Question

Please, help. find the steady state vector for the Matrix (1st row: 0.2, 0.3, 0.5; 2nd row: 0.3, 0.6, 0.1; 3rd row : 0.5, 0.1, 0.4)

Answer 1

\[\left[\begin{matrix}0.2 & 0.3 & 0.5 \\ 0.3 & 0.6 & 0.1\\0.5 & 0.1 & 0.4\end{matrix}\right]\] I stuck at factor the characteristic equation of lamda
I got \[\lambda^3 - 12\lambda^2+9\lambda+110 =0\]
and \[(\lambda-10)(\lambda^2 -2\lambda-11)\]
Anyone helps me check whether mine is right or wrong and if it's right, why cannot I factor to find the other eigenvalues exceed 10?

Answer 2

you should get one of the lambdas= 1
so it looks like you have the wrong characteristic equation.

Answer 3

I follow the formula to take lamda, \[\lambda^3 - (trace)\lambda^2 + (A11+A22+A33)\lambda- \det (A)\]

Answer 4

my pro claims that the matrix is symmetry but I don't know whether it helps or not

Answer 5

I used some online tools, but this looks like the characteristic equation:
-x^3 + 6/5x^2 - 9/100x - 11/100, which factors as:
http://www.wolframalpha.com/input/?i=factor%28+-x%5E3+%2B+6%2F5x%5E2+-+9%2F100x+-+11%2F100%29.

Answer 6

-1/100 (-1+x) (-11-20 x+100 x^2)

Answer 7

jump is getting a lambda=1 which is a good sign.

Answer 8

I'll be back in 30 minutes sorry everybody, it's emergency situation , forgive me for that

Answer 9

although the original matrix is Markov, we can take 10 times to get the new matrix at integers and find out the eigenvectors. those are same.

Answer 10

Hello, Hoa - the fraction in that characteristic polynomial is unimportant, it's just a constant factored out: (-1+x) (-11-20 x+100 x^2). I don't think you can scale a matrix before taking the determinant: since det(xA-lambdaI) is not equal to det(a-lambdaI).

Answer 11

so, my new one is\[\left[\begin{matrix}2 & 3 & 5 \\ 3 & 6 &1\\5 & 1 & 4\end{matrix}\right]\] and base on this I make lamda equation, eigenvalue is 1/10 of the new one

Answer 12

@jumping_mouse . sure. the det (A -lamda I) != det (A) but the way to get the former is what I did.

Answer 13

Okay, I see what you are up to. You have a lambda = 10, so that value is your lambda = 1, right? And the associated eigenvector is (1,1,1), right?

Answer 14

yes.

Answer 15

but the eigenvector (1,1,1) should come from the eigenvalue which is NOT 1

Answer 16

@jumping_mouse take a look at http://www.youtube.com/watch?v=Sf91gDhVZWU. the pro teaches me how to get characteristic equation from the original A

Answer 17

you should be getting
x^3 - 12*x^2 + 9*x + 110=0
for your char equat.

Answer 18

yes, exactly what I have

Answer 19

for you "new" matrix. The x-10 corresponds to lambda=1 in the original problem

Answer 20

yes

Answer 21

your "new" lambdas are 10, x= -2sqrt(3) + 1, x = 2sqrt(3) + 1
the lambdas for the original are 1/10 of these

Answer 22

you mean the leftover is solved by x = (-b +-sqr(b^2 -4ac)) / 2a? formula of quadratic equation?

Answer 23

you are solving
\[ (\lambda-10)(\lambda^2 -2\lambda-11) = 0 \]
which means
\[ \lambda-10=0 \text{ or } \lambda^2 -2\lambda-11=0\]

Answer 24

I know the steady state comes from the eigenvalue 1. but how to solve for that?

Answer 25

as you prove, the other eigenvalue do not give out the steady state because they are !=1

Answer 26

use the eigenvalue e.g. 10
and subtract it from the main diagonal of the "new" matrix
now find the null space of the resultant matrix

for other eigenvalues, do the same thing: subtract 2 sqrt(3)+1 from the main diagonal, and use elimination to find the null space. This will be the eigenvector.

Answer 27

let me try and tell you my result. Can I reduced to row echelon form to get easier

Answer 28

Can I reduced to row echelon form to get easier. Yes that is exactly what you do

Answer 29

I did not have the result as my pro's one. I get vector steady is (409,25,23) while his is (1/3,1/3,1/3) :(

Answer 30

did you row reduce
-8 3 5
3 -4 1
5 1 -6
?

Answer 31

R2+R3+R1 -->R1 get 0, 0, 0
R2 stay there 3 , -4, 1
2R2 -R3 get R3 1, -9, 8

Answer 32

R1 replace R3 1, -9, 8
1/23(-3 R3 + R2) 0, 1, 1
0, 0 0

Answer 33

oh, oh, I got it now. thanks a TON Phi

Answer 34

no, just 1,1,1 not 1/3

Answer 35

yes, the eigenvector can be scaled by anything. so 1/3 1/3 1/3 is the same as 1 1 1

for the steady state of your problem you need more information, some initial condition
(such as the states sum to 1) to pick out 1/3 1/3 1/3

Answer 36

oh, yes, I know what it means. since we have eigenvector is 1, 1, 1 the steady state is (1/ 1+1+1, 1/1+1+1 and 1/1+1+1) it means )1/3,1/3,1/3) my god. eigenvector of eigenvalue 1 gives out steady state by that formula . hiiiiiiihi! thanks a lot

Answer 37

Why, I don't deserve medal , phi/

Answer 38

I stuck everywhere. If I don't have you guide, how can I get that result?

Answer 39

anyway, I'm happy now, very very happy:)