Suppose I am considering the following matrix-valued equation:
\begin{align} A = B + C,\\ \text{where } A,B,C \in \mathbb{R}^{n\times n} \end{align}
My aim is:
Given $B$, I want to find a particular $C$ such that I can ensure that $A$ obtains a particular set of eigenvalues.
Initially $A$ and $B$ do not share any common eigenvalues.
My question is:
Is it possible to "reach" every single set of eigenvalues for $A$, from $B$, using $C$? If $A$ is to remain symmetric, positive-definite is it possible to find some minimal restrictions on this additive $C$ matrix such that this is possible (e.g. $C$ only needs to be diagonal? Or $C$ only needs to be rank 1).
Yes, though the recipe I have in mind may not be very satisfactory. Assuming $B$ is diagonalizable, diagonalize it as $PDP^{-1}$, then take $C=PEP^{-1}$ where $E$ is diagonal with $E_{ii}=\lambda_i-D_{ii}$, and $\lambda_i$ is the desired $i$th eigenvalue. If $B$ is not diagonalizable, first perturb it so that it becomes diagonalizable (rolling the perturbation into the final $C$) and then apply the same recipe.
In terms of possible assumptions on what classes of $C$'s you can use, that's a bit of an ambitious question; it would be easier to answer with particular assumptions rather than asking such a broad question.