1. ## Equilibrium

I don't understand how the following equation is done:

Given is a matrix that represents how coins are circulating each day:
P: 60% stay at P, 10% go to Q, 30% go to R
Q: 10% go to P, 80% stay at Q, 10% go to R
R: 10% go to P, 20% go to Q, 70% stay at R

C: ((0.6,0.1,0.3),(0.1,0.8,0.1),(0.1,0.2,0.7))

Calculate the equilibrium division, as the number of days reaches infinity
I understand how to calculate the characteristic polynomial, which is
-λ^3+2.1λ^2-1.4λ+0.3
I understand how to calculate the eigenvalues and -vectors, which are
λ = 1 : (4,9,7)
λ = 0.6 : (0,-1,1)
λ = 0.6 : (-1,-1,2)
I understand how to calculate the transition matrix and the inverse (which is simply the eigenvectors)
T = (((1/20),(25/20),(-16/20)),((1/20),(-15/20),(4/20)),((1/20),(5/20),(4/20)))
T^-1 = ((4,9,7),(0,-1,1),(-1,-1,2))
But now comes the part I do not understand:
lim C^n = T^-1 * ((1,0,0),(0,0,0),(0,0,0)) * T
n ->
I get how the matrix is formed (because when you near infinity, all values below 1 get reduced to 0). But when you multiply by this matrix, you will get a matrix like:
((a,b,c),(0,0,0),(0,0,0)).

(((4/20),(9/20),(7/20)),((4/20),(9/20),(7/20)),((4/20),(9/20),(7/20)))
I dont understand how they can multiply by that matrix and still get that answer.
Any help would be much appreciated.

2. Originally Posted by viva
I don't understand how the following equation is done:

I understand how to calculate the characteristic polynomial, which is

I understand how to calculate the eigenvalues and -vectors, which are

I understand how to calculate the transition matrix and the inverse (which is simply the eigenvectors)

But now comes the part I do not understand:

I get how the matrix is formed (because when you near infinity, all values below 1 get reduced to 0). But when you multiply by this matrix, you will get a matrix like:
((a,b,c),(0,0,0),(0,0,0)).

I dont understand how they can multiply by that matrix and still get that answer.
Any help would be much appreciated.

If A is the transition matrix and x the equilibrium distribution then x=Ax or, depending how this has been defined by your course x=xA.

So the equilibrium distribution is an eigen vector of the transition matrix corresponding to the eigen value 1, normalised so that the sum of its components is 1.

CB

3. I'm confused. So they don't calculate anything using this?
lim C^n = T^-1 * ((1,0,0),(0,0,0),(0,0,0)) * T
n ->
You just take the eigenvector of eigenvalue 1 and normalise it?
Then why go through all the hassle of getting T and T^-1?

4. The reason you have to fully diagonalize the matrix C is that you're trying to find an arbitrary power of C. That's difficult to do without diagonalizing it. All the eigenvalue and eigenvector stuff is there to allow you to compute the arbitrary power of C. Once you have that, you can take the limit, and then re-multiply by T and T inverse in order to get the final result.

[EDIT]: See CB's post below for a correction.

5. I still think I don't fully get it, so I made a practice question:

I know how to do a, b and c. But not d and e. This is my answer, please tell me where I go wrong:

6. Originally Posted by Ackbeet
The reason you have to fully diagonalize the matrix C is that you're trying to find an arbitrary power of C. That's difficult to do without diagonalizing it. All the eigenvalue and eigenvector stuff is there to allow you to compute the arbitrary power of C. Once you have that, you can take the limit, and then re-multiply by T and T inverse in order to get the final result.
I'm sorry but if the intention is to find the equilibrium solution then you do not need to find the limit of the n-th power of the transition matrix as n goes to infinity. If (one or more) equilibrium solutions exist then they are eigenvectors of the transition matrix corresponding to eigenvalue 1.

The only problem would be if the eigenspace of the eigenvalue 1 was of dimension greater than 1, in which case you would need to determine which eigenvector corresponds to the initial distribution, for which one of the most obvious method would be to compute the limit of the powers of the transition matrix etc. That is not the case here the multiplicity of the eigenvalue 1 is 1 and its eigenvector is unique up to an arbitrary multiple which is resolved as we want an eigenvector who's elements sum to 1 (since this is the case of the initial state and probability is conserved in this problem).

Now read the question, it asks nothing of us other than to calculate the equilibrium distribution. So doing stuff other than computing eigenvalues and vectors and making the appropriate observations about the behaviour of the process might be nice but gets you no extra marks for a lot of extra work, and indicates a lack of understanding of what is really going on.

CB

7. Originally Posted by CaptainBlack
I'm sorry but if the intention is to find the equilibrium solution then you do not need to find the limit of the n-th power of the transition matrix as n goes to infinity. If (one or more) equilibrium solutions exist then they are eigenvectors of the transition matrix corresponding to eigenvalue 1.

The only problem would be if the eigenspace of the eigenvalue 1 was of dimension greater than 1, in which case you would need to determine which eigenvector corresponds to the initial distribution, for which one of the most obvious method would be to compute the limit of the powers of the transition matrix etc. That is not the case here the multiplicity of the eigenvalue 1 is 1 and its eigenvector is unique up to an arbitrary multiple which is resolved as we want an eigenvector who's elements sum to 1 (since this is the case of the initial state and probability is conserved in this problem).

Now read the question, it asks nothing of us other than to calculate the equilibrium distribution. So doing stuff other than computing eigenvalues and vectors and making the appropriate observations about the behaviour of the process might be nice but gets you no extra marks for a lot of extra work, and indicates a lack of understanding of what is really going on.