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Consider the following continuous-time state space representation of the form:

$\frac{d}{dx}x(t) = Ax(t)+Bu(t), \quad y(t)=Cx(t), \quad t\in \mathbb{R}^{+}$

$A=\begin{bmatrix}-1&3&0&0\\-3&-1&0&0\\0&0&0&3\\0&0&-3&0 \end{bmatrix} \quad B = \begin{bmatrix}0\\1\\0\\0 \end{bmatrix} \quad C=\begin{bmatrix}1&0&0&1 \end{bmatrix}$

The corresponding eigenvalues are: $-1+3i, \ -1-3i, \ 0+3i \ \text{and} \ 0-3i$.

The answer states that this system is Lyaponov stable. But I'm wondering why.

Is it because the Jordan blocks of the eigenvalues with zero real-part are $1$x$1$. Because this matrix is in Jordan Form?

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    $\begingroup$ Your $A$ matrix in not in Jordan form, but in its real Jordan form, so you have to look at the size of each full real Jordan block. $\endgroup$ – Kwin van der Veen Nov 21 '18 at 14:28
  • $\begingroup$ I thought that a 4x4 real jordan form had an identity matrix in the upper right corner? like this:$\begin{bmatrix}-1&3&1&0\\-3&-1&0&1\\0&0&0&3\\0&0&-3&0 \end{bmatrix}$ $\endgroup$ – user463102 Nov 21 '18 at 17:00
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    $\begingroup$ Yes, only your example is not a real Jordan block, because then both 2x2 matrices on the diagonal need to be the same. $\endgroup$ – Kwin van der Veen Nov 21 '18 at 18:54
  • $\begingroup$ Thank you. I did not know that a right upper 2x2 identity only appears if both 2x2 matrices on the diagonal are the same. $\endgroup$ – user463102 Nov 22 '18 at 7:44
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Write $I_n$ for a $n \times n$ identity matrix.

Take $P = (1/2)I_4$ and $V(x) = x^T P x$, which is clearly positive definite. Now calculate the directional derivative:

$$ \begin{align} \dot{V}(x) &= \dot{x}^T P x + x^T P \dot{x} \\ &= x^T A^T P x + x^T P A x \\ &= x^T(A^T P + P A) x \\ &= x^T Q x, \end{align} $$

insert $A$ and $P$ and derive $Q$:

$$ \begin{align} Q = A^T P + P A &= \begin{bmatrix} -1 & -3 & 0 & 0 \\ 3 & -1 & 0 & 0 \\ 0 & 0 & 0 & -3 \\ 0 & 0 & 3 & 0 \end{bmatrix} \frac{1}{2} I_4 + \frac{1}{2} I_4 \begin{bmatrix} -1 & 3 & 0 & 0 \\ -3 & -1 & 0 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & -3 & 0 \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} . \end{align} $$

So your directional derivative is $\dot{V}(x) = -x_1^2 - x_2^2$, which is negative semi-definite. Therefore, the system is Lyapunov stable (but not asymptotically Lyapunov stable).

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    $\begingroup$ In this case it might be even easier to just look at the eigenvalues, because they all have an algebraic multiplicity of one. Only if the algebraic multiplicity of an eigenvalue with zero real part is higher than one, you also need to check its geometric multiplicity. $\endgroup$ – Kwin van der Veen Nov 21 '18 at 14:37

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