Consider the following setting. We have a symmetric matrix $A \in \mathbb{R}^{n \times n}$ with eigenvalues $\lambda_1, \dotsc, \lambda_n \in \mathbb{R}$ and corresponding eigenvectors $v_1, \dotsc, v_n \in \mathbb{R}^n$. It is well-known that the eigenvectors form an orthogonal basis of $\mathbb{R}^n$.
Now, suppose we have a vector $u \in \mathbb{R}^n$ that is an approximation of one of the eigenvectors, say of $v_k$, in the sense that $$ u^Tv_k \approx 1\quad \text{and} \quad u^T v_j \approx 0 \,, \text{for } j \neq k\,. $$ We assume $u$ is normalised, i.e. $\|u\|_2 = 1$. Now define the matrix $\tilde{A}$ by $$ \tilde{A} := A - i(I - uu^T)\,, $$ where $i$ denotes the imaginary unit and $I$ is the identity matrix of appropriate size (note that $I - uu^T$ is the projection onto the orthogonal complement of $u$, but I don't know if this is important). The matrix $i(I - uu^T)$ is a complex symmetric matrix of rank $n-1$.
My question is now: Can we say anything about the eigenvalues of $\tilde{A}$ assuming the eigenvectors and -values of $A$ and $I - uu^T$ are known?
Intuitively (and this is also what numerical examples show), I expect the eigenvalues to be raised into the complex plane with the eigenvalue corresponding to the eigenvector $v_k$ barely being raised (because we subtract from $A$ the projector onto the orthogonal complement of $u$ which is an approximation of $v_k$, so eigenvalues corresponding to vectors in the orthogonal complement of $v_k$ should be affected the most by this perturbation and this are exactly the eigenvalues $\lambda_j\,, j \neq k$, due to the pairwise orthogonality of the eigenvectors).
How can I make this intuition more quantitative?
You can write $$ \tilde{A} := A - i(I - v_kv_k^T) -i(v_kv_k^T -uu^T)\,. $$ You know $A - i(I - v_kv_k^T)$ exactly and $-i(v_kv_k^T -uu^T)$ is small.
Kato p. 94-95 may be relevant.