Eigenvalue perturbation

Remarks

So $\lambda(\varepsilon) = \lambda_0 + \varepsilon \, \dfrac{y^* B x}{y^* x} + O(\varepsilon^2)$. To obtain the coefficient, differentiate $A(\varepsilon) x(\varepsilon) = \lambda(\varepsilon) x(\varepsilon)$ at $0$ and multiply on the left by $y^*$; the terms with $x'(0)$ cancel.

Example: for a Hermitian $A$ one may take $y = x$, so $\lambda'(0) = x^* B x / x^* x$. For $A = \operatorname{diag}(1, 2)$ and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, with $\lambda_0 = 1$ and $x = y = e_1$, this gives $\lambda'(0) = 0$. Indeed the eigenvalues of $A + \varepsilon B$ are $\frac{3}{2} \pm \sqrt{\frac{1}{4} + \varepsilon^2}$, and the smaller one is $1 - \varepsilon^2 + O(\varepsilon^4)$.

Non-example: for $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, the eigenvalue $0$ is not simple. With $B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$, the eigenvalues of $A + \varepsilon B$ are $\pm \sqrt{\varepsilon}$, which are not analytic in $\varepsilon$ at $0$. Here the right eigenvector $x = e_1$ and the left eigenvector $y = e_2$ satisfy $y^* x = 0$, so the formula for $\lambda'(0)$ has no meaning.