Re: Determining the root of the characteristic equation
- To: mathgroup at smc.vnet.net
- Subject: [mg113470] Re: Determining the root of the characteristic equation
- From: Christofer Bogaso <bogaso.christofer at gmail.com>
- Date: Sat, 30 Oct 2010 04:36:52 -0400 (EDT)
Thanks Daniel, as you said, using NSolve[] function I got the roots of the
characteristic equation. This is great!!! However my job is not over here as
I need to calculate the absolute value (i.e. modulus) of all roots.
I have tried with *Abs[NSolve....* function. However I could not get the
numbers at all. Would you please guide me what is the correct way to do
that?
Thanks,
On Fri, Oct 29, 2010 at 9:09 PM, Daniel Lichtblau <danl at wolfram.com> wrote:
> 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,
>>
>
> Just solve for det(matrix) = 0.
>
> Example:
> n = 5;
> a1 = RandomInteger[{-10,10}, {n,n}];
> a2 = RandomInteger[{-10,10}, {n,n}];
> mat = IdentityMatrix[n] - a1*z - a2*z^2;
>
> Here we solve for that determinant being zero.
>
> Solve[Det[mat]==0, z]
>
> Could instead use NSolve if you just want numeric solutions.
>
> Daniel Lichtblau
> Wolfram Research
>