Let $A$ be a normal matrix.
As shown for example in these notes (link to pdf, see pagg. 120 and 121), given a matrix $B$ with a single non-zero entry and some parameter $\varepsilon\in\mathbb R$, the spectrum of $A+\varepsilon B$ is the same as that of $A$ up to $\mathcal O(\epsilon)$ terms.
This is easily seen by expressing $A$ in its eigenvectors basis, in which it's diagonal. Then, if $B$ has a single off-diagonal element, then $\sigma(A+\varepsilon B)=\sigma(A)$. If, instead, $B$ has a single element on the diagonal, then clearly $\sigma(A+\varepsilon B)=\sigma(A)+\mathcal O(\varepsilon)$.
In the above link is stated that this holds true also for a generic matrix $B$, not just a single-entry one. How is this more general case proved?
We know that $\sigma(A+\epsilon B)=\sigma(A)+\mathcal O(\epsilon)$. Write $B$ as sum over its components: $B=\sum_{ij}B_{ij}e_i e_j^*$.
Then, $\sigma(A+\epsilon B)=\sigma\left(A+\sum_{ij}\epsilon B_{ij} e_i e_j^*\right)$, thus we can just apply $\sigma(A+\epsilon e_i e_j^*)=\sigma(A)+\mathcal O(\epsilon)$ $N^2$ times to obtain the result.