Forward-looking difference equation: closed-form solution to $\sum_{j=0}^{\infty}A^jB\Lambda^j$?

Question

Assume $$Z_t = A\mathbb{E}_tZ_{t+1} + B\nu_t$$ and $$\nu_t = \Lambda \nu_{t-1} + e_t$$

Where $A, B$, and $\Lambda$ are conformable square matrices and $e_t$ are IID and mean zero. $Z_t$ and $V_t$ are vectors.

Solving the first equation ahead yields $$Z_t = A^2\mathbb{E}_tZ_{t+2} + (AB\Lambda + B)v_t$$ $$ = A^k\mathbb{E}_tZ_{t+k} + \sum_{j=0}^{k-1}A^jB\Lambda^j v_t$$ $$ = \bigg(\sum_{j=0}^{\infty}A^jB\Lambda^j\bigg)v_t$$

Assuming the first term $A^k\mathbb{E}_tZ_{t+k}\rightarrow 0$ as $k\rightarrow\infty$

Is there a closed form solution for $\sum_{j=0}^{\infty}A^jB\Lambda^j$ ?

It looks like a geometric series, but I'm unsure of the solution give the matrix $B$ sandwiched between.

If it helps, in this case the matrix $\Lambda$ is diagonal.

Answer 1

Note that $\Lambda$ can be written as $$\Lambda=diag\{\lambda_1,\cdots,\lambda_n\}$$ and therefore $$\Lambda^j=diag\{\lambda_1^j,\cdots,\lambda_n^j\}$$ or equivalently $$\Lambda^j=\sum_{i=1}^n{\lambda_i^j e_ie_i^T}$$ where $e_i$ is the $i$-th column of the identity matrix. Then

$$\sum_{j=1}^{\infty}{A^jB\Lambda^j}=\sum_{j=1}^{\infty}{A^jB\sum_{i=1}^n{\lambda_i^j e_ie_i^T}}=\sum_{i=1}^n{\left(\sum_{j=1}^{\infty}{\lambda_i^jA^j}\right)Be_ie_i^T}=\sum_{i=1}^n{\left(I-\lambda_iA\right)^{-1}Be_ie_i^T}$$

as long as $\|\lambda_iA\|< 1$ for all $i=1,\cdots,n$.