Let $$A:= \begin{pmatrix} a&b_1&b_1&\cdots&b_1 \\ b_2&a&b_2&\cdots&b_2 \\ b_3&b_3&a&\cdots&b_3 \\ \vdots&\cdots&&&\vdots \\b_m&b_m&b_m&\cdots&a \end{pmatrix}$$ be a $m\times m$ matrix such that all elements are (strictly) positive and $a+b_1+\cdots+b_m<1$ with $b_1+b_2+\dots+b_m<a$. WLOG assume $0<b_1\leq b_2\leq\dots \leq b_m$.
By Perron-Frobenius theorem, there exists a (strictly) positive eigenvalue $g>0$ with algebraic multiplicity 1 and an associated positive eigenvector.
I want to show that $$\|A^t v\|_1 \leq g^t \|v\|_1$$ for $t\in \mathbb N$ where $\|\cdot\|_1$ is the 1-norm and $v:=(b_1,b_2,\dots,b_m)$.
What I have tried:
I was able to prove this for the case of $m=2$ by finding all the eigenvalues and diagonalizing $A$ to express $A^t$ explicitly. Also, I found that the characteristic polynomial $\phi(x)$ of $A$ is $\phi(x)=f(a-x)$ where $$f(y):=y^m-\sum_{i>j}b_i b_j y^{m-2}+2\sum_{i>j>k}b_ib_jb_ky^{m-3}- \dots+(-1)^{m-1}(m-1)b_1\dots b_m .$$
Any help is appreciated.
I changed notation slightly and put the $b_i$'s in a vector $\mathbf b$. If we let diagonal matrix $D := \text{diag}\big(\mathbf b\big)$, then
$A = D\mathbf {11}^T - D + aI$ and consider similar matrix $C$, where
$D^\frac{1}{2}CD^\frac{-1}{2}= A$ and in particular
$C = D^\frac{1}{2}\mathbf {11}^TD^\frac{1}{2} - D^\frac{1}{2}(I)D^\frac{1}{2} + aI = D^\frac{1}{2}\big(\mathbf {11}^T - I\big)D^\frac{1}{2} + aI$
which is real symmetric
now for the main problem
$\big \Vert A^k \mathbf b\big \Vert_1 $
$= \mathbf 1^T A^k \mathbf b$
$= \mathbf 1^T \big(A^k\big) D \mathbf 1$
$= \mathbf 1^T \big(D^\frac{1}{2}C D^\frac{-1}{2}\big)^kD \mathbf 1 $
$= \mathbf 1^T \big(D^\frac{1}{2}C^k D^\frac{-1}{2}\big)D \mathbf 1 $
$= \mathbf 1^T D^\frac{1}{2}C^k D^\frac{1}{2}\mathbf 1 $
$= \text{trace}\Big( C^{k} D^\frac{1}{2} \mathbf 1\mathbf 1^TD^\frac{1}{2}\Big)$
$\leq \Big\Vert C^{k} D^\frac{1}{2} \mathbf 1\mathbf 1^TD^\frac{1}{2}\Big\Vert_{S_\infty}$
$\leq \Big\Vert C^{k}\Big \Vert_{S_\infty}\cdot \Big\Vert D^\frac{1}{2} \mathbf 1\mathbf 1^TD^\frac{1}{2}\Big\Vert_{S_\infty}$
$= g^k \cdot \text{trace}\Big(D^\frac{1}{2} \mathbf 1\mathbf 1^TD^\frac{1}{2}\Big)$
$= g^k \cdot \text{trace}\Big(\mathbf 1^TD\mathbf 1 \Big)$
$= g^k \cdot \text{trace}\Big(\mathbf 1^T\mathbf b \Big)$
$= g^k \cdot \big \Vert \mathbf b \big \Vert_1$
as desired
The inequalities follow because
1.) $\big(C^{k} D^\frac{1}{2} \mathbf 1\mathbf 1^TD^\frac{1}{2}\big)$ is a rank one matrix, so the trace gives its sole non-zero eigenvalue (in fact its Perron root) which is bounded above by it maximal singular value -- given by the Schatten infinity norm (also known as the Operator 2 norm)
2.) the Schatten norms are submultiplicative (and in the case here of the Operator 2 norm, this is directly provable by Cauchy Schwarz)
3.) $\big(D^\frac{1}{2} \mathbf 1\mathbf 1^TD^\frac{1}{2}\big)$ is positive semidefinite and rank one, so its maximal singular value is given by its sole non-zero eigenvalue which is given by its trace