solve the matrix

soni85 · 2014-02-19 17:46:04

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)

bobbym · 2014-02-19 20:10:03

Hi;

Welcome to the forum.

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

Bob · 2014-02-19 20:58:12

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.

Bob

soni85 · 2014-02-19 21:02:52

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)

Bob · 2014-02-19 21:06:14

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

soni85 · 2014-02-19 21:37:22

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

Bob · 2014-02-19 23:42:11

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

soni85 · 2014-02-20 00:04:26

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

Bob · 2014-02-20 00:32:47

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 (2014-02-20 01:06:09)

soni85 · 2014-02-20 17:17:01

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

Bob · 2014-02-20 20:41:26

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

soni85 · 2014-02-21 01:43:25

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)

Bob · 2014-02-21 04:41:01

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.

Bob

bobbym · 2014-02-21 05:04:40

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.

Math Is Fun Forum

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

solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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

Re: solve the matrix

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