To establish the sign of the real part of the eigenvalues of a real square matrix $A$, we usually try to find a symmetric positive definite matrix $P$ verifying the matrix inequality $A^\top P + PA \prec 0$.
For example, let $$A=\left( \begin{array}{cc} 0 & -1 \\ 1 & -1 \end{array}\right)$$ and we look for a solution $P\succ0$ such that $$A^\top P+PA = \left( \begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right).$$
As the solution $$P=\left( \begin{array}{cc} 1.5 & -0.5 \\ -0.5 & 1 \end{array}\right)$$ is positive definite, then this implies that $A$ has eigenvalues with negative real part.
Can we generalize this result for any dissipative matrix?
I will use the definition that a matrix $A$ defines is dissipative if and only if $A+A^T$ does not have positive eigenvalues. This is equivalent to saying that $A+A^T$ is Hurwitz stable. Therefore, one can characterize dissipative matrices $A$ in terms of the existence of a positive definite matrix $P$ such that $(A+A^T)P+P(A+A^T)\prec0$.