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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.

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  • $\begingroup$ Are you interested in the case $\lambda \to 0$ or $\lambda \to \infty$ here? $\endgroup$ Commented Jul 19, 2020 at 8:10
  • $\begingroup$ @FedericoPoloni : $\lambda \to \infty$ $\endgroup$
    – Rajesh D
    Commented Jul 19, 2020 at 8:11

2 Answers 2

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Re your answer:

Statement 1 follows from Weyl's inequalities 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 $\|M^{-1}e\| \leq \|M^{-1}\|\|e\|$.

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  • $\begingroup$ $M$ is positive definite, as $A+C$ is psd. $\endgroup$
    – Rajesh D
    Commented Jul 19, 2020 at 8:21
  • $\begingroup$ In question I had stated $A+C$ is known to be $psd$ so $M$ is a pd. Hope that solves the problem. $\endgroup$
    – Rajesh D
    Commented Jul 19, 2020 at 8:22
  • $\begingroup$ @RajeshDachiraju Updated the counterexample. $\endgroup$ Commented Jul 19, 2020 at 8:23
  • $\begingroup$ Your example still assumes $\rho(M) = 0$ which is not possible. $\endgroup$
    – Rajesh D
    Commented Jul 19, 2020 at 8:25
  • $\begingroup$ $M$ is positive definite. $\endgroup$
    – Rajesh D
    Commented Jul 19, 2020 at 8:25
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My proof (some statements are without proof, for which I need references)

Let $X$ be any $n\times n$ square matrix and Let $\rho_1(X), \rho_2(X), \rho_3(X),...\rho_i(X),...\rho_n(X)$ denote the eigenvalues in descending order, of the matrix $X$.

Statement : 1* $$\rho_1(A+C) = n+O(1/\lambda)$$ $$\rho_i(A+C) = 0+O(1/\lambda),i = 2,3...n$$

Let $M = A+C+\frac{1}{\lambda^2}I_n$

Statement : 2 $$\rho_1(M) = n+O(\frac{1}{\lambda})$$ $$\rho_i(M) = 0+O(\frac{1}{\lambda}),i = 2,3,...n$$

statement : 3 $$\rho_1(M^{-1}) = O(\lambda),i = 1,2,3,...n-1$$ $$\rho_{n-1}(M^{-1}) = O(1)$$

statement : 4* The highest Eigen value of $M^{-1}$ is $O(\lambda)$, hence $b = M^{-1}e = O(\lambda)$, which means $b_i = O(\lambda), i = 1,2,3...n$

Statements 1 and 4, I have no proofs or references, which I need some help. Statements 2 and 3 are true and sources readily available.

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