Multiplicative factorization of stochastic growth time series--solving for an eigenfunction/eigenvector I'm trying to understand the tools used/presented in Lars Hansen's Econometrica paper "Dynamic Valuation Decomposition within Stochastic Economies." In a part in the paper, Hansen introduces a long-term approximation. This approximation first requires us to solve for a principle eigenfunction of a kind of conditional expectation operator. As an exercise, I'm trying to solve the following example.
$\newcommand{\E}{\mathbb E}$
Suppose that
$$
X_{t+1} = A X_t + B W_{t+1}
$$
where A has stable eigenvalues and $\{W_{t+1} : t = 0,1,... \}$ is an iid sequence of multivariate standard normally distributed random vectors. Suppose that
$$
\log M_{t+1} - \log M_t = D \cdot X_t + F \cdot W_{t+1}.
$$
I want to show that there exists a solution to the equation
$$
\E \left [ \frac{M_{t+1}}{M_t} e(X_{t+1}) \mid X_t = x \right ] = \exp(\eta) e(x) \tag{1}
$$
for $\log e(x) = H \cdot x.$ I also want to compute that value of $\eta$.

Progress:

The beginning of the problem is straightforward. Using the properties of log-normal random variables, we can calculate from (1)
\begin{align}
\exp\{ (D' + H'A) x + \frac 12 (F' + H' B)'(F' + H'B)\} &amp;= \exp\{\eta + H' x\} \\
(D' + H'(A - I)) x + \frac 12 (F' + H'B)'(F' + H'B) &amp;= \eta.
\end{align}
This is probably a simple question from here. It seems clear that a solution exists, and I have solved for $\eta.$ But from what I understand, solving this problem involves making a "guess" that $e$ was linear in $x$. I don't know what $H$ is. Thus, I need to solve for $H$ and $\eta$. How would I do this?

Note: This problem is related to this problem about the "Decomposition of an additive functional into a Martingale part and other."
The progress given so far appears to be correct. Finishing this problem just requires us to argue that $H$ must be chosen so that the equation
$$
(D' + H'(A - I)) x + (F + H'B)'(F + H'B) = \eta
$$
holds for all $x$. Thus, we must have $D' + H'(A - I) = 0$ and, consequently,
$H' = D'(I-A)^{-1}$. This implies that $\eta = (F' + H'B)'(F' + H'B)$.