Re: Determining the root of the characteristic equation for time
- To: mathgroup at smc.vnet.net
- Subject: [mg113472] Re: Determining the root of the characteristic equation for time
- From: "Nasser M. Abbasi" <nma at 12000.org>
- Date: Sat, 30 Oct 2010 04:37:14 -0400 (EDT)
- References: <iae7h3$7g9$1@smc.vnet.net> <4CCAA5F6.1000808@12000.org> <AANLkTimFd7SjjZos6btMCK3Zwfy-UfVwkS0YeneQDsZ6@mail.gmail.com>
- Reply-to: nma at 12000.org
On 10/29/2010 8:06 AM, Christofer Bogaso wrote:
> Dear Nasser,
>
> Thank you so much for your help. Here I was trying with my own example:
>
> Alpha1 = {{0.07}, {0.17}};
> Beta1 = {{1, -4}};
> Gamma11 = {{0.24, -0.08}, {0, -0.31}};
> Gamma12 = {{0, -0.13}, {0, -0.37}};
> Gamma13 = {{0.20, -0.06}, {0, -0.34}};
> PI1 = Alpha1.Beta1;
> A1 = IdentityMatrix[2] + PI1 + Gamma11;
> A2 = Gamma12 - Gamma11;
> A3 = Gamma13 - Gamma12;
> A4 = -1*Gamma13;
> Clear[z];
> Abs[Solve[Det[IdentityMatrix[2] - A1*z - A2*(z^2) - A2*(z^3) - A2*(z^4)]
> ==0, z]]
>
> While I executing the last line, I got strage figures (please see the
> attachment on my workings):
>
May be changing the last line above with the following is what you meant
to do?
Abs[z/.Solve[Det[IdentityMatrix[2]-A1*z-A2*(z^2)-A2*(z^3)-A2*(z^4)]==0,z]]
> \!\({{Abs[z -> Root[\(-1.`\) + 1.63`\ #1 +
> 0.31`\ -0\ #1 - 0.7068`\ #1\^2 - 0.3325`\ -0\ #1\^2 - \
> 0.1632`\ #1\^3 + 0.0643`\ -0\ #1\^3 - 0.1632`\ #1\^4 +
> 0.07869999999999999`\ -0\ #1\^4 + 0.07679999999999998`\
> #1\^5 \
> + 0.033099999999999984`\ -0\ #1\^5 + 0.0432`\ -0\ #1\^6 + 0.0288`\ -0\ #1\^7
> \
> + 0.0144`\ -0\ #1\^8&, 1]]}, {Abs[z -> Root[\(-1.`\) + 1.63`\ #1 + 0.31`\
> -0\
> \ #1 - 0.7068`\ #1\^2 - 0.3325`\ -0\ #1\^2 - 0.1632`\ #1\^3 + 0.0643`\ -0\
> #1\
> \^3 - 0.1632`\ #1\^4 + 0.07869999999999999`\ -0\ #1\^4 +
> 0.07679999999999998`\ #1\^5 + 0.033099999999999984`\ -0\ \
> #1\^5 + 0.0432`\ -0\ #1\^6 + 0.0288`\ -0\ #1\^7 + 0.0144`\ -0\ #1\^8&,
> 2]]}, {Abs[
> z -> Root[\(-1.`\) + 1.63`\ #1 +
> 0.31`\ -0\ #1 - 0.7068`\ #1\^2 - 0.3325`\ -0\ #1\^2 - \
> 0.1632`\ #1\^3 + 0.0643`\ -0\ #1\^3 - 0.1632`\ #1\^4 +
> 0.07869999999999999`\ -0\ #1\^4 + 0.07679999999999998`\
> #1\^5 \
> + 0.033099999999999984`\ -0\ #1\^5 + 0.0432`\ -0\ #1\^6 + 0.0288`\ -0\ #1\^7
> \
> + 0.0144`\ -0\ #1\^8&, 3]]}, {Abs[z -> Root[\(-1.`\) + 1.63`\ #1 + 0.31`\
> -0\
> \ #1 - 0.7068`\ #1\^2 - 0.3325`\ -0\ #1\^2 - 0.1632`\ #1\^3 +
> 0.0643`\ -0\ #1\^3 - 0.1632`\ #1\^4 +
> 0.07869999999999999`\ -0\ #1\^4 + 0.07679999999999998`\
> #1\^5 \
> + 0.033099999999999984`\ -0\ #1\^5 + 0.0432`\ -0\ #1\^6 + 0.0288`\ -0\ #1\^7
> \
> + 0.0144`\ -0\ #1\^8&, 4]]}, {Abs[z -> Root[\(-1.`\) + 1.63`\ #1 + 0.31`\
> -0\
> \ #1 - 0.7068`\ #1\^2 - 0.3325`\ -0\ #1\^2 - 0.1632`\ #1\^3 + 0.0643`\ -0\
> #1\
> \^3 - 0.1632`\ #1\^4 +
> 0.07869999999999999`\ -0\ #1\^4 + 0.07679999999999998`\
> #1\^5 \
> + 0.033099999999999984`\ -0\ #1\^5 + 0.0432`\
> -0\ #1\^6 + 0.0288`\
> -0\ #1\^7 + 0.0144`\ -0\ #1\^8&, 5]]}, {Abs[
> z -> Root[\(-1.`\) + 1.63`\ #1 + 0.31`\ -0\ #1 - 0.7068`\ \
> #1\^2 - 0.3325`\ -0\ #1\^2 - 0.1632`\ #1\^3 +
> 0.0643`\ -0\ #1\^3 - 0.1632`\ #1\^4 +
> 0.07869999999999999`\ -0\ #1\^4 + 0.07679999999999998`\
> #1\^5 \
> + 0.033099999999999984`\ -0\ #1\^5 + 0.0432`\ -0\ #1\^6 + 0.0288`\ -0\ #1\^7
> \
> + 0.0144`\ -0\ #1\^8&,
> 6]]}, {Abs[z -> Root[\(-1.`\) + 1.63`\ #1 + 0.31`\ -0\ #1 -
> \
> 0.7068`\ #1\^2 - 0.3325`\ -0\ #1\^2 - 0.1632`\ #1\^3 + 0.0643`\ -0\ #1\^3 -
> \
> 0.1632`\ #1\^4 +
> 0.07869999999999999`\ -0\ #1\^4 + 0.07679999999999998`\
> #1\^5 \
> + 0.033099999999999984`\ -0\ #1\^5 + 0.0432`\ -0\ #1\^6 + 0.0288`\
> -0\ #1\^7 + 0.0144`\ -0\ #1\^8&, 7]]}, {Abs[z -> \
> Root[\(-1.`\) + 1.63`\ #1 + 0.31`\ -0\ #1 - 0.7068`\ #1\^2 -
> 0.3325`\ -0\ #1\^2 - 0.1632`\ #1\^3 +
> 0.0643`\ -0\ #1\^3 - 0.1632`\ #1\^4 + 0.07869999999999999`\
> -0\ #1\^4 + 0.07679999999999998`\ #1\^5 + 0.033099999999999984`\ -0\ \
> #1\^5 + 0.0432`\ -0\ #1\^6 + 0.0288`\ -0\ #1\^7 + 0.0144`\ -0\ #1\^8&,
> 8]]}}\)
>
> My intention is to get the absolute values for all root including complex
> root. I am expecting exactly one root must be "1".
>
> I am sorry as my question looks very simple. However forgive me as I am very
> new on Mathematica and trying to switch from R.
>
> Thanks and regards,
>
> On Fri, Oct 29, 2010 at 4:16 PM, Nasser M. Abbasi<nma at 12000.org> wrote:
>> 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
>>
>
--Nasser