# Stability with eigenvalues' real part equal to zero

I know, from Lyapunov criteria, that a system is stable (not asymptotically) if the system has eigenvalues with negative real part or it has eigenvalues with real part equals to zero, but in this case the algebraic multiplicity must be equal to the geometric multiplicity. I don't understand why there's this condition when the real part is equal to zero.

## 1 Answer

If the algebraic multiplicity of an eigenvalue is greater than the geometric multiplicity, then the system matrix is not diagonalizable and there are vectors which are not linear combinations of the eigenvectors of the matrix.

If you interpret the Lyapunov matrix as a "generalized potential energy function" (but not necessarily the real physical potential energy of the system) it should be clear why having a possible motion of the system which is not included in subspace spanned by the eigenvectors is going to be a problem, when trying to prove or disprove stability.

• Thank you for the response. I understand but my teacher didn't specify that the algebraic multiplicity must be equal to the geometric when the real part is negative. – th3gr3yman Dec 1 '19 at 17:39