Let $b = [b\_1,b\_2,b\_3,...b\_n]^T$
$A = [a\_{ij}]\_{n \times n}$ such that $a\_{i,j} = 1\forall 1\le i,j \le n$


$C = [c\_{ij}(\lambda)]\_{n \times n}$ such that $c\_{ij}(\lambda) = O(\frac{1}{\lambda})$


$A+C$ is known to be a symmetric positive semi definite matrix


$I\_n$ is an $n\times n$ Identity matrix


$e = [e\_1,e\_2,e\_3,...e\_n]^T$


$$b = (A+C+\frac{1}{\lambda^2}I\_n)^{-1}e$$


I am trying to find an asymptotic expansion for $b\_i$


Show that $b\_i = O(\lambda)$.


For example If I take matrix $C$ as all zeros, $A$ is given as all ones, and the case $n=2$ I get $$b\_i = O(\lambda)$$


Is there any good argument to generalize for any $n$, that $b\_i = O(\lambda)$?


PS: The elements of all matrices are non negative, except matrix $e$, which can
take negative values.



Asymptotic expansion involving a matrix equation
Re [your answer](https://mathoverflow.net/a/365990/1898):


Statement 1 follows from [Weyl's inequalities](https://en.wikipedia.org/wiki/Weyl%27s_inequality#Weyl%27s_inequality_about_perturbation) and $\rho\_1(C) = O(\lambda^{-1})$.


$2 \implies 3$ seems problematic; what if $C = \frac{1}{\lambda^3}I$, which should be possible given the definition of big-O? Then $\rho\_1(M^{-1}) \sim \lambda^2 \not\in O(\lambda)$.


Statement 4 seems just the matrix norm inequality $\|M^{-1}e\| \leq \|M^{-1}\|\|e\|$.


