Let $T_0$ be the transition matrix of a purely clustered Markov chain, i.e. a direct sum of $k \geq 2$ irreducible stochastic matrices. Given a perturbation $T(x) = T_0 + xE$ of $T_0$ such that $T(x)$ is also stochastic, how small must $x$ be in order for us to recover the indices of the direct summands of $T_0$? We give a simple algorithm based on the orthogonal projection matrix onto the left or right singular subspace corresponding to the $k$ smallest singular values of $I - T(x)$ which allows for exact recovery all clusters when $x = O\left(\frac{\sigma_{n - k}}{||E||_2\sqrt{n_1}}\right)$ and approximate recovery of a single cluster when $x = O\left(\frac{\sigma_{n - k}}{||E||_2}\right)$, where $n_1$ is the size of the largest cluster and $\sigma_{n - k}$ the $(k + 1)$st smallest singular value of $T_0$.

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关于马尔可夫链中簇的基于投影的恢复的注释

令$T_0$ 为纯聚类马尔可夫链的转移矩阵，即$k \geq 2$ 不可约随机矩阵的直接和。给定 $T_0$ 的扰动 $T(x) = T_0 + xE$ 使得 $T(x)$ 也是随机的，$x$ 必须有多小才能恢复 $ 的直接被加数的指数T_0$？我们给出了一种基于正交投影矩阵到对应于 $I - T(x)$ 的 $k$ 最小奇异值的左或右奇异子空间的简单算法，它允许在 $x = O\left 时精确恢复所有簇(\frac{\sigma_{n - k}}{||E||_2\sqrt{n_1}}\right)$ 和当 $x = O\left(\frac{\sigma_{ n - k}}{||E||_2}\right)$，其中 $n_1$ 是最大簇的大小，$\sigma_{n - k}$ 是 $(k + 1)$st 最小的单数$T_0$ 的价值。