Consider the following Sylvester equation, where each of the known coefficient matrices ($A$, $B$, $C$) is symmetric positive definite and has dimensions $n \times n$

\begin{align*} C = A^TXA + B^TXB. \end{align*}

In my case, the coefficient matrices are such that the equation has a unique real solution which is also symmetric.

I have been trying to prove that the solution must also be positive definite and I am struggling. The struggle makes me think that this is actually not true.

I would be grateful for either a pointer to related literature, a way to prove my hypothesis or a counterexample. If it is not true in general, are there any known conditions on $A$ and $B$ which make this true?


1 Answer 1


This is not true. For example, if $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}, B = \begin{bmatrix} 4 & 0 \\ 0 & 1 \end{bmatrix}, C = \begin{bmatrix} 17 & 16 \\ 16 & 17 \end{bmatrix} $$ then each of $A$, $B$, and $C$ is symmetric and positive definite. However, it is straightforward to check that the unique solution to the Sylvester equation is $$ X = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}, $$ which is not positive (semi)definite.


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