MATH319 Slides

130 A solution of Lyapunov’s equation

Corollary

Suppose that A is a real square matrix such all its eigenvalues are in the open left half plane {λ𝐂:λ<0}. Then for all positive definite P, there exists a positive definite K such that

AK+KA=-P.

Note that exp(tA)Pexp(tA) is positive definite by exercise A3.2. Indeed exp(tA)Pexp(tA)Y,Y=Pexp(tA)Y,exp(tA)Y is positive and continuous for t>0 and Y0. Hence

K=0exp(tA)Pexp(tA)𝑑t

is also positive definite. [MATLAB gives K=lyap(A,P).]