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

Question

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?

Answer 1

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.

Answer 2

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.

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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.