So I am interested in finding how many $X = X^\top$ there are that solve

$$ A^\top X + X\,A + Q - X\,B\,R^{-1}B^\top X = 0 \tag{1} $$

with $A,Q,P \in \mathbb{R}^{n \times n}$, $B\in\mathbb{R}^{n \times m}$, $R\in\mathbb{R}^{m \times m}$, $Q = Q^\top \succeq 0 $ and $R = R^\top \succ 0$.

I know that if the pair $(A, B)$ is stabilizable, then there should be at least two solutions; one positive definite and the other one negative definite. In MATLAB the positive definite solution can be found numerically using care(A,B,Q,R). By negating $A$ and $X$ in $(1)$ the negative definite solution can be obtained numerically using -care(-A,B,Q,R).

I also seemed to be able to consistently find another solution by iteratively solving the following Sylvester equation

$$ A^\top X_{n+1} + X_{n+1}\,A + Q - X_{n}\,B\,R^{-1}B^\top X_{n+1} = 0 \tag{2} $$

so assume that $X_{n}$ is constant when solving for $X_{n+1}$. The initial value for $X_{0}$ did not seemed to matter to which value it eventually converged to. Using this iterative method I was also able to find another solution by instead solving for the inverse of $X$. So solving the following Sylvester equation each iteration instead

$$ X^{-1}_{n+1}\,A^\top + A\,X^{-1}_{n+1} + X^{-1}_{n}\,Q\,X^{-1}_{n+1} - B\,R^{-1}B^\top = 0 \tag{3} $$

and in the end take the inverse of the final $X^{-1}_n$. Using the negation trick, as done when using care, did not yield additional solutions when iterating using either $(2)$ or $(3)$.

For example for

$$ A = \begin{bmatrix} 0 & 1 \\ 2 & -3 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad Q = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, \quad R = 1 $$

I found the following four solutions using the techniques described above

$$ X_1 = \begin{bmatrix} 15.8672 & 4.2361 \\ 4.2361 & 1.4127 \end{bmatrix}, \quad X_2 = \begin{bmatrix} -3.8672 & 4.2361 \\ 4.2361 & -7.4127 \end{bmatrix}, $$

$$ X_3 = \begin{bmatrix} -1.2553 & -0.2361 \\ -0.2361 & 0.2447 \end{bmatrix}, \quad X_4 = \begin{bmatrix} 13.2553 & -0.2361 \\ -0.2361 & -6.2447 \end{bmatrix}. $$

When $n = 1$ then equation $(1)$ is just a quadratic equation and should therefore only have at most two unique solutions. But for $n\geq2$ I have shown that there can be at least four solutions. But can there be more and does this number increase as $n$ gets bigger? Would removing the constraint on $Q$ and $R$ that they need to be positive (semi-)definite change this?

  • 1
    $\begingroup$ I suggest you take a look into Willems' classical paper "Least Squares Stationary Optimal Control and the Algebraic Riccati Equation" which in my opinion anyone interested in control should have read. For you chapter 5 is particularly relevant (although I recommend reading the whole paper). In chapter 5, you will also find an example (Example 2) for a 2-D ARE with infinitely many solutions. $\endgroup$
    – Nukular
    Nov 5, 2017 at 10:07
  • $\begingroup$ ... sorry, I meant chapter 6 $\endgroup$
    – Nukular
    Nov 5, 2017 at 10:48

2 Answers 2


Ok, actually this started off as a comment but now that I've read my own paper suggestion in more detail, I can even give you an answer.

You find the answer to your question in Willems' "Least Squares Stationary Optimal Control and the Algebraic Riccati Equation", Remark 18:

  • There exist infinitely many solutions if and only if the minimal polynomial of $A^+$ (which is the closed-loop matrix when using the stabilizing solution of the ARE for feedback) is nonequal to its characteristic polynomial.

  • Otherwise, there exist at most $2^n$ solutions. There may be less, namely if $A^+$ has complex and/or repeated eigenvalues

  • For single-input systems, a sufficient condition for finiteness of the number of solutions is controllability

For more details and the proofs I refer to Willems. And I really recommend reading this amazing paper which is in my opinion among the best control papers of all time.

  • $\begingroup$ I assume you are referring to this paper? $\endgroup$ Nov 5, 2017 at 17:40
  • $\begingroup$ Yes, I hope you can get access to it somehow...? $\endgroup$
    – Nukular
    Nov 5, 2017 at 20:39

If minimality (hence controllability and observability) is assumed then all solutions of the Riccati equation is given by the eigenspace of the associated Hamiltonian.

So if you rewrite the equation

$$A^\top X + X\,A + Q - X\,B\,R^{-1}B^\top X = 0$$


$$\begin{bmatrix}X &I\end{bmatrix} \begin{bmatrix}A &-BR^{-1}B^\top \\-Q & -A^\top\end{bmatrix} \begin{bmatrix}I \\X\end{bmatrix} = 0 $$ the solutions are the positive definite, negative definite and indefinite combinations of the eigenvectors of the hamiltonian. That's where the $2^n$ in Nukular's answer comes from. Since Hamiltonians have a symmetric eigenvalue spectrum with respect to the imaginary axis if there are positive definite solutions there are also negative definite solutions. If there are eigenvalues on the imaginary axis then ARE admits no solutions (or infinite solutions depending on your timezone).

  • $\begingroup$ If the Hamiltonian has $2\,n$ unique (symmetric) eigenvalues, wouldn't that imply there are $2\,n$ choose $n$ solutions (which is more than $2^n$ for $n>1$). These solutions don't all have to be symmetric, but it is not obvious to me why there would be at most $2^n$ symmetric solutions. $\endgroup$ Jan 5, 2021 at 15:45

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