Sensitivity of the second eigenvector (mixing time) in the Google markov transition matrix to changes in $\alpha$

Question

I am reading materials on PageRank seen here: https://arxiv.org/pdf/2207.02296, and here: https://www.stat.uchicago.edu/~lekheng/meetings/mathofranking/ref/langville.pdf. Given the walk matrix $\mathbf{M}$ depicting a Markov chain, some $0 < \alpha < 1$, and $n\times1$ probability vector $\mathbf{p}$, we can write the stationary distribution with: \begin{align*} \mathbf{p}^T = \mathbf{p} \biggl(\alpha\mathbf{M} + (1-\alpha)\frac{\mathbf{e}\mathbf{e}^T}{n}\biggr). \end{align*} where $\mathbf{e}$ is the all ones vector. We know the second eigenvalue $\lambda_2$ of this weighted sum walk matrix $\alpha\mathbf{M} + (1-\alpha)\frac{\mathbf{e}\mathbf{e}^T}{n}$ is the mixing time.

Now I want to determine the change in $\lambda_2$ as a function of change in $\alpha$. That is to say given perturbed $\alpha$: $0 < \alpha + \Delta \alpha <1$, then what is the $\lambda_2$ of this system: \begin{align*} \mathbf{p}^T = \mathbf{p} \biggl((\alpha + \Delta \alpha) \mathbf{M} + (1-\alpha - \Delta \alpha)\frac{\mathbf{e}\mathbf{e}^T}{n}\biggr). \end{align*}

I saw a similar post here: Given a perturbation of a symmetric matrix, find an expansion for the eigenvalues, but it's unclear if it answers my question.

Answer 1

A general method is to perturbatively expand the characteristic polynomial: $$ \chi(\lambda) = \det(\lambda I-T) \\ T = \pi+\alpha(M-\pi) \quad \pi = \frac{ee^T}n $$ Using the differential of the determinant: $$ \det(A+dB) = \det(A)+\text{Tr}(C^TdB)+o(dB) $$ with $C$ the cofactor matrix of $A$, you get: $$ d\chi = \text{Tr}(C^T(M-\pi))d\alpha $$ with $C(\lambda)$ the cofactor matrix of $\lambda I-T$. Thus, if your eigenvalue is not degenerate, i.e. $\chi'(\lambda)\neq 0$, the first order perturbation is given by: $$ d\lambda = -\frac{\text{Tr}(C^T(M-\pi))}{\chi'(\lambda)}d\alpha +o(d\alpha) = -\frac{\text{Tr}(C^TM)-\frac{e^TCe}n}{\text{Tr}(C)}d\alpha +o(d\alpha) $$ This holds in particular for the second eigenvalue $\lambda_2$. Numerically, this formula is not efficient as it requires computing the cofactor matrix.

You can generalise the previous result if your second eigenvalue is degenerate. The perturbation will generically lift the degeneracy. Quantitatively, if it is a root of order $p$ so that $\chi^{(p)}(\lambda)\neq0$, you have $p$ roots: $$ d\lambda = \left(-\frac{\text{Tr}(C^T(M-\pi))}{\chi^{(p)}(\lambda)}d\alpha\right)^{1/p}+o(d\alpha^{1/p}) $$ and are all multiple of each other of a $p$-th root of unity.

On a terminology note, $t_r=\frac1{\ln\lambda_2}$ is the relaxation time. This is different from the mixing time: $$ t_m(\varepsilon) = \min \left\{ t \geq 0 : \max_{1\leq i\leq n}\frac12\sum_{j=1}^n|(T^n)_{ij} - p_j| \leq \varepsilon \right\} $$ with $p$ the stationary distribution: $$ pT = p $$ Not only do the definitions differ, but they can have very different scalings. In a lot of applications, you have a cutoff phenomenon where $t_r\ll t_m$. Intuitively, $t_m$ is the time to reach the stationary state from a deterministic initial condition, while $t_r$ is the decorrelation time once you already are in the stationary state