* arrange the resulting equation and the second equation of {eq}`lag-lqdp-eq2` into the form

$$

L\ \begin{pmatrix}x_{t+1}\cr \mu_{t+1}\cr\end{pmatrix}\ = \ N\ \begin{pmatrix}x_t\cr \mu_t\cr\end{pmatrix}\

L\ \begin{bmatrix}x_{t+1}\cr \mu_{t+1}\cr\end{bmatrix}\ = \ N\ \begin{bmatrix}x_t\cr \mu_t\cr\end{bmatrix}\

,\ t \geq 0,

$$ (eq:systosolve)

where

$$

L = \ \begin{pmatrix}I & BQ^{-1} B^\prime \cr 0 & A^\prime\cr\end{pmatrix}, \quad N = \

\begin{pmatrix}A & 0\cr -R & I\cr\end{pmatrix}.

L = \ \begin{bmatrix}I & BQ^{-1} B^\prime \cr 0 & A^\prime\cr\end{bmatrix}, \quad N = \

\begin{bmatrix}A & 0\cr -R & I\cr\end{bmatrix}.

$$

When $L$ is of full rank (i.e., when $A$ is of full rank), we can write

system {eq}`eq:systosolve` as

$$

\begin{pmatrix}x_{t+1}\cr \mu_{t+1}\cr\end{pmatrix}\ = M\ \begin{pmatrix}x_t\cr\mu_t\cr\end{pmatrix}

\begin{bmatrix}x_{t+1}\cr \mu_{t+1}\cr\end{bmatrix}\ = M\ \begin{bmatrix}x_t\cr\mu_t\cr\end{bmatrix}

$$ (eq4orig)

where

$$

M\equiv L^{-1} N = \begin{pmatrix}A+B Q^{-1} B^\prime A^{\prime-1}R &

-B Q^{-1} B^\prime A^{\prime-1}\cr -A^{\prime -1} R & A^{\prime -1}\cr\end{pmatrix}.

M\equiv L^{-1} N = \begin{bmatrix}A+B Q^{-1} B^\prime A^{\prime-1}R &

-B Q^{-1} B^\prime A^{\prime-1}\cr -A^{\prime -1} R & A^{\prime -1}\cr\end{bmatrix}.

$$ (Mdefn)

+++

It helps to introduce a $(2n \times 2n)$ matrix

$$

J = \begin{pmatrix}0 & -I_n\cr I_n & 0\cr\end{pmatrix}.

J = \begin{bmatrix}0 & -I_n\cr I_n & 0\cr\end{bmatrix}.

$$

The rank of $J$ is $2n$.

y_{t+1} = M y_t

$$ (eq658)

where $y_t = \begin{pmatrix}x_t\cr \mu_t\cr\end{pmatrix}$.

where $y_t = \begin{bmatrix}x_t\cr \mu_t\cr\end{bmatrix}$.

Consider a **triangularization** of $M$

$$

V^{-1} M V= \begin{pmatrix}W_{11} & W_{12} \cr 0 & W_{22}\cr\end{pmatrix}

V^{-1} M V= \begin{bmatrix}W_{11} & W_{12} \cr 0 & W_{22}\cr\end{bmatrix}

$$ (eqn:triangledecomp)

where

Write equation {eq}`eq6510` as

$$

\begin{pmatrix}y^\ast_{1t}\cr y^\ast_{2t}\cr\end{pmatrix}\ =\ \left[\begin{matrix} W^t_{11} &

W_{12, t}\cr 0 & W^t_{22}\cr\end{matrix}\right]\quad \begin{pmatrix}y^\ast_{10}\cr

y^\ast_{20}\cr\end{pmatrix}

\begin{bmatrix}y^\ast_{1t}\cr y^\ast_{2t}\cr\end{bmatrix}\ =\ \left[\begin{matrix} W^t_{11} &

W_{12, t}\cr 0 & W^t_{22}\cr\end{matrix}\right]\quad \begin{bmatrix}y^\ast_{10}\cr

y^\ast_{20}\cr\end{bmatrix}

$$

where $y^\ast_t = V^{-1} y_t$, and in particular where

the inverse of $V,$ it follows that

$$

(V^{21} \ V^{22})\quad \begin{pmatrix}V_{11}\cr V_{21}\cr\end{pmatrix} = 0

(V^{21} \ V^{22})\quad \begin{bmatrix}V_{11}\cr V_{21}\cr\end{bmatrix} = 0

$$

which implies