We will denote \(\mathbb S_n,\,\mathbb S_n^+,\mathbb S_n^{++}\) to be the set of all real symmetric matrices, the set of all positive semi-definite matrices and the set of all positive definite matrices respectively. For any two matrices \(A,\,B\in \mathbb S_n\), we call \(A \geq B\) (\(A > B\)) if \(A - B\) is positive semi-definite (positive definite).

Let \(A\in \mathbb R^{n\times n}\) and \(C \in \mathbb R^{m\times n}\) be two matrices. Further define \(Q \in \mathbb S_n\) and \(R\in\mathbb S_n\) to be two positive definite matrices. The DARE function \(g: \mathbb S_n^+\rightarrow\mathbb S_n^+\) is defined as

\begin{equation} \label{eq:riccati1} g(X) \triangleq \left[\left(A X A' + Q\right)^{-1} + C'R^{-1}C\right]^{-1}. \end{equation}

From Equation \eqref{eq:riccati1}, we can see that \(g\) is monotonically increasing with respect to \(X\).

Using the matrix inversion lemma, we can write \(g\) as

\begin{equation} \label{eq:riccati2} g(X) = h(X) - h(X) C' \left(C h(X) C'+R\right)^{-1}Ch(X), \end{equation}

where \(h(X) = AXA' + Q\). For simplicity, we will define

\begin{equation} \tilde g(X) = X - X C' \left(C X C'+R\right)^{-1}CX. \end{equation}

Hence, \(g = \tilde g \circ h\).

We only need to show that \(\tilde g\) is concave since \(h\) is linear with respect to \(X\). Consider the following function

\begin{align} \label{eq:linear} \phi (K,X) &\triangleq (I-KC) X (I-KC)' + KRK'\\ & = X - KCX - XC'K' + K(CXC'+R)K'. \end{align}

By completing the square, \(\phi(K,X)\) can be rewritten as

\begin{equation} \label{eq:linear2} \phi(K,X) = \tilde g(X) + (K-K_*) (CXC'+R)^{-1}(K-K_*)', \end{equation}

where \(K_* = XC'(CXC'+R)^{-1}\). Therefore,

\begin{align} \label{eq:riccatilinear} \tilde g(X) = \min_{K} \phi(K,X). \end{align}

Since \(\phi(K,X)\) is linear with respect to \(X\) for a fixed \(K\), \(\tilde g(X)\) is concave with respect to \(X\).

It is well known that if the matrix pair \((A,C)\) is detectable, then the equation \(X = g(X)\) has a unique positive definite solution, which we will denote as \(X_*\).

Let \(X_0 = 0\) and denote \(X_{k+1} = g(X_k)\). Clearly \(X_0 \leq X_*\) and \(X_0 \leq X_1\). Now apply \(g\) on both sides of the inequalities, we have \[ g(X_0) \leq g(X_*),\,g(X_0)\leq g(X_1). \] Notice that \(g(X_0) = X_1,\,g(X_1) = X_2\) and \(g(X_*) = X_*\). Therefore, we have the following chain of inequalities \[ X_0 \leq X_1 \leq X_2 \leq \dots \leq X_*. \]

Since \(\{X_k\}\) is an increasing sequence with an upper bound, \(\{X_k\}\) has a limit, which we can denote as \(X_\infty\). Now by the fact that \(g\) is continuous, \(X_\infty = g(X_\infty)\). Finally, be the fact that \(X_*\) is the unique fixed point, we have that \(X_\infty = X_*\).

Similar trick can be applied to the case where we can find an \(X_1 = g(X_0) \leq X_0\). In that case, we know that \(X_*\) must be no greater than \(X_0\).

Therefore, we can establish the equivalency of the following statements:

1. \(X_* \leq P\)
2. There exists an \(X_0 \leq P\), such that \(X_1 = g(X_0) \leq X_0\).

We already proved that Statement 2 implies 1. On the other hand, if Statement 1 holds, then \(X_* = g(X_*)\leq X_*\).

Now if we define \(S = X_0^{-1}\), then \(X_0 \leq P\) is equivalent to \(S \geq P^{-1}\). Moreover, \(g(X_0)\leq X_0\) is equivalent to

\begin{align} \label{eq:fixpointequivalence} (A X A' + Q)^{-1} + C'R^{-1}C \geq S \end{align}

Using matrix inversion lemma on the LFS of \eqref{eq:fixpointequivalence}, we have \[ Q^{-1} - Q^{-1}A(S + A'Q^{-1}A)^{-1}A'Q^{-1} + C'R^{-1}C\geq S, \] which is equivalent to

\begin{align} \label{eq:fixpointequivalence2} Q^{-1} - S + C'R^{-1}C \geq Q^{-1}A(S + A'Q^{-1}A)^{-1}A'Q^{-1}. \end{align}

Now using Schur's complement, \eqref{eq:fixpointequivalence2} is equivalent to the following linear matrix inequality:

\begin{align} \label{eq:lmi} \begin{bmatrix} Q^{-1} - S + C'R^{-1}C&Q^{-1}A\\ A'Q^{-1}& S+A'Q^{-1}A \end{bmatrix} \geq 0. \end{align}

As a result, the condition that \(X_* \leq P\) is equivalent to \eqref{eq:lmi} and \(S\geq P^{-1}\). Notice that the LFS of \eqref{eq:lmi} is linear with respect to \(C'R^{-1}C\). This fact can leveraged for sensor placement problems.

Consider the following functions:

\begin{align} \label{eq:linearfractional} f_1(x) = \frac{a_1x+b_1}{c_1x+d_1}, f_2(x) = \frac{a_2x+b_2}{c_2x+d_2}. \end{align}

Then

\begin{align} \label{eq:linearfractionalcomposition} f_1(f_2(x)) &= \frac{a_1 \frac{a_2x+b_2}{c_2x+d_2} + b_1}{c_1 \frac{a_2x+b_2}{c_2x+d_2} + d_1}\\ &=\frac{(a_1a_2+b_1c_2)x + a_1b_2 + b_1d_2}{(c_1a_2+d_1c_2)x + c_1b_2 + d_1d_2}. \end{align}

One important observation is that the coefficients on the numerator and the denominator can be computed via matrix multiplication:

\begin{align} \begin{bmatrix} a_1&b_1\\ c_1&d_1 \end{bmatrix}\times\begin{bmatrix} a_2&b_2\\ c_2&d_2 \end{bmatrix} = \begin{bmatrix} a_1a_2+b_1c_2&a_1b_2+b_1d_2\\ c_1a_2+d_1c_2&c_1b_2+d_1d_2 \end{bmatrix}. \end{align}

We will now move to the case where \(X\) is a matrix instead of just a scalar. Let us consider the following matrix: \[ \Phi = \begin{bmatrix} \Phi_{11}&\Phi_{12} \\ \Phi_{21}&\Phi_{22} \end{bmatrix} \in \mathbb R^{2n\times 2n}, \] where each \(\Phi_{ij}\in \mathbb R^{n\times n}\). Define the following operator

\begin{align} \label{eq:homographic} F(\Phi,X) \triangleq (\Phi_{11}X+\Phi_{12})(\Phi_{21}X+\Phi_{22})^{-1}. \end{align}

Similarly for any \(\Phi_1,\,\Phi_2 \in \mathbb R^{2n\times 2n}\), we have

\begin{align} \label{eq:homographiccomposition} F(\Phi_1,F(\Phi_2,X)) = F(\Phi_1\Phi_2,X). \end{align}

We now establish the link between linear fractional transformations and Riccati equations. We will assume that \(A\) is invertible. We know that

\begin{align} \label{eq:riccatidecompose1} AXA' &= F\left(\begin{bmatrix} A & 0\\ 0 & (A')^{-1} \end{bmatrix}, X\right),\\ \label{eq:riccatidecompose2} (X + Q)^{-1} &= F\left(\begin{bmatrix} 0 & I\\ I & Q \end{bmatrix}, X\right),\\ \label{eq:riccatidecompose3} (X + C'R^{-1}C)^{-1} &= F\left(\begin{bmatrix} 0 & I\\ I & C'R^{-1}C \end{bmatrix}, X\right). \end{align}

As a result, using \eqref{eq:homographiccomposition}, we have

\begin{align} \label{eq:riccatihomo} g(X) = F(\Phi_g,X), \end{align}

where

\begin{align} \label{eq:Phimatrixg} \Phi_g \triangleq \begin{bmatrix} A&Q(A')^{-1} \\ C^TR^{-1}CA&(I+C^TR^{-1}CQ)(A')^{-1} \end{bmatrix}. \end{align}

Hence, \(g^{(n)}(X) = F(\Phi_g^n,X)\).

We will simply state several key theorems in this paper.

Consider the following function \(\delta: \mathbb S_n^{++}\times \mathbb S_n^{++}\rightarrow \mathbb R\) defined as

\begin{align} \label{eq:metric} \delta(X,Y) = \sqrt{\sum_{i=1}^n \left(\log \lambda_i\right)^2}, \end{align}

where \(\lambda_1,\dots,\lambda_n\) are the eigenvalues of \(XY^{-1}\). One can prove that it is a metric on the space of \(\mathbb S_n^{++}\). The main property of this metric is that it is invariant under conjugacy and inversion, i.e.,

\begin{align} \label{eq:invariance} \delta(X^{-1},Y^{-1}) =\delta(X,Y) = \delta (AXA',AYA'). \end{align}

Moreover, if a positive semi-definite matrix \(P \geq 0\) and the smallest eigenvalue of \(P\) is \(\beta\), then

\begin{align} \label{eq:contraction} \delta(X+P,Y+P)\leq \frac{\alpha}{\alpha+\beta}\delta(X,Y) \leq \delta(X,Y). \end{align}

where \(\alpha\) is the largest eigenvalue of \(X\) and \(Y\).

Therefore, from \eqref{eq:riccatidecompose1}, \eqref{eq:riccatidecompose2} and \eqref{eq:riccatidecompose3} we know that

\begin{align} \label{eq:riccaticontraction} \delta(g(X),g(Y))\leq \delta(X,Y). \end{align}

It can be further proved that if \((A,C)\) is observable, then there exists a \(k\) and \(0 < \rho < 1\), such that

\begin{align} \label{eq:riccaticontraction2} \delta(g^{(k)}(X),g^{(k)}(Y))\leq\rho \delta(X,Y). \end{align}