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  "body": "\n## Definition\n\nLet $n, m \\ge 1$.\nLet $A$ be a real $n \\times n$ [[matrix|0VD9ZSWWJG]] and $B$ a real $n \\times m$ matrix.\nLet $Q$ be a real [[symmetric|0VR671PHE5]] [[positive semidefinite|01MMVW8CKC]] $n \\times n$ matrix.\nLet $R$ be a real symmetric [[positive definite|0KX66CZMBB]] $m \\times m$ matrix.\nWrite $M^T$ for the [[transpose|09DXWYRD43]] of $M$.\n\n1. The *continuous-time algebraic Riccati equation* is\n$$A^T X + XA - XBR^{-1}B^T X + Q = 0$$\nfor an unknown real symmetric $n \\times n$ matrix $X$.\n2. The *discrete-time algebraic Riccati equation* is\n$$X = A^T X A - A^T X B (R + B^T X B)^{-1} B^T X A + Q$$\nfor an unknown real symmetric positive semidefinite $n \\times n$ matrix $X$.\n\n## Remarks\n\nA solution $X$ of the continuous equation is *stabilising* if every eigenvalue of $A - BR^{-1}B^T X$ has negative real part.\nIf the pair $(A, B)$ is stabilisable and the pair $(Q^{1/2}, A)$ is detectable, the continuous equation has a unique positive semidefinite solution, and it is stabilising.\nThe optimal feedback of the linear-quadratic regulator is then $u = -R^{-1}B^T X x$.\nExample: for $n = m = 1$, $A = 0$ and $B = Q = R = 1$, the continuous equation is $1 - X^2 = 0$, with solutions $X = 1$ and $X = -1$.\nOnly $X = 1$ is stabilising, since $A - BR^{-1}B^T X = -X$.\nFor $B = 0$ the continuous equation becomes the [[Lyapunov equation|92GVPZH9FT]] $A^T X + XA + Q = 0$, which is linear and not of Riccati type.\n",
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