Does ${\rm Tr}(ABC) = {\rm Tr}(ABC')$ imply ${\rm Tr}(C) = {\rm Tr}(C')$ for matrix polynomials?

Question

As the title says, if for (not necessarily invertible) matrices $A,B,C,C'$ if we have

$$ Tr(ABC) \;\; =\;\; Tr(ABC') $$

does this imply $Tr(C) = Tr(C')$? The only thing I can imagine doing is subtracting one side from the other to obtain $Tr(AB(C-C')) = 0$, but how can we argue that the trace of $C-C'$ is zero? Any thoughts or suggestions are appreciated.

Edit

I need to drastically change what I wrote here: Suppose $A, C,$ and $C'$ are in fact nonzero matrix polynomials with $B$ invertible. We can therefore write $A =A(\lambda), C = C(\lambda), C' = C'(\lambda)$ for a variable $\lambda$.

Answer 1

Why should it? Take $A=B=0,$ and take any $C,C'$ you like.

Edit: Regarding your edit, instead let $B$ be $I_2$ (the $2\times2$ identity matrix), $A$ be any non-zero $2\times2$ matrix such that $A^\dagger=-A$ (which can be considered a constant matrix polynomial), $C$ be any non-zero $2\times2$ matrix polynomial you like, and $C'=C+I_2.$ Clearly, $C,C'$ do not have equal trace, but you should be able to show that $AB(C-C')$ has trace $0$.

Answer 2

Certainly not. What if $Tr(AB)=0$? Or even worse, what if $AB=0$?

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For your edited question: Take $B$ to be the identity matrix. You essentially asking if $$Tr(AC)=0$$ does it imply $Tr(C)=0$, which again, is far from possible.