Question

let matrix M =(1st row: 0.7, 0.2,0.2; 2nd row: 0,0.2,0.4; 3rd row: 0.3,0.6,0.4) find the only one steady state vector and what is M^infinitive. Please, help

Answer 1

\[\left[\begin{matrix}0.7 & 0.2 & 0.2 \\ 0 & 0.2 & 0.4\\0.3 & 0.6 & 0.4\end{matrix}\right]\]
find steady state vector and \\[M^\infty \]

Answer 2

is it something like solving
\[[x,y,z]\left[\begin{matrix}0.7 & 0.2 & 0.2 \\ 0 & 0.2 & 0.4\\0.3 & 0.6 & 0.4\end{matrix}\right]=[x,y,z]\]

??

Answer 3

no, it's stochastic matrix

Answer 4

eigenvectors and the steady state after infinitive changing

Answer 5

i guess i am no use then, but i thought the steady state vector was found by solving that equation

Answer 6

I know how to find them out but i confused between 2 concepts, one way, steady state comes from eigenvalue =1, and can stop there to take eigenvector and steady state vector from that. another way is finding out all of eigenvalue, find out eigenvectors respect to them, put everything in formula to get M^infinitive and that is steady state. I don't know when going this way, when going that way. too confused

Answer 7

you can diagonalize matrix M
\[ M = S \Lambda S^{-1} \]
where S's columns are the eigenvectors and Λ is a diagonal matrix with the eigenvalues on its diagonal.
\[ M M = S \Lambda S^{-1}S \Lambda S^{-1} = S \Lambda^2 S^{-1} \]
or in general
\[ M^n= S \Lambda^n S^{-1} \]

A diagonal matrix to the nth power is the values on its diagonal to the nth power
here, the first eigenvalue will be 1, and all others less than 1.
that means as n-> infinity, only the first eigenvalue remains.

\[ M^{\infty} \]will have the steady-state eigenvector in its columns
(scaled so that each column sums to 1)

Answer 8

so, when the question is finding out steady state, I just stop at eigenvalue 1 and have the eigenvector and steady state vector then. If the question is what is M^infinitive, I have to find all of them to put associate eigenvectors into the matrix, in that matrix, there will be a column vector associate with eigenvalue 1 and pick it out if the next question is find steady state, is it right?. I am sorry for my English. I am not native one.
You must spend a lot of time to figure out what it is to help me. Thanks a lot.

Answer 9

in other words, steady state is a small part of M^infinitive stage. I can skip if the question doesn't ask me to go that far and I cannot if the question ask me to go to M^infinitive. am I right?

Answer 10

If the question is what is M^infinitive, I have to find all of them ?

no, we need only the steady state eigenvector


A =
0.7000 0.2000 0.2000
0 0.2000 0.4000
0.3000 0.6000 0.4000

A^100 =
0.4000 0.4000 0.4000
0.2000 0.2000 0.2000
0.4000 0.4000 0.4000

the eigenvector for lambda= 1 is
v1=
0.6667
0.3333
0.6667

if we make it so the components add to 1 we get
v1/sum(v1)
0.4000
0.2000
0.4000

which is the column of A^100 (not quite infinity, but big enough)

Answer 11

hey, how can you get A^100 without calculate S ,S^-1, and that the process of taking all of them. is it right?

Answer 12

I used matlab

Answer 13

LOL....LOL...

Answer 14

and it probably used eigenstuff to do the computation

Answer 15

I know, but just for now only, not for my test day. I have no computer then.

Answer 16

I just used it to show that the steady state eigenvector makes up the columns of M^infty

Answer 17

Ok, I got you now. Thanks a Ton