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.



Asymptotic behavior of a matrix equation and its eigenvalues
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 $a,b,c,d,\dots$.


