IndentityMatrix[...] > is singular (that is, has zero determinant. In your example these would > be functions of k. In simplified form I get > > Out[8]//InputForm= > {-1 + I, -1 - I, (-Cosh[k] - Sin[k] - > Sqrt[Cosh[k]^2 - 2*Cosh[k]*Sin[k] + Sin[k]^2 + 4*Cos[k]*Sinh[k]])/2, > (-Cosh[k] - Sin[k] + Sqrt[Cosh[k]^2 - 2*Cosh[k]*Sin[k] + Sin[k]^2 + > 4*Cos[k]*Sinh[k]])/2} > > Daniel Lichtblau > Wolfram Research > > Thank you for your fast response. I apologies for not making myself clear. The warm up example comes from a differential equation eigenvalue problem rather than a simple matrix problem. The differential equation is given below and results in the matrix that I gave previously. Thus starting from the beginning we have a differential equation and boundary conditions which will only be satisfied for certain values of, k, which I take to be the eigenvalues. The problem is then to find these eigenvalues and the associated eigensolutions. (My actual problem involves sets of differential equations of this form.) The warm-up differential equation is > > In[103]:= > DSolve[Derivative[4][y][x] - k^4*y[x] == 0, y[x], x] > > Out[103]= > {{y[x] -> C[2]/E^(k*x) + E^(k*x)*C[4] + C[1]*Cos[k*x] + C[3]*Sin[k*x]}} > > I have translated this solution to the equivalent form > > In[104]:= > e1 = A1*Cos[k*x] + A2*Sin[k*x] + A3*Cosh[k*x] + A4*Sinh[k*x]; > > The boundary conditions are > > In[107]:= > bc = {0 == (e1 /. x -> 0), 0 == (1/k^2)*(D[e1, {x, 2}] /. x -> 0), 0 == (e1 /. x -> 1), > 0 == (1/k)*(D[e1, x] /. x -> 1)}; > > Which give rise to a set of linear equations whose coefficient matrix may be found from > > In[108]:= > e2 = Normal[CoefficientArrays[bc, {A1, A2, A3, A4}]] > > Out[108]= > {{0, 0, 0, 0}, {{-1, 0, -1, 0}, {1, 0, -1, 0}, {-Cos[k], -Sin[k], -Cosh[k], -Sinh[k]}, > {Sin[k], -Cos[k], -Sinh[k], -Cosh[k]}}} > > The values of k that satisfy this equation I am calling eigenvalues. Each value of {A1, A2, A3, A4} which is associated with each eigenvalue I am calling an eigenvector. We thus come to the matrix which I defined previously from which my eigenvalues and vectors must be extracted. > > In[109]:= > e3 = e2[[2]] > > Out[109]= > {{-1, 0, -1, 0}, {1, 0, -1, 0}, {-Cos[k], -Sin[k], -Cosh[k], -Sinh[k]}, > {Sin[k], -Cos[k], -Sinh[k], -Cosh[k]}} > Okay. I had considered that you might be doing some sort of Sturm-Liouville problem, but the setup looked too much like a standard linear algebra eigensystem. So we begin anew with the matrix and its determinant. mat = {{-1, 0, -1, 0}, {1, 0, -1, 0}, {-Cos[k], -Sin[k], -Cosh[k], -Sinh[k]}, {Sin[k], -Cos[k], -Sinh[k], -Cosh[k]}}; det = Det[mat]; For the task at hand you can use NullSpace numerically after first extracting the root in k. In[130]:= root = FindRoot[det==0, {k,6,8}] Out[130]= {k -> 7.06858} In[132]:= Chop[NullSpace[mat /. root]] Out[132]= {{0, -0.999999, 0, 0.00120412}} But there is a symbolic approach as well. We set up the symbolic expression mat.vec==0 with vec a vector of indeterminates. We algebraicize by converting trigs and hyperbolics to ordinary variables and adding subsidiary defining relations e.g. Sin[k]->sk, Cos[k]->ck, and we have sk^2+ck^2==1. Variations on this idea involve different algebraic substitutions. For example we could convert to exponentials, or use the standard rational parametrization of the trig and hyperbolic functions. vars = Array[x,Length[mat]]; subs = {Cos[k]->ck,Sin[k]->sk,Cosh[k]->chk,Sinh[k]->shk}; polys1 = mat . vars /. subs; polys2 = {ck^2+sk^2-1, chk^2-shk^2-1}; detpoly = Det[mat]/.subs; We could "normalize" the eigenvector with a relation that the sum of squares of its components must be unity. As it will lead to a simpler result, we instead insist that a particular component be unity. This must be done carefully as it is is not really a "generic" approach. So we use the earlier numeric computation to choose one of the components, the second one, say, that did not vanish. In[144]:= InputForm[polys = Join[polys1,polys2,{x[2]-1,detpoly}]] Out[144]//InputForm= {-x[1] - x[3], x[1] - x[3], -(ck*x[1]) - sk*x[2] - chk*x[3] - shk*x[4], sk*x[1] - ck*x[2] - shk*x[3] - chk*x[4], -1 + ck^2 + sk^2, -1 + chk^2 - shk^2, -1 + x[2], 2*ck*shk - 2*chk*sk} In[146]:= InputForm[soln = vars /. Solve[polys==0, vars]] Out[146]//InputForm= {{0, 1, 0, -(chk*ck) + shk*sk}} Daniel Lichtblau Wolfram Research

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