maximum eigenvalue of rank 1 update matrix I am stuck with the following problem. The ingredients are:

*

*$A=diag(a_1,\dots,a_n)$, $a_i \in (0,1)$ and $a_1+\dots+a_n=1$,

*$B=diag(b_1,\dots,b_n)$, $b_i \in (0,1)$ and $b_1<b_2<\dots<b_n$,

*the entries of the matrices $A$ and $B$ are related by $\sum\limits_{i=1}^n a_i {b_i \over 1- b_i}<1$.

Denote ${\bf b}^T=(\sqrt{a_1} b_1,\dots,\sqrt{a_n}b_n)$, $c={1 \over \sum_{i=1}^n a_i (1-b_i)}$. The problem is to prove that the maximum eigenvalue of the matrix $D=B+ c {\bf b} {\bf b}^{T}$ is less than 1.
I've been studying the steady-state distribution of the two-dimensional Markov chain. Its steady-state vector turned out to be ${\vec p}_k={\vec p}_1 D^{k-1}$, $k \ge 1$. As usual, ${\vec p}_1$ is found from the normalization condition, which requires evaluation of $\sum_{k=0}^\infty D^k$. To make it $(I-D)^{-1}$, the aforementioned proof is required. My numerous numerical expreiments show that the maximum eigenvalue is $<1$. But I cannot prove it.
What I've tried and failed:

*

*interative methods (like power iteration) don't help since the power $D^k$ does not have a nice expression.

*Gershgorin theorem does not help, since some entries of $D {\vec 1}$ may be $>1$.

*the matrix $c {\bf b} {\bf b}^{T} $ is of rank 1 and has one eigenvalue $\rho=c \sum_{i=1}^n {a_ib_i^2}$. This matrix is symmetric.

*the eigenvalues of $D$ are all different and do satisfy the well-known secular equation $f(\lambda)=0$, where:
$$
f(\lambda)=1+ c \sum_{i=1}^n {a_i b_i^2 \over b_i - \lambda}.
$$

*denote the roots of $f(\lambda)=0$ by $\lambda_1,\dots,\lambda_n$. According to the theory, the roots satisfy the inequalities:
$$
b_1<\lambda_1<b_2<\lambda_2<\dots<b_n<\lambda_n.
$$
This does not help since the upper bound for the maximum eigenvalue $\lambda_n$ remains unknown. The theory says that $\lambda_i=b_i+m_i \rho$, where $m_i \in (0,1)$ and $\sum_{i=1}^n m_i=1$. But I can't prove that $\lambda_n=b_n+m_n\rho<1$.

*$\sum\limits_{i=1}^n a_i {b_i \over 1- b_i}<1$ is equivalent to $1<\sum\limits_{i=1}^n {a_i \over 1-b_i}<2$. Since $\sum_i (u_i/v_i) \sum_i (u_iv_i) \ge 1$ we have $\sum\limits_{i=1}^n a_i (1-b_i)>{1 \over2}$ and thus $1<c<2$.

*one of the classical inequalities gives
$$
\sum_{i=1}^n a_i (1-b_i)+ (1-b_1)(1-b_r) \sum_{i=1}^n {a_i \over 1-b_i} \le 2- b_1-b_r,
$$
and thus
$$
\sum_{i=1}^n a_i (1-b_i) \le 2- b_1-b_n - (1-b_1)(1-b_n) = 1-b_1 b_n <1,
$$
and thus, instead of $1<c<2$ we have ${1 \over 1-b_1 b_n}<c<2$. Moreover, since $\sum_{i=1}^n a_i (1-b_i)>{1\over 2}$, from the last inequality we get
${1\over 2} \le 1-b_1 b_n$ or $b_1 b_n < {1\over 2}$. Since $b_1<b_2<\dots<b_n$ we have
$$
b_1 b_i < {1\over 2}.
$$
Unfortunately, all this did not help me to make a step towards the proof and I would be grateful for any useful suggestions.
 A: Seems that I have found the solution. It is given by the Example 7.10.3 on page 620 in here link.
Specifically, if a matrix $H$ is non-negative then its maximum eigenvalue, say $\rho(H)$, satisfies $\rho(H)<r$ if and only if $(rI-H)^{-1}$ exists and $(rI-H)^{-1}\ge 0$.
Let us consider the matrix $(I-D)^{-1}$. Due to Sherman–Morrison formula
this inverse is equal to
$$
(I-D)^{-1}=(I-B)^{-1} + c {(I-B)^{-1} {\bf b} {\bf b}^{T} (I-B)^{-1} \over 1 - c {\bf b}^T (I-B)^{-1} {\bf b} }
= \underbrace{(I-B)^{-1}}_{\ge 0 \, component-wise\!\!\!\!\!\!\!\!} + {c \over \alpha} (I-B)^{-1} \underbrace{{\bf b} {\bf b}^{T}}_{\ge 0} (I-B)^{-1}, 
$$
and exists if and only if $\alpha \neq 0$. By direct inspection the value of $\alpha$ is equal to
$$
\alpha=1-c\sum_{i=1}^n {a_i b_i^2 \over 1-b_i}
$$
and clearly $\alpha\neq 0$, and thus $(I-D)^{-1}$ exists.
Now notice that $(I-D)^{-1}\ge 0$ if ${c\over \alpha}>0$. The value of $\alpha$ can be rewritten:
$$
\alpha=1-c\sum_{i=1}^n {a_i (b_i-1+1)^2 \over 1-b_i}=
c \left (2-\sum_{i=1}^n {a_i \over 1-b_i} \right )
=
c \left (1- \sum_{i=1}^n {a_i b_i \over 1-b_i} \right ).
$$
Thus ${c\over \alpha}>0$ if ${1 \over \left (1- \sum\limits_{i=1}^n {a_i b_i \over 1-b_i} \right )}>0$, which is equivalent to $\sum\limits_{i=1}^n {a_i b_i \over 1-b_i}<1$.
Thus we can use the statement of the Example 7.10.3 and thus the maximum eigenvalue of $D$ is less than 1.
A: what you want to do is show
$D-I=\big(B-I)+ c {\bf b} {\bf b}^{T}\prec \mathbf 0$
via congruence transform, this is the same as showing
$-I + c \big(I-B\big)^\frac{-1}{2}{\bf b} {\bf b}^{T}\big(I-B\big)^\frac{-1}{2}\prec \mathbf 0$
or that the trace of that rank 1 matrix is $\lt 1$.  Rescaling by $c^{-1}$ gives the criterion
$\sum_{i=1}^n a_i\frac{ b_i^2}{1-b_i}  =\text{trace}\Big( \big(I-B\big)^\frac{-1}{2}{\bf b} {\bf b}^{T}\big(I-B\big)^\frac{-1}{2}\Big)\lt c^{-1}=\sum_{i=1}^n a_i(1-b_i) $
if you subtract the LHS from the RHS this is equivalent to proving
$0\lt \sum_{i=1}^n a_i\frac{(1-b_i)^2- b_i^2}{1-b_i}$  which holds because
$0$
$\lt 1 -\sum_{i=1}^n a_i\frac{b_i}{1-b_i}$
$= \big(\sum_{i=1}^n a_i\frac{1-b_i}{1-b_i}\big) - \sum_{i=1}^n a_i\frac{b_i}{1-b_i} $
$= \sum_{i=1}^n a_i\frac{1-2b_i}{1-b_i} $
$=\sum_{i=1}^n a_i\frac{(1-b_i)^2- b_i^2}{1-b_i}$
where the inequality is bullet point 3 in the original post
