Given some matrix $A = \begin{bmatrix} a & x & y \\ x & b & z \\ y & z & c \end{bmatrix} | \sum_{i \in \text{row}_j(A)}i = 1$, under which conditions does there exist some vector $\vec{e}$ representing an equilibrium state such that $\vec{e} = A\vec{e}$?
Is there a general solution to $e$, given $A$?
Given $A\vec{x} = \lambda \vec{x}$, $\lambda$ is an eigenvalue of $A$ and $\vec{x}$ is an eigenvector of $A$. An eigenvector of $A$ is a vector which is transformed into a scalar multiply of itself when transformed by $A$.
If the only real eigenvalue of $A$ is $1$, then the eigenvector of $A$ must be the equilibrium state in a discrete dynamic system: given that $\vec{e} = A\vec{e}$, after any number of timesteps, $\vec{e}$ will always be $\vec{e}$.
MIT's Linear Algebra Notes
Berkeley's Linear Algebra Notes