This is a follow-up to this question.
Consider the following 2-dimensional system $$ \dot{x}(t) = A(t)x(t) \quad x(0)\in\mathbb{R}^2, $$ where $A(t)$ is a 2-dimensional time-varying matrix. Suppose that the origin of the above system is an unstable equilibrium.
Now consider the following "perturbed" system $$ \dot{z}(t) = (A(t)+\delta(t)C)z(t) \quad z(0)\in\mathbb{R}^2, $$ where $C$ is a constant $2\times 2$ matrix and $\delta(t)$ is a zero-mean periodic function of $t$.
My question: Is the origin of the "perturbed" system unstable for every choice of $C$ and $\delta(t)$ as above?
As in my previous question, my feeling is that the answer is in the negative; finding an explicit counterexample does not seem easy though.
The answer would be no, namely a known counter example would be a Kapitza pendulum. This is an inverted pendulum whose base is oscillating vertically. When linearizing at the origin the dynamics can be written as
$$ \begin{bmatrix} \dot{\theta} \\ \ddot{\theta} \end{bmatrix} = \left( \begin{bmatrix} 0 & 1 \\ \alpha & -\beta \end{bmatrix} + \sin(\omega\,t) \begin{bmatrix} 0 & 0 \\ \gamma & 0 \end{bmatrix} \right) \begin{bmatrix} \theta \\ \dot{\theta} \end{bmatrix}, $$
where $\alpha,\beta>0$ thus $A(t)$, while actually not time varying, is unstable. For certain choices for $\omega$ and $\gamma$ this system can be made stable. For example the system is stable when using $\beta=\omega=\gamma=1$ and $\alpha=0.2$.