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Prove the Lyapunov equation

  1. Apr 20, 2013 #1
    The matrix [itex]\mathbf{B}[/itex]satifies the following Lyapunov equation
    [tex]\begin{gathered}\mathbf{A}^{T}\mathbf{B}\end{gathered}+\mathbf{BA}=-\mathbf{I}[/tex]
    prove that necessary and sufficient condition generating a symmetric and positive determined [itex]\mathbf{B}[/itex]is that all of the eigen values of [itex]\mathbf{A}[/itex]should be negative.
    (Hints: rewritten [itex]\mathbf{A}[/itex]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. jcsd
  3. Apr 25, 2013 #2
    The sufficiency can be obtained by considering
    [itex] B=\int_{0}^{∞} e^{A^τ t} Q e^{A t} dt [/itex]

    Inserting into the Lyapunov equation gives
    [itex] 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 [/itex]
    since the eigenvalues of [itex]A[/itex] are negative. Now just let [itex]Q=I[/itex].
     
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