Re: Determining the root of the characteristic equation for time

*To*: mathgroup at smc.vnet.net*Subject*: [mg113464] Re: Determining the root of the characteristic equation for time*From*: "Nasser M. Abbasi" <nma at 12000.org>*Date*: Sat, 30 Oct 2010 04:35:44 -0400 (EDT)*References*: <iae7h3$7g9$1@smc.vnet.net>*Reply-to*: nma at 12000.org

On 10/29/2010 3:26 AM, Christofer Bogaso wrote: > Hello all, > > I am a new user of Mathematica and would appreciate if somebody help > me with some hints how can I use Mathematica to solve a typical > problem. > > Suppose I have a matrix polynomial for "z" (which is a scalar) like: > > Determinant of (Identity_matrix[5] - A1*z - A2*(z^2)) = 0 > > Here A1 and A2 are 2 separate given square matrices of length 5. > > I have faced this problem while solving the characteristic polynomial > of a multivariate time series model. I am aware of the capability of > Mathematica for it's symbolic computation. > > It would really be helpful if somebody helps me how to do that using > Mathematica. > > Thanks and regards, > may be something like this: In[59]:= Clear[z]; n=5; A1=Table[RandomReal[1],{i,5},{j,5}]; A2=Table[RandomReal[1],{i,5},{j,5}]; Solve[Det[IdentityMatrix[n]-A1*z-A2*(z^2)]==0,z] Out[63]= {{z->-2.4021738242972597},{z->-1.340898151721915},{z->-0.8241488489166163-0.2913103890152045 I},{z->-0.8241488489166163+0.2913103890152045 I},{z->0.2868840290088525-1.8095777731669567 I},{z->0.2868840290088525+1.8095777731669567 I},{z->0.30173526317565597},{z->1.0027788027642381-0.527105585145346 I},{z->1.0027788027642381+0.527105585145346 I},{z->2.005527612079525}} --Nasser