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If we have a linear system, represented in State Space and the A matrix is singular (det(A) == 0), can we expect any special properties from the system?

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  • $\begingroup$ Is your system discrete time or continuous time? $\endgroup$ – Jazzmaniac Nov 18 '14 at 12:06
  • $\begingroup$ @Jazzmaniac, continuous time, I think. This is a question to me by a friend, that I couldn't tackle. He's studying some quantum mechanics and just asked about linear systems. $\endgroup$ – Vorac Nov 18 '14 at 13:53
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If the system matrix $A$ is singular then it has at least one eigenvalue equal to $0$. Since the eigenvalues of $A$ represent the poles of the transfer function of the system, there is at least one pole at $s=0$, which makes the system unstable (or marginally stable if there is a single pole at $s=0$, but in any case not asymptotically stable).

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If the system is continuous, the system will have an infinitely many fixed points, which is the null space of $A$. If the geometric multiplicity of 0 eigenvalue is less than its algebraic multiplicity, i.e. the dimension of the null space is less than the dimension of the eigenspace of 0, the system is unstable. If they are equal the system is marginally stable.

If the system is discrete, some of the modes will be finite. This means some states of the homogenous system will be 0 within a finite amount of time for any initial condition.

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