$\dot{\mathbf{y}}=A\mathbf{y}$ stability and geometric/algebraic multiplicities

Question

Why $\dot{\mathbf{y}}=A\mathbf{y}$ is stable only when algebraic and geometric multiplicities are the same for every eigenvalue with zero real part?

Answer 1

Because the order 1 system for the linear ODE $y'''=0$ has growing solutions $y=x$ and $y=x^2$.