Given a diagonal matrix $A$ with diagonal elements $a_1,\dotsc,a_n$ what can we say about the eigenvalues of the perturbation $A+E(t)$ satisfying $E(t) = O(t^k)$ so that there exists a constant $K>0$ so that $\|E(t)\|\leq Kt^k$.
Now the determinant of $\det(A+E(t)-\zeta I)$ is a sum of the form
$$p(\zeta) = \sum_j \left(\prod_{i=l_j}^{n}(a_{\pi_j(i)}-\zeta+O(t^k))O(t^{k\cdot l_j})\right)$$
where $\pi_j$ denotes permutations on $\{1,\dotsc,n\}$.
I would like to say that the roots of $p$ with respect to $\zeta$, here denoted $(b_i)_{i=1}^{n}$, should satisfy
$$|a_{i}-b_i| = O(t^k)$$
however I don't know how to prove this or if it is true?
By expanding the product $\prod_{i=l_j}^{n}(a_{\pi_j(i)}-\zeta+O(t^k))$ into a sum of the form
$$\sum \prod(a_{\tilde{\pi}(i)}-\zeta)O(t^{k \tilde{l}_j})$$
i get an even bigger sum and this does not seem to help me at all so I am kind of stuck at this point.
Here are a couple hints. I'll come back and fill in more details once you've had a chance to look at them.
A necessary condition for $ \Vert E(t) \Vert = \mathcal{O}(t^k) $ as $t\to0$ is that each entry also goes to zero like $t^k$.
Gershgorin's circle theorem says that all eigenvalues are contained within the union of disks centered at the diagonal elements of a matrix, where each disk has radius equal to the sum of magnitude of the off diagonal elements in the corresponding row. What can you tell about the row sums and diagonal elements of $A+E(t)$?