Problem with understanding theorem on Riccati Equation. `The matrices $A,B,C,D,X$ are real, square, $n \times n$.
I have trouble understanding theorem 7.1.2 from Lancaster & Rodman "Algebraic Riccati Equations".
The part that I understand is as follows.
A matrix $X$ solves equation $XBX + XA -DX - C = 0$ iff we can find a matrix $Z$ such that we have $ \left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] \left[ \begin{array}{c} I \\ X \end{array} \right] = \left[ \begin{array}{c} I \\ X \end{array} \right] Z$.
The part I don't understand is this.
Denote $T = \left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] $.
Consider the Jordan decomposition of $T$: $T= W J W^{-1}$.
Denote a matrix containing a subset of Jordan chains of $T$ (a subset of columns of $W$ such that each chain is either completely included or completely excluded) as as $V = \left[ \begin{array}{c} Y \\ Z \end{array} \right]$. Denote the corresponding sub-matrix of $J$ as $J_s$. Assume furthermore that $Y$ is invertible.
We have $TV = VJ_s$, i.e. $ \left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] \left[ \begin{array}{c} Y \\ Z \end{array} \right] = \left[ \begin{array}{c} Y \\ Z \end{array} \right] J_s  $.
Now the bit where I get lost is Theorem 7.1.2, which seems to imply (maybe I understand it wrong) the following.
$ \left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] \left[ \begin{array}{c} Y \\ Z \end{array} \right] = \left[ \begin{array}{c} Y \\ Z \end{array} \right] J_s  \quad \Longrightarrow \quad \exists Z'.\;  \left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] \left[ \begin{array}{c} I \\ ZY^{-1} \end{array} \right] = \left[ \begin{array}{c} I \\ ZY^{-1} \end{array} \right] Z'$
Edit: Thanks to the answer, I now see that a good choice of $Z'$ is $Y J_s Y^{-1} $.
 A: It might help if you quoted Theorem 7.1.2 for those of us who don't have Lancaster and Rodman, but it seems pretty clear: just take $Z' = Y J Y^{-1}$
(this $Z'$ is, of course, not supposed to be the transpose of $Z$).
Of course $Y$ must be invertible for this to make sense.
A: $J$ is not the Jordan block in this context. If it was, there would be dimension mismatch. Since $T$ is $2n \times 2n$, so is $V$. Therefore, $Y$ or $Z$ could not be square.
I believe the authors tried to prove this: Suppose the Riccati equation has a solution. We need to show that there exists matrices $Y$, $Z$ and $J$ with $Y$ is invertible, such that
$$\left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] \left[ \begin{array}{c} Y \\ Z \end{array} \right] = \left[ \begin{array}{c} Y \\ Z \end{array} \right] J$$
Rewriting the equations
$$\begin{align}
AX + BY &= YJ \\
Y^{-1}AX + Y^{-1}BY &= J \\
\\
CY + DZ &= ZJ \\
 &= ZY^{-1}AY + ZY^{-1}BZ \\
C + DZY^{-1} &= ZY^{-1}A  + ZY^{-1}BZY^{-1}
\end{align}$$
Hence, $ZY^{-1}$ is a solution to the Riccati equation, so these matrices exist.
