I have a markov chain matrix with probabilities as such, on finding the steady state..
(x,y,z) * [0.9 0.1 0] = (x,y,z)
[0 0.9 0.1]
[.05 .05 .9]
Knowing that x + y + z = 1
How can i know the values of x,y and z ?
I came up with these 3 equations but got lost :
0.9x + 0y + 0.05z = x
0.1x + 0.9y + 0.05z = y
0x + 0.1y + 0.9z = z
Could be simple substitution and elimination maths but i can't succeed in finding the values.
Let $A=\begin{bmatrix} 0.9 & 0.1 & 0 \\ 0 & 0.9 & 0.1 \\ 0.05 & 0.05 & 0.9 \end{bmatrix}$. You want to find some non zero vector $v$ such that $v^T A = v^T$.
in other words, you are looking for an element of $\ker (A^T-I)$.
$A^T-I = \begin{bmatrix} -0.1 & 0 & 0.05 \\ 0.1 & -0.1 & 0.05 \\ 0 & 0.1 & -0.1 \end{bmatrix}$. Since the null space of $B$ and $\lambda B$ are the same for $\lambda \neq 0$, we look at $B=20(A^T-I) = \begin{bmatrix} -2 & 0 & 1 \\ 2 & -2 & 1 \\ 0 & 2 & -2 \end{bmatrix}$ instead. If $Bv = 0$, then the first line gives $v_1$ in terms of $v_3$, and the second line gives $v_2$ in terms of $v_3$, hence we can set $v_3 = 1$ and compute a vector $v\neq 0$ satisfying $A^T v = v$.