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We have a matrix valued function $A:\mathbb{R}_+\to \mathbb{R}^{m\times m}$. It is known that $A(\lambda)$ is a positive definite matrix for all $\lambda\in\mathbb{R}_+$ Denoting $\rho_i(A(\lambda))$ the $i^{th}$ eigenvalue of the matrix $A(\lambda)$, we know that asymptotically each eigenvalue of the matrix $A(\lambda)$ is of the form $$\rho_i(A(\lambda)) = \beta_i\lambda + \alpha_{i0} + \frac{\alpha_{i1}}{\lambda} + \frac{\alpha_{i2}}{\lambda^2} + ...$$ where $\beta_i>0,\alpha_{i0},\alpha_{i1},\alpha_{i2,...\in\mathbb{R}}$ are constants independent of $\lambda$.

Now there is a $m\times 1$ matrix $L$ whose elements are some contant reals(independent of $\lambda)$. The matrix $m\times 1$ $c(\lambda)$ is given as $$c(\lambda) = A(\lambda)L$$

Can we show that each element of the matrix $c$, that is say $c_j(\lambda)$ is of the form $$c_j(\lambda) = \gamma_j\lambda + \kappa_{j0} + \frac{\kappa_{j1}}{\lambda} + \frac{\kappa_{j2}}{\lambda^2}+...$$ where $\gamma_j,\kappa_{j0},\kappa_{j1},\kappa_{j2},...\in \mathbb{R}$ are constants independent of $\lambda$

PS : At the end of the day, I am interested in the asymptotic behavior as $\lambda\to\infty$

My attempt

As $A(\lambda)$ is pd, the eigenvectors of this matrix form a basis for $\mathbb{R}^m$. So expanding the constant matrix $L$ along this basis we have $$L = a_1e_1(\lambda) + a_2e_2(\lambda) + ...+a_me_m(\lambda)$$ where $e_1(\lambda),e_2(\lambda),..e_m(\lambda)$ are the eigenvectors of the matrix $A(\lambda)$ and $a_1,a_2,...a_m \in \mathbb{R}$ are some constants.

Substituting this in equation $c = A(\lambda)L$ we get $$c = \sum_{i=1}^ma_1\rho_i(A(\lambda))e_i(\lambda)$$ So it follows from here that $$c_j(\lambda) = \gamma_j\lambda + \kappa_{j0} + \frac{\kappa_{j1}}{\lambda} + \frac{\kappa_{j2}}{\lambda^2}+...$$

unless the eigenvectors $e_i(\lambda)$ behave strange. All we know is $\|e_i(\lambda)\|_2 = 1$. I am not sure if this is enough to complete the proof or there need to be any more assumptions.

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Of course, without assumptions on the behavior of the eigenvectors, your desired conclusion will not hold. E.g., for $t:=\lambda$, let $$A(t):=\left( \begin{array}{cc} 2+\cos t & \sin t \\ \sin t & 2-\cos t \\ \end{array} \right),\quad L:=\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right).$$ The eigenvalues of $A(t)$ are $3$ and $1$ for all $t$, whereas $$AL=\left( \begin{array}{c} 2+\cos t \\ \sin t \\ \end{array} \right),$$ and $\sin t$ cannot be represented as $at+b+c/t+d/t^2+\cdots$ for any constants $a,b,c,d,\dots$.

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  • $\begingroup$ @losifPinelis : Thanks for the example. Looks like the eigenvectors should converge elementwise. $\endgroup$ – Rajesh D Jul 31 '20 at 16:28

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