I have a problem when I deal with a transition matrix question. To explain it briefly, my problem is whether the second smallest eigenvalue of $P^T(I-ee^T)P$ is not less than $1$, where $P$ is a square matrix meeting some conditions.
Suppose $A$ is an $n\times n$ symmetric matrix with all elements positive. Let $d_i$ be the sum of the $i-$th row of $A$ and suppose $d_i>0$. Let $D=\text{diag}(d_1,d_2,\dots,d_n)$ and $B=D^{-1}A$. Then we know that $B$ is the row-normalized matrix of $A$ and the sum of any row of $B$ equals $1$. $B=D^{-1}A$ is similar to $D^{-1/2}AD^{-1/2}$, which is also a symmetric matrix. Therefore the eigenvalues of $B$ are all real. $B$ is also a stochastic matrix, hence the spectral radius $\rho(B)=1$.
Let $e=[1,1,\dots,1]^T/\sqrt{n}$ be an $n\times 1$ vector. Now we consider a matrix $P=(1+s)I_{n\times n}-sB,\ s>0$. From the discussion above, we can obtain that the eigenvalues of $P$ are all between $1$ and $1+2s$. $I-ee^T$ is a semi-positive definite matrix and not a full-rank matrix, so $P^T(I-ee^T)P$ has an eigenvalue $0$ and all the other eigenvalues are larger than $0$. Now I want to figure out whether the second smallest eigenvalue of the matrix $P^T(I-ee^T)P$ is not less than $1$.
I use Matlab to randomly generate $A$ thousands of times and the proposition is true for all the generated $A$. Could anyone find out a counter-example or prove this proposition?
Update (Sept.19):
Now I have a way to handle this problem, but stuck halfway. I will illustrate it in the following part, which may help fix the problem.
This passage (13) shows a relation among the eigenvalues of $M+N$, $M$, and $N$, where $M$ and $N$ are symmetric matrices. To be more specific, suppose the eigenvalues of $M$ are $\alpha_1\geq\dots\geq\alpha_n$, the eigenvalues of $N$ are $\beta_1\geq\dots\geq\beta_n$, and the eigenvalues of $M+N$ are $\gamma_1\geq\dots\geq\gamma_n$. Then we have from the passage that $$\max_{i+j=n+k}(\alpha_i+\beta_j)\leq\gamma_k.$$ We denote $M=P^TP$, $N=-P^Tee^TP$, and $M+N=P^T(I-ee^T)P$, then $N$ has $n-1$ eigenvalues of $0$ and one negative eigenvalue. Therefore, by using the result from the passage, we have $$\gamma_{n-1}\geq\max(\alpha_{n-1}+\beta_{n},\alpha_{n}+\beta_{n-1})\geq\alpha_n.$$ We expect $\alpha_n\geq 1$, but unfortunately, despite the fact that the eigenvalues of $P$ are all between $1$ and $1+2s$, there is nothing to do with $P^TP$. The smallest eigenvalue of $P^TP$ may be smaller than $1$.