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