I am studying nonnegative matrices and their use in modelling Markov chains. I am curious about the nature of the sequence of partial sums of a spectral decomposition for an irreducible nonnegative matrix, $A$. The matrix $A$ satisfies the Perron-Frobenius theorem, and thus there exists a dominant eigenvalue $\rho_0$ such that $|\rho| > |\rho_i| \ \forall i>0$. The matrix is diagonalizable and hence can be written as a spectral decomposition, $$ A_{ji} = \sum_m \rho_m u_i^{(m)} v_j^{(m)} $$ where I denote $u^{(m)}, v^{(m)}$ as the $m^{\text{th}}$ left and right eigenvectors associated to $\rho_m$. I use the normalization condition $\sum_i v_i = 1$ and $\sum_i u_i v_i = 1$. The sum $\sum_j \left(A^N\right)_{ji}$ is the state probability distribution after $N$ steps. This leads to sums such as $$ \sum_j A_{ji} = \sum_j \sum_m \rho_m u_i^{(m)} v_j^{(m)} = \sum_m \rho_m u_i^{(m)} $$
For certain matrices $A^{(c)}$ and $\{A^{(k)}\}$, I have proven the following inequality $$ \sum_m \frac{\rho_m^{(c)}}{\rho_0^{(c)}} u_i^{(c)(m)} \geq \sum_m \left( \prod_k \frac{\rho_m^{(k)}}{\rho_0^{(k)}} u_i^{(k)(m)}\right)^{\alpha_k}. $$ I would like to use this to show that $$ u_i^{(0)(c)} \geq \prod_k\left( u_i^{(k)(0)}\right)^{\alpha_k}$$ which I have demonstrated with numerous experiments.
Essentially I would like to "extract" the $m=0$ term from either side of the proven inequality, and I am unsure if this is even possible. The problems are: $u_i^{(m)}$ can be complex for $m\neq 0$ (by Perron-Frobenius theorem), but I know that the entire sum is real (because it is related to a probability). I also know that $|\rho| > |\rho_i| $ as mentioned but I'm not sure how I can use it.
My best idea from what I've tried thus far: $$ \left( \sum_m \frac{\rho_m}{\rho_0} u_i^{(m)} \right)^{N} = \left(u_i^{(0)} + \sum_{m>1} \frac{\rho_m}{\rho_0} u_i^{(m)} \right)^{N} = u_i^{(0)N}\left(1 + \sum_{m>1} \frac{\rho_m}{\rho_0} \frac{u_i^{(m)}}{u_i^{(0)}} \right)^{N} $$ So somehow I'd like to show that $\sum_{m>1} \frac{\rho_m}{\rho_0} \frac{u_i^{(m)}}{u_i^{(0)}}$ is "small" perhaps, so that I can apply this expansion above and retain the $m=0$ inequality.
Note: Due to the probability interpretations and positivity of $\rho_0 u_i^{(0)}$, the sum $\sum_{m>1} \frac{\rho_m}{\rho_0} \frac{u_i^{(m)}}{u_i^{(0)}}$ is real!
Any thoughts/ideas are greatly appreciated!