# Bound on the ratio of top 2 eigenvalues

Let $$P$$ be a $$n \times n$$ stochastic matrix such that $$P_{ij}=\tau$$ if $$i \neq j$$ and $$P_{ii} = 1 - (n-1)\tau$$ where $$0<\tau < \frac{1}{n}$$.
It is clear that the largest eigenvalue of $$P$$ is 1, and the second largest eigenvalue is $$(1-n\tau)$$, hence $$\frac{\lambda_{2}}{\lambda_{1}} = 1-n\tau \leq 1 - 2\tau.$$ Let $$D$$ be a $$n \times n$$ diagonal matrix such that $$D_{ii} \geq 1$$ for all $$i$$. Consider the matrix $$PD$$ and let $$\lambda_{1}',\lambda_{2}'$$ be the top two eigenvalues. Prove that $$\frac{\lambda_{2}'}{\lambda_{1}'} \leq 1-2\tau.$$ I have verified that it's true for $$n=2,3$$ by brute force calculations. Also using Horn's inequalities I can find a bound which is much worse. Thanks

I claim that $$\lambda_2'/\lambda_1'\leqslant 1-\frac{2\tau}{1-(n-2)\tau}$$ which is bit stronger than you ask for. This is sharp as the example $$D_{11}=D_{22}=1\gg \max(D_{33},\dots,D_{nn})$$ shows.
Of course it is not important that $$D_{ii}\geqslant 1$$, only that $$D_{ii}>0$$ (since everything is homogeneous in $$D$$). We may suppose that all $$D_{ii}$$'s are distinct, since $$\lambda_1'$$ and $$\lambda_2'$$ are continuous with respect to $$D$$. Denote $$D_{ii}=d_i$$, $$\beta=1-n\tau$$, $$\rho=\beta/\tau$$. Then we should prove that $$\lambda_2'/\lambda_1'\leqslant \rho/(\rho+2)$$.
Rewriting the eigenvector equation in the form $$(\tau J+\beta I)Dx=\lambda x$$, for the coordinates $$x_i$$ of the eigenvector $$x$$ with eigenvalue $$\lambda$$ we get $$\beta d_ix_i+\tau(\sum_j d_j x_j)=\lambda x_i$$. Denote $$\mu=\sum d_j x_j$$. If $$\mu=0$$, we get $$x_i(\lambda-\beta d_i)=0$$ for all $$i$$, so $$\lambda$$ must be equal to certain $$\beta d_j$$ and $$x_i=0$$ for $$i\ne j$$. But this contradicts to $$\mu=0$$. So $$\mu\ne 0$$, we have $$x_i=\tau \mu (\lambda-\beta d_i)^{-1}$$ and $$\sum d_i(\lambda-\beta d_i)^{-1}=(\tau \mu)^{-1}\sum d_i x_i=\tau^{-1}.$$ This equation (for $$\lambda$$) has $$n$$ roots in total. If $$d_1, by continuity there is unique root between $$\beta d_{i-1}$$ and $$\beta d_i$$ and one root on the interval $$(\beta d_n,\infty)$$.
By homogeneity we may suppose that $$\lambda_1'=1$$. Then $$\beta d_n<1$$ and $$\lambda_2'\in (\beta d_{n-1},\beta d_n)$$. Denote $$\beta d_i(1-\beta d_i)^{-1}=\theta_i$$, we have $$\sum \theta_i=\rho$$ and $$\beta d_i=\theta_i(1+\theta_i)^{-1}$$. Assume on the contrary that $$\lambda_2'>\rho/(\rho+2)$$, then also $$\theta_n/(1+\theta_n)=\beta d_n>\rho/(\rho+2)$$, $$\theta_n>\rho/2$$. Denoting $$m=\lambda_2'$$ we get $$\rho=\sum \frac{\beta d_i}{m-\beta d_i}=\sum \frac{\theta_i}{m-(1-m)\theta_i}=- \frac{\theta_n}{m-(1-m)\theta_n}+\sum_{i=1}^{n-1} \frac{\theta_i}{m-(1-m)\theta_i}\leqslant \\ - \frac{\theta_n}{m-(1-m)\theta_n}+\frac{\sum_{i=1}^{n-1} \theta_i}{m-(1-m)\sum_{i=1}^{n-1} \theta_i}= \frac{\theta_n}{m-(1-m)\theta_n}+\frac{\rho -\theta_n}{m-(1-m)(\rho-\theta_n)}=:f(m)$$ (the last denominator is positive since $$m/(1-m)>\rho/2>\rho-\theta_n$$.) The function $$f(m)$$ is decreasing between $$\alpha:=(\rho-\theta_n)/(1+\rho-\theta_n)$$ and $$\beta:=\theta_n/(1+\theta_n)$$. We have $$\alpha<\rho/(\rho+2)<\beta$$ and $$m\in (\rho/(\rho+2),\beta)$$. Therefore $$\rho\leqslant f(m)\leqslant f(\rho/(\rho+2))=-(\rho+2)$$, a contradiction.