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Let's say I have the following LTI system: $$\dot{x}(t) = \mathbf{A} x(t) + \mathbf{B} u(t)$$ I need to somehow show the following is true or false (proof):

This system is controllable if and only if for any $\mu$ such that $(−\mu\mathbf{I − A})$ is Hurwitz, there exists a unique positive definite solution $\mathbf W$ to the Lyapunov equation:

$$\mathbf A \mathbf W + \mathbf W \mathbf A^T − \mathbf B\mathbf B^T = −2\mu\mathbf W$$

Can anyone see the proof? I tried replacing $\mathbf A$ in the Lyapunov equation with $(−\mu\mathbf I − \mathbf A)$, and then adding the $−2\mu\mathbf W$, to the left side, but the $\mu$'s end up cancelling out, so I'm not sure. I'd really appreciate any help.

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  • $\begingroup$ What is your definition of controllable? $\endgroup$ – Peter K. Apr 22 '16 at 2:06
  • $\begingroup$ Well normally I would say that the controllability matrix (which has the form R =[B, AB, A^2B, ....A^(n-1)B]) is full rank. $\endgroup$ – John Alperto Apr 22 '16 at 2:14
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The required step you were looking for seems to be grouping each $\mu W$ with $A$. Denote the inverse of $W$ with $P$. Then pre and post multiply the Lyapunov equation with P. Then

$$ 0 = (\mu I +A)W + W(\mu I + A)^T - BB^T $$ From here you can negate the equation and use the textbook arguments about the grammians of controllability. Hurwitz property implies the positive definiteness of $W$ and vice versa.

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Not really a direct help, but this paper might provide some insight, particularly the paragraph below.

enter image description here

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