Controlling the spectrum of $S + \mu \vec{v} \vec{v}^\dagger$ by changing $\mu$. Suppose we have a hermitian matrix $S \in \mathbb{C}^{N \times N}$. For $\mu \in \mathbb{R}$ and $\vec{v} \in {\mathbb{C}^N}$ , let
$$ \tag{1} S(\mu) = S + \mu \cdot \vec{v} \vec{v}^{\dagger}.$$
The claim is that for any $s \in \mathbb{R}$ there exists $\mu$ such that the following eigenvalue equation is true:
$$ \tag{2} S(\mu) \vec{u} = s \vec{u}.$$
Specifically, $\mu(s)$ is given as follows:
$$\tag{3} \mu(s) = \left( \sum_{n=1}^N \frac{\langle\vec{s}_n, \vec{v} \rangle^2}{s-s_n} \right)^{-1}.$$
In the above, $\vec{s}_n$ and $s_n$ are eigenvectors and non-degenerate eigenvalues of $S$.
How do I derive Eq (3) ?
 A: First, note that if $s = s_n$ for any $n$, we can just take $\mu = 0$, so we only need to do the $s \neq s_n$ case.
Since $S$ is Hermitian, we let $S = U \Lambda U^*$ be its eigendecomposition, with $U$ unitary having columns $\{\mathbf{s}_n\}$.
Now, we need to solve the equation
\begin{align*}
0
= \det(S + \mu vv^* - sI)
= \det(\Lambda - sI + \mu (U^* v) (U^*v)^*).
\end{align*}
For clearer notation, we'll write $\tilde{v} := U^*v$, and factor out the diadonal $\Lambda - sI$ to get
\begin{align*}
0
&= \det(\Lambda - sI) \det (I + (\Lambda - sI)^{-1} \mu \tilde{v} \tilde{v}^*).
\end{align*}
Since $s \neq s_n$ for all $n$, we have that $\det(\Lambda - sI) \neq 0$, so we only need to solve the latter term being zero.
Moreover, determinants have the important property that $\det(I + AB) = \det(I + BA)$ for any matrices $A, B$, such that $AB$ and $BA$ are both square (this comes from the fact that $AB$ and $BA$ have the same eigenvalues, except if one of then has more rows and columns, it has enough additional eigenvalues with value 0 to make up this difference).
So we have now reduced the problem to solving
\begin{align*}
0
&= \det(I + \mu \cdot \tilde{v}^* (\Lambda - sI)^{-1} \tilde{v}) \\
&= 1 + \mu \cdot \tilde{v}^* (\Lambda - sI)^{-1} \tilde{v} \\
&= 1 + \mu \sum_{n=1}^N \frac{|\tilde{v}_n|^2}{s_n - s}, \\
\mu
&= \Bigl(\sum_{n=1}^N \frac{|\tilde{v}_n|^2}{s - s_n}\Bigr)^{-1}.
\end{align*}
But recall that $\tilde{v} = U^* v$, so that $\tilde{v}_n = \langle \mathbf{s}_n, v\rangle$, which yields
$$
\mu
= \Bigl(\sum_{n=1}^N \frac{|\langle \mathbf{s}_n, v\rangle|^2}{s - s_n}\Bigr)^{-1}.
$$
I realise that this has an extra square exponent when compared to your eq. (3), but I believe that this is the correct expression.
