monotonicity of eigenvalues with respect to perturbation Let $A$ be a matrix, $B$ be a diagonal matrix, and $\epsilon$ be a scalar. Are eigenvalues of $A+\epsilon B$ monotonic with respect to $\epsilon$ ?
 A: I suppose both $A$ and $B$ are Hermitian, otherwise $A+\epsilon B$ can possess nonreal eigenvalues, so that you cannot speak of monotonicity.
Let $\lambda_k(X)$ denotes the $k$-th smallest eigenvalue of a Hermitian matrix $X$. When $B$ is positive semidefinite and $\epsilon\ge0$, we have $\lambda_k(A)\le\lambda_k(A+\epsilon B)$ (this is a consequence of Weyl's inequality) and hence $\lambda_k(A+\epsilon B)$ is monotonic in $\epsilon$. The similar holds when $B$ is negative definite.
When $B$ is indefinite, eigenvalues of $A+\epsilon B$ may no longer be monotonic functions of $\epsilon$. For a counterexample, consider the matrix
$$
A+\epsilon B:= \pmatrix{1&\sqrt{2}\\ \sqrt{2}&2} + \epsilon\pmatrix{-1\\ &2} = \pmatrix{1-\epsilon&\sqrt{2}\\ \sqrt{2}&2+2\epsilon}.
$$
Its characteristic polynomial is $\lambda^2 - (\epsilon+3)\lambda - 2\epsilon^2$. The discriminant of this quadratic function is $(\epsilon+3)^2+8\epsilon^2$, which is always positive. Therefore, the paths of the two eigenvalues of $A+\epsilon B$ do not cross and the smaller eigenvalue $\lambda_\min$ of $A+\epsilon B$ is always given by
$$
2\lambda_\min(\epsilon) = (\epsilon+3) - \sqrt{ 9\epsilon^2 + 6\epsilon + 9 }.
$$
It is easy to verify that $\lambda_\min'(0)=0$ and $\lambda_\min''(\epsilon)<0$. Hence $\lambda_\min(\epsilon)$ is not monotonic --- it attains local maximum at $\epsilon=0$.
