Is it true that the state-space realization above is minimal if and only if there exists a real $\alpha$ for which the following two matrix equations have symmetric and positive definite solutions P and W? Prove or give counter example.

$A^{T}P + PA + \alpha P= -BB^{T}$

$AW + WA^{T} + \alpha W= -C^{T}C$

I know that a system is minimal if and only if it is controllable and observable. My first instinct was to use the PBH test for controllability and observability but the form of the equations does not lend itself to that test. Thoughts?

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    $\begingroup$ Think Grammians & Lyapunov equations and write the equations as $(A+{\alpha \over 2})^T + P (A+{\alpha \over 2}) + BB^T=0$, etc, and choose $\alpha$ such that the resulting matrix is Hurwitz. $\endgroup$ – copper.hat Dec 17 '16 at 19:14
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    $\begingroup$ Note that $(A,B)$ is cc. iff $(A-\lambda I ,B)$ is cc. for any $\lambda$. $\endgroup$ – copper.hat Dec 17 '16 at 19:21
  • $\begingroup$ cc. = completely controllable. $\endgroup$ – copper.hat Dec 17 '16 at 19:50
  • $\begingroup$ Did you figure it out? $\endgroup$ – copper.hat Dec 17 '16 at 20:45
  • $\begingroup$ @copper.hat Suppose (A,B) is cc. Then By the controllability Gramian we know $BB^{T}$ is positive definite iff all eigenvalues of A have negative real parts. Let P = $\int_0^\infty e^{(A+\alpha/2)^{T}t}BB^{T}e^{(A+\alpha/2)t}$_dt_ so the left side factors and reduces to $e^{(A+\alpha/2)^{T}t}BB^{T}e^{(A+\alpha/2)t}$ evaluated at $t=0$ and $\infty$. Then select $\alpha$ such that eigenvalues of A have negative real parts. Then the exponential goes to $0$ and the LHS reduces to the RHS. $\endgroup$ – Kal Dec 17 '16 at 21:07

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