# Continuous Markov chain, finding the stationary distribution

I am given a system of the form $x'(t)=Ax(t)$ where $A\in M_{3}(\mathbb{R})$ is a diagonalizable matrix and an initial condition $x(0)=\begin{pmatrix}1/3\\ 2/3\\ 0 \end{pmatrix}$ and I am being asked to find the stationary distribution.

How can I find the stationary distribution ?

I have calculated the eigenvales and eigenvectors of $A$ and then solved the ODE by finding a basis for the solution space and choosing a linear combination that would satisfy the initial condition to obtain $x(t)$.

How to I proceed ?

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Solve $Az=0$. Any solution $z$ with nonnegative coordinates summing to $1$ is a stationary distribution.
So in general: $z$ is a stationary distribution iff $Az=0$ and $Z$'s coordinates summ to $1$ ? in particular I think that the number of stationary distributions is the geometric multiplicity of $0$ (that is an eigenvalue if such $z$ exist), correct ? thank you for your help! – Belgi Oct 24 '12 at 9:54