Prescribed solutions to a Riccati-type matrix equation

Question

I would like to find a matrix (or ideally all matrices) $\Sigma$ such that $$ (1) \quad P + (P \Sigma) + (P \Sigma)^T + (P \Sigma H) P^{-1} (P \Sigma H)^T \leq 0, $$ where '$\leq$' denotes negative semi-definiteness and $P$ and $H$ are prescribed symmetric positive definite square matrices. This is a similar to the Riccati equation $$ (2) \quad Q + (XA) + (XA)^T - (XB) R^{-1} (XB)^T \leq 0. $$ In fact (1) is obtained from (2) by letting \begin{align} P &= Q = X = -R \\ \Sigma &= A \\ \Sigma H &= B. \end{align} Most literature about the Riccati equation seems to be focused on finding $X$ given $A$, $B$ and $R$, or in my case, finding $P$ given $\Sigma$ and $H$. But what can be said about the converse? Given $P$ and $H$, does a $\Sigma$ exist such that (1) is satisfied? If so, how can it be found?

Answer 1

Let $Q=HP^{-1}H$ and $X=Q^{1/2}P\Sigma Q^{1/2}$. Left- and right-multiply your equation $(1)$ by $Q^{1/2}$, we get $Q^{1/2}PQ^{1/2} + X + X^T + XX^T\le0$, i.e. $$ I - Q^{1/2}PQ^{1/2} \ge (I+X)(I+X)^T.\tag{3} $$ Since the RHS of $(3)$ is positive semidefinite, the equation is solvable only when the spectral norm/radius of $Q^{1/2}PQ^{1/2}$ is at most $1$. If this is the case and we orthogonally diagonalise $I-Q^{1/2}PQ^{1/2}$ as $USU^T$, then any matrix of the form $X=UDV^T-I$, where $V$ is any real orthogonal matrix and $D$ is any diagonal matrix that satisfies $0\le D\le S^{1/2}$, is a solution to $(3)$, so that $\Sigma=P^{-1}Q^{-1/2}XQ^{-1/2}$ solves your equation.