# Why the Markov chain transition matrix doesn't have an eigenvalue $\lambda=1$?

Given an instance of the transverse Ising model, I am trying to sample from the Boltzmann probability distribution $$\mu(s)=\frac{e^{-E(s)/T}}{Z}$$ ($$Z$$ is the partition function, $$s$$ is a spin configuration) using Markov chains. Consider the following transition matrix P representing the MC process: $$P(s'|s)=A(s'|s)Q(s'|s)$$ where $$Q$$ is the proposal strategy and $$A$$ is the acceptance probability:$$A(s'|s)=min(1, \frac{\mu(s')Q(s|s')}{\mu(s)Q(s'|s)})$$ Given that $$Q$$ is symmetric $$(Q(s'|s)=Q(s|s'))$$, we have: $$A(s'|s)=min(1, \frac{\mu(s')}{\mu(s)})$$

$$P$$ satisfies the detailed balance condition: $$P(s'|s)\mu(s)=P(s|s')\mu(s')$$

Given a certain instance of the model, the spectrum of the $$Q$$ matrix always features an eigenvalue $$\lambda=1$$. However, this is not the case for $$P$$ whose eigenvalues are always $$<1$$. Being $$P$$ the transition matrix, isn't it supposed to always have an eigenvalue $$\lambda=1$$?

• As you say, if $P$ is a stochastic matrix (i.e., its row sums are equal to 1), then it must have an eigenvalue equal to 1. Indeed, by definition $P\mathbf{1} = \mathbf{1}$, so $\mathbf{1}=(1,\dots,1)$ is an eigenvalue with eigenvector 1. Why do you say $P$ doesn't have such an eigenvalue? If you've determined this numerically, something is wrong with your code. The Perron-Frobenius theorem is relevant here. Also, note that while many people here will familiar with bra-ket notation, not everyone will be, so you might want to define your terms.
– snar
Commented Feb 10, 2023 at 15:59
• @snar thank you very much for your comment, I'll add more details such that the question is clear regardless of the reader's background. Commented Feb 10, 2023 at 16:12
• @snar The problem is that in my simulations $Q$ turns out to be a stochastic matrix, but because I have to include the acceptance probability $A$ (since $P(s'|s)=Q(s'|s)A(s'|s)$), $P$ is not stochastic. I don't understand where is the mistake. Commented Feb 10, 2023 at 16:44

\begin{align*} P(s'|s) &= A(s'|s) Q(s'|s) \tag{1}\\ P(s'|s)\mu(s) &= P(s|s')\mu(s') \tag{2}. \end{align*}
Most likely, OP is thinking of the Metropolis-Hastings algorithm, for which (2) is true, so long as one remembers to include the fact that if a proposal $$s'$$ is rejected, then $$s' = s$$. In words, the algorithm used says: pick a point according to $$Q$$, and then accept or reject it according to $$A$$; if rejected, continue with the previous point $$s$$. Thus, no matter whether the sample is accepted or rejected, the probability that $$s$$ moves to somewhere is 1.
If indeed OP meant (1), then I challenge OP to prove (2) while maintaining $$\sum_{s\in S} \mu(s) = 1$$. Here, $$S$$ is the state space.
The correct definition of the Metropolis-Hastings transition kernel (for which (2) holds) is $$P(s'|s) = A(s'|s)Q(s'|s) + \delta_{\{s\}}(s') \sum_{s''\in S} (1-A(s''|s))Q(s''|s).$$ From this correct representation of $$P$$ it follows immediately that \begin{align*} \sum_{s' \in S} P(s' | s) &= \sum_{s'\in S} A(s'|s)Q(s'|s) + \sum_{s' \in S} \delta_{\{s\}}(s') \sum_{s''\in S} (1-A(s''|s))Q(s''|s) \\ &= \left(\sum_{s'\in S} \delta_{\{s\}}(s')\right)+\left(\sum_{s'\in S} A(s'|s)Q(s'|s)\right) -\left(\sum_{s'\in S} \delta_{\{s\}}(s')\right)\left(\sum_{s''\in S} A(s''|s)Q(s''|s)\right) \\ &= 1 \end{align*} where we used the facts that for any $$s,s'$$, $$\sum_{s'\in S} Q(s'|s) =1, \quad \sum_{s'\in S} \delta_{\{s\}}(s') = 1$$ and that $$s''$$ is just a dummy variable (to cancel out the two sums over $$A\times Q$$).