# Asymptotic behavior of a matrix equation and its eigenvalues

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.

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 $$a,b,c,d,\dots$$.