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 $b_i = O(\lambda)$?

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.

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 that $b_i = O(\lambda)$?

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 $b_i = O(\lambda)$?