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