Let's talk about the long time limit. I mean let's talk about when

$$ |\pi \rangle = \lim_{t\to \infty} \exp(t G) | \hat 1 \rangle = \lim_{t \to

\infty} \exp(t \widetilde G) | \hat 1 \rangle $$

We have already stated that

$$ L = A - D $$

$$ \widetilde{L} = A^\top - \overline{D} $$

and when $G = LQ^{-1}$ and $\widetilde{G}=L \overline{Q}^{-1}$ for non-negative

$Q$, $overline{Q}$ diagonal, then $G$, $\overline G$ generate valid stochastic processes.

Special cases include the [random walk normalised Lapacian](http://en.wikipedia.org/wiki/Laplacian_matrix#Random_walk_normalized_Laplacian)

meaning $Q$ would be the degree matrix of the graph.

Moreover, we know that for $L$ and $\tilde L$ that $| \hat 1 \rangle$ is a

valid eigenstate with eigenvalue $0$. It is a general property of a graph

Lapacian that the all ones vector $| \hat 1 \rangle$ is an eigenstate of

highest (zero) eigenvalue. If the graph is strongly connected, the all ones

vector is the unique vector with this highest eigenvalue. If the graph is not

strongly connected, then we need to know about the

* [Algebraic connectivity](http://en.wikipedia.org/wiki/Algebraic_connectivity)

> Note also that if there are k disjoint connected components in the graph, then

this vector of all ones can be split into the sum of k independent $\lambda = 0$

eigenvectors of ones and zeros, where each connected component corresponds to an

eigenvector with ones at the elements in the connected component and zeros

elsewhere.

For now, I want to assume that the graph is strongly connected, and worry about

other cases later.

We want to consider the case where both $G$ and $\widetilde{G}$ share the same

stationary state $| \pi \rangle$ - in other words when

$$ LQ^{-1} | \pi \rangle = \widetilde{L} \overline{Q}^{-1} | \pi \rangle $$

We know that $LQ^{-1} | \pi \rangle = Q | \hat 1 \rangle = 0$ and so

$$ (A^\top - \overline{D}) \overline{Q}^{-1} Q | \hat 1 \rangle = 0 $$

However, we know that $(A^\top - \overline{D}) | \hat 1 \rangle = 0$ and that

$$ (A^\top - \overline{D}) \leq 0 $$

So when $\overline{Q}^{-1}Q | \hat 1 \rangle = | \hat 1 \rangle$ we have a valid

eigenstate of the Lapacian $(A^\top - \overline{D})$. We further know that $Q$,

$\overline{Q}$ are diagonal, and so the relationship gives

$$ \overline{Q}_{ij}^{-1} Q_{jk} = \delta_{ik} $$

and so we determine from the uniqueness of inverses that $\overline{Q} = Q$.

But this is not interesting, since every Lapacian has $L |\hat 1 \rangle = 0$

and so there is nothing special about the fact that $LQ^{-1}$ has $|\pi

\rangle$ as its steady state, for multiple $L$. In other words, we need to dig a bit deeper than this to find something that relates to the structure of the graph/system.

$$ |\pi \rangle = \lim_{t\to \infty} \exp(t G) | \hat 1 \rangle = \lim_{t \to

\infty} \exp(t \widetilde G) | \hat 1 \rangle $$

We have already stated that

$$ L = A - D $$

$$ \widetilde{L} = A^\top - \overline{D} $$

and when $G = LQ^{-1}$ and $\widetilde{G}=L \overline{Q}^{-1}$ for non-negative

$Q$, $overline{Q}$ diagonal, then $G$, $\overline G$ generate valid stochastic processes.

Special cases include the [random walk normalised Lapacian](http://en.wikipedia.org/wiki/Laplacian_matrix#Random_walk_normalized_Laplacian)

meaning $Q$ would be the degree matrix of the graph.

Moreover, we know that for $L$ and $\tilde L$ that $| \hat 1 \rangle$ is a

valid eigenstate with eigenvalue $0$. It is a general property of a graph

Lapacian that the all ones vector $| \hat 1 \rangle$ is an eigenstate of

highest (zero) eigenvalue. If the graph is strongly connected, the all ones

vector is the unique vector with this highest eigenvalue. If the graph is not

strongly connected, then we need to know about the

* [Algebraic connectivity](http://en.wikipedia.org/wiki/Algebraic_connectivity)

> Note also that if there are k disjoint connected components in the graph, then

this vector of all ones can be split into the sum of k independent $\lambda = 0$

eigenvectors of ones and zeros, where each connected component corresponds to an

eigenvector with ones at the elements in the connected component and zeros

elsewhere.

For now, I want to assume that the graph is strongly connected, and worry about

other cases later.

We want to consider the case where both $G$ and $\widetilde{G}$ share the same

stationary state $| \pi \rangle$ - in other words when

$$ LQ^{-1} | \pi \rangle = \widetilde{L} \overline{Q}^{-1} | \pi \rangle $$

We know that $LQ^{-1} | \pi \rangle = Q | \hat 1 \rangle = 0$ and so

$$ (A^\top - \overline{D}) \overline{Q}^{-1} Q | \hat 1 \rangle = 0 $$

However, we know that $(A^\top - \overline{D}) | \hat 1 \rangle = 0$ and that

$$ (A^\top - \overline{D}) \leq 0 $$

So when $\overline{Q}^{-1}Q | \hat 1 \rangle = | \hat 1 \rangle$ we have a valid

eigenstate of the Lapacian $(A^\top - \overline{D})$. We further know that $Q$,

$\overline{Q}$ are diagonal, and so the relationship gives

$$ \overline{Q}_{ij}^{-1} Q_{jk} = \delta_{ik} $$

and so we determine from the uniqueness of inverses that $\overline{Q} = Q$.

But this is not interesting, since every Lapacian has $L |\hat 1 \rangle = 0$

and so there is nothing special about the fact that $LQ^{-1}$ has $|\pi

\rangle$ as its steady state, for multiple $L$. In other words, we need to dig a bit deeper than this to find something that relates to the structure of the graph/system.