I have a theorem on differentiabiliy of eigenvalues that says :
Let $\lambda_{\star}$ be a simple eigenvalues of A and let $v_{\star}$ be its right eigenvector and $u_{\star}$ its left eigenvector.
Then for $\varepsilon$ small enough, the matrix $A(\varepsilon) = A + \varepsilon \cdot E$ has an eigenvalue $\lambda(\varepsilon)$ close to $\lambda_{\star}$. And $\lambda(\varepsilon)$ is differentiable around $0$ and we have $$\lambda(\varepsilon) = \lambda_{\star} + \varepsilon \frac{u_{\star}^{H}E v_{\star}}{u_{\star}^{H}v_{\star}} + O(\varepsilon^{2})$$
The $3$ matrices I perturbed are : $$\begin{pmatrix} -1 & 0 & 0\\ 0& e^{-i\pi/3}& 0\\ 0& 0 & e^{i\pi/3} \end{pmatrix}$$ $$\begin{pmatrix} -1 & 20 & 200\\ 0& e^{-i\pi/3}& 1\\ 0& 0 & e^{i\pi/3} \end{pmatrix}$$ $$\begin{pmatrix} -1 & 200 & 2\cdot10^{6}\\ 0& e^{-i\pi/3}& 1\\ 0& 0 & e^{i\pi/3} \end{pmatrix}$$
So I calculated every eigenvalues of A and I perturbed A with a random matrix E of norm $1$ and re-calculated its eigenvalues. The circles I drew are centered with the corresponding eigenvalue and his radius is $\kappa\cdot\varepsilon$ with $\kappa = \frac{u_{\star}^{H}E v_{\star}}{u_{\star}^{H}v_{\star}}$ like the Taylor development in the previous theorem.
But as you can see in the figure below, the eigenvalues stay in the circle according to the theorem for the first matrix, but for matrices $2$ and $3$, for $2$ eigenvalues, some of the perturbed eigenvalues are going outside which is strange because the theorem takes into account every $\kappa$ for each eigenvalues so it is not supposed to go outside each circle.
Of course we have $O(\varepsilon^{2})$ in the Taylor development but as you can see in the first plot, we have $\varepsilon = 5\cdot 10^{-1}$ and the perturbed eigenvalues are not going outside any circles.
And this is a theorem so it's true so at a moment there is a numerical instability or too much rounding errors.. but where ?
There must a thing with the condition of the perturbed matrix but it's never taken into account in this theorem ?
If you are genuinely using $\epsilon=0.5$ then $\epsilon^2=0.25$ and you should expect the points to exist within a circle of radius $a\epsilon+b \epsilon^2$ for some constants $a, b$ dependant on the matrix. The fact that for the first matrix all the points lie inside a circle of radius $a\epsilon$ implies that $b\leq 0$. Because $\epsilon$ is so large these corrections have a large impact.