How does the singularity of a system matrix affect the system's stability?

Question

What can be said about system stability, given a singular system matrix below? \begin{align} A = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -0.0095 & -0.0095 & -0.0095 & -0.0095\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -0.2335 & 0 & 0.2335 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.2335 & 0 & -0.2335 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -0.0098 & 0.0098 & -0.0098 & 0.0098 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -0.0393 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -0.0393 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -0.0393 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -0.0393 \end{bmatrix} \end{align}

The characteristic equation is given by $\begin{vmatrix}sI-A\end{vmatrix}$, and this determinant ist non-zero even for the singular matrix above. Using Matlab I can compute roots(poly(A)) which will tell me the roots of characteristic equation as: $\begin{bmatrix} 1 & 0.1574 & 0.0093 & 0.0002 & 0.0000 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$, so the system is generally unstable as it has only right hand side roots. But does singularity of $A$ determine this instability in a way? What other problems can singularity of system matrix cause?

Answer 1

Edited*Below is true only for diagonalizable matrices)*

If the matrix is singular (which is equivalent to saying that it has at least one eigenvalue 0), it means that perturbations in the kernel (i.e. space of vectors x for which Ax=0) of this matrix do not grow, so the system is neutrally stable in the subspace given by the kernel.

If it has ANY positive eigenvalue(s), it is unstable because any perturbation with nonzero component in the subspace of corresponding eigenvector(s) will grow to infinity.

So in your case since you have positive eigenvalues, the matrix being singular doesn't really change anything.

On the other hand, if you had only negative eigenvalues and also zero eigenvalues, then the presence of zero eigenvalues will change the stability from Asymptotic stability to neutral stability.

Asymptotic stability: All perturbations go to 0 given enough time.

Neutral stability: All perturbations are bounded for all time.

Answer 2

Stability is determined by the eignvalues. If you see what means singular matrix to your eigenvalues, you have the connect to stability.

Note: just to be coscious be careful not to confuse things with so called singular value of a matrix this is something different.