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Re: Determining the root of the characteristic equation

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  • Subject: [mg113470] Re: Determining the root of the characteristic equation
  • From: Christofer Bogaso <bogaso.christofer at>
  • 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

On Fri, Oct 29, 2010 at 9:09 PM, Daniel Lichtblau <danl at> 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

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