Problem with understanding theorem on Riccati Equation.

Question

`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} $.

Answer 1

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.

Answer 2

$J$ 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.