# Vorticity Matrix for Markov chain

I have a markov chain with $$Q(u,v)$$ as transition probability matrix and $$\pi(u)$$ as stationary distribution defined on state space $$\Omega$$. The dimension of matrix $$Q$$ is $$nxn$$ and vector $$\pi$$ is $$1xn$$.

I need to construct a vorticity matrix $$\Gamma (u,v)$$ of dimension $$nxn$$ which has below properties

1. $$\Gamma$$ is skew symmetric matrix i.e, $$\Gamma (u,v) = -\Gamma (v,u) \quad ,\forall \, u,v \in \Omega$$

2. Row sum of $$\Gamma$$ is zero for every row i.e, $$\sum_v \Gamma (u,v) = 0 \quad ,\forall \, u \in \Omega$$

3. Third property is, $$\Gamma(u,v) > -\pi (v)Q(v,u) \quad ,\forall \, u,v \in \Omega$$

My question is : How to construct vorticity matrix $$\Gamma (u,v)$$ which satisfies above three properties? I need to construct at least one such matrix.

Is there any systematic way to build such matrices

NOTE: Transition probability matrix $$P$$, and stationary distribution $$\pi$$ has below properties

Row sum of $$P$$ is one for each row, $$\sum_v P(u,v)=1 \quad ,\forall \, u \in \Omega$$ $$\pi$$ is probability distribution hence, $$\sum_v \pi(v) = 1$$ Stationary distribution condition for $$\pi$$, $$\sum_u \pi(u) P(u,v) = \pi(v) \quad ,\forall \, v \in \Omega$$

• Changing third property to $\Gamma \geq \epsilon + \dots$ for some small $\epsilon$, you can use LP for this. – Eugene Aug 4 '19 at 19:42

One can verify that the following matrix $$\Gamma$$ has the desired properties: $$\Gamma = [\pi]Q - Q^{\top}[\pi]$$, where $$[\pi]$$ is a diagonal matrix with diagonal elements being the stationary distribution. This construction relates closely to the reversibility of a Markov chain as far as I know.

• $\Gamma = [\pi]Q - Q^{\top}[\pi]$ , turns out to be zero matrix for simple random walk in which neighbour node is chosen uniformly $$Q = \left( \begin{matrix} 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 1 & 0 \\ \frac{1}{3} & 0 & \frac{1}{3} & 0 & \frac{1}{3} \\ 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \end{matrix} \right)$$ $$\pi = \left( \begin{matrix} \frac{2}{10} & 0 & 0 & 0 & 0 \\ 0 & \frac{2}{10} & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{10} & 0 & 0 \\ 0 & 0 & 0 & \frac{3}{10} & 0 \\ 0 & 0 & 0 & 0 & \frac{2}{10} \end{matrix} \right)$$ – user3322017 Jun 6 '20 at 10:16
• Since I can't add image of graph in comments, I am giving vertices and edges for graph. Vertex set V = {1,2,3,4,5}. Edge set E = { (1,2), (1,4), (2,5), (4,5), (3,4) }. All edges are undirected – user3322017 Jun 6 '20 at 10:19
• That means your Q matrix represents a reversible Markov chain. In this case, $\Gamma$ is indeed zero. Actually, one checks whether a Markov chain is reversible by checking if $[\pi]Q=Q^\top[\pi]$. I am not aware of how to construct a non-zero $\Gamma$ matrix in this case. – user121747 Jun 12 '20 at 22:16