Prove the Lyapunov equation

1. Apr 20, 2013

jarvisyang

The matrix $\mathbf{B}$satifies the following Lyapunov equation
$$\begin{gathered}\mathbf{A}^{T}\mathbf{B}\end{gathered}+\mathbf{BA}=-\mathbf{I}$$
prove that necessary and sufficient condition generating a symmetric and positive determined $\mathbf{B}$is that all of the eigen values of $\mathbf{A}$should be negative.
(Hints: rewritten $\mathbf{A}$in the Jordan normal form, one can easily prove the proposition)
But I still cannnot figure it out with the hints!Waiting for your excellent proof!

2. Apr 25, 2013

ekkilop

The sufficiency can be obtained by considering
$B=\int_{0}^{∞} e^{A^τ t} Q e^{A t} dt$

Inserting into the Lyapunov equation gives
$AB + BA^{T} = A \int_{0}^{∞} e^{A^τ t} Q e^{A t} dt + \int_{0}^{∞} e^{A^τ t} Q e^{A t} dt A^{T} = \int_{0}^{∞} \frac{d}{dt} (e^{A^τ t} Q e^{A t}) dt = [e^{A^τ t} Q e^{A t}]_{0}^{∞} = -Q$
since the eigenvalues of $A$ are negative. Now just let $Q=I$.