# Is the spectrum of $AB+C$ the same as $BA+C$?

It is known that for matrices $$A$$ and $$B$$, $$AB$$ and $$BA$$ are isospectral. However, I am not sure if this holds for a when these terms are in a sum, i.e. if $$AB+C$$ is isospectral to $$BA+C$$. My intuition tells me that I shouldn't expect these to have the same eigenvalues, since eigenvalues of a sum generally isn't a nice function of the eigenvalues of the summands. If this is not the case, what restrictions can we put on $$A$$, $$B$$, and $$C$$ to make this true?

Let $$A=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\qquad B=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},\qquad C=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$$ Then $$AB+C=\begin{pmatrix} 2 & 0 \\ 0 & 0\end{pmatrix},\text{ while } BA+C=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$ If you want to look to put additional constraints on $$A$$, $$B$$ and $$C$$ so that $$AB+C$$ and $$BA+C$$ have the same spectrum, a first reflex might be to impose that $$A$$, $$B$$ and $$C$$ are symmetric or self-adjoint. Xander Henderson's discussion shows however that this leads nowhere. The so-called "Weyl inequalities" allow one to bound the eigenvalues of a sum of symmetric matrices in terms of the eigenvalues of the summands.
Pick your two favorite non-commuting square matrices $$A$$ and $$B$$, and set $$C = -AB$$. Then $$AB - C = 0$$, while it is almost surely the case that $$BA - C \ne 0$$ and, moreover, has nontrivial eigenvalues. As a silly example, take
$$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \qquad\text{and}\qquad B = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}.$$ Then, with $$C=-AB$$, $$BA + C = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - \begin{pmatrix} 3 & 1 \\ 7 & 3\end{pmatrix} = \begin{pmatrix} 1 & 5 \\ -6 & -1\end{pmatrix},$$ which is non-singular.
Addressing the last part of your question, i.e. "what restrictions can we put on $$A$$, $$B$$, and $$C$$ to make this true?": the answer is likely going to be disappointing, i.e. the set of triples $$\{A,B,C\}$$ such that $$\operatorname{spec}(AB + C) = \operatorname{spec}(BA+C)$$ is likely going to be precisely the set of triples $$\{A,B,C\}$$ such that $$\operatorname{spec}(AB + C) = \operatorname{spec}(BA+C).$$ I doubt that you are going to find a better (i.e. more useful or interesting) characterization.