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#1 2014-02-19 17:46:04

soni85
Member
Registered: 2014-02-19
Posts: 6

solve the matrix

hi, i have this equation


where,
I- identity matrix. (n*n)
Q-(n*n) matrix

so,

If u=2 and xp=1;
then,


(G(1) - (n*1) matrix ,depends on 'u'; It has 0.5 in u-1 and u+1 th position. (count starts from 0); P(xp)  - (1*n) matrix ,depends on xp ; It has 0.5 in xp-1 and xp+1 th position.)

If xp < u then the E(xp) is constant, if xp > u then E(xp) is linearly decreasing. But is it possible to get a final simple equation with notations u , xp and n(matrix size)?

Last edited by soni85 (2014-02-19 21:32:24)

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#2 2014-02-19 20:10:03

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,364

Re: solve the matrix

Hi;

Welcome to the forum.

What is E(xp)? Do you have an example?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#3 2014-02-19 20:58:12

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,380

Re: solve the matrix

hi soni85

Welcome to the forum.

I- identity matrix. (n*n)   AND Q-(n*n) matrix  IMPLIES inverse is n*n
P-(1*n) matrix ...
G(1)-(1*n) matrix, ....

Did you mean that G is n*1 ?

Sizes:  (1*n)(n*n)(1*n)  ???

I assume that Q and G are fully known, and that P and X are not (other than the O.5s)

So

will result in a known matrix n*1 let's say R

The problem is then

If, say, n = 5, then these are too many variables to produce a unique solution.

And where did 'u' come from ?

I suspect I'm misunderstanding your problem.  Would you please post an example with specific values to make this clearer ?  Thanks.  smile

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#4 2014-02-19 21:02:52

soni85
Member
Registered: 2014-02-19
Posts: 6

Re: solve the matrix


let say u=2 and xp=1;
then

   will give a column matrix with first three elements constant and the rest will be in linear form.


Q will be like this,


G(1) depends on 'u'; It has 0.5 in u-1 and u+1 th position. (count starts from 0);
P(xp)  depends on xp ; It has 0.5 in xp-1 and xp+1 th position.

Last edited by soni85 (2014-02-20 00:03:22)

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#5 2014-02-19 21:06:14

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,380

Re: solve the matrix

Sorry, I'm still confused about this.

Please post the entire equation, not just a bit.  And still 'u' has appeared out of nowhere ???

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#6 2014-02-19 21:37:22

soni85
Member
Registered: 2014-02-19
Posts: 6

Re: solve the matrix

I've modified my question. Hope its clear now.

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#7 2014-02-19 23:42:11

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,380

Re: solve the matrix

Have a look at this:

http://www.mathsisfun.com/algebra/matri … lying.html

Particularly look at the section on what size of matrix may be multiplied by what size of matrix.

There is also a matrix calculator on MIF.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#8 2014-02-20 00:04:26

soni85
Member
Registered: 2014-02-19
Posts: 6

Re: solve the matrix

sorry about that... i was confused while typing. I've changed it now.

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#9 2014-02-20 00:32:47

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,380

Re: solve the matrix

I've changed G to a 4 by 1 column matrix.

So we have

xp = 1 and u = 2, determining the position of the first 0.5 in the row or column,  and


E will be a 1 by 1 matrix.

You want a formula for E that depends on u and xp.  Is that right ?

if xp > u then E(xp) is linearly decreasing.

What does this mean?  Surely E is determined as number whatever u and xp are ?
Bob

Last edited by bob bundy (2014-02-20 01:06:09)


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#10 2014-02-20 17:17:01

soni85
Member
Registered: 2014-02-19
Posts: 6

Re: solve the matrix

Yes thats right.
If xp > u then E(xp) linearly decreases as xp increases.

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#11 2014-02-20 20:41:26

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,380

Re: solve the matrix

I am going to replace 'xp' by 'v' to make the notation easier to follow. And I will replace  (I-Q)-¹ by a single matrix R

The value of v picks out the values of R from row v and from row v+2, adds them and multiplies by 0.5

The value of u picks out column u and column u+2, adds them and multiplies by 0.5.

Everything else is a zero, so just 4 elements of R are picked out.

The result is

where 

is the element in the ith row and jth column of R

I have no idea what to do when xp>u as you have not said the exact linear dependency rule.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#12 2014-02-21 01:43:25

soni85
Member
Registered: 2014-02-19
Posts: 6

Re: solve the matrix

thanks for your effort. Yes finally E is based on four values. So if i change v i'll get a different E value for the same u. What i need is an another simpler equation for E, not with all these matrices. I know it from simulations that  E=2(n-v) if v > u and E=2(n-u) if v < u. But is it possible to get these results from the equation given in the question?  I saw a paper with similar problem. They've solved it using recurrence relations.
Title: "the random walk between a reflecting and an absorbing barrier". (I couldn't use the url)

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#13 2014-02-21 04:41:01

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,380

Re: solve the matrix

hi soni85

I had a look at that link and couldn't see any connection with what you've been asking about.  What are your simulations about?

Please go back to the beginning and describe what this is about and how you arrived at the matrix equation.  There may be other members who will recognise this problem then.  smile

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#14 2014-02-21 05:04:40

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,364

Re: solve the matrix

I saw a paper with similar problem. They've solved it using recurrence relations.

It does not look like your Q came from an absorbing chain. For one thing it is not stochastic. If you derived it from a random walk please provide the original problem.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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