TITLE: How to prove $\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$?
QUESTION [30 upvotes]: there is a similar thread here Coordinate-free proof of $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$?, but I'm only looking for a simple linear algebra proof.

REPLY [9 votes]: For any couple $(A,B)$ of $n\times n$ matrices with complex entries, the following identity holds:
  $$ \operatorname{Tr}(AB) = \operatorname{Tr}(BA).$$

Proof. Assuming $A$ is an invertible matrix, $AB$ and $BA$ share the same characteristic polynomial, since they are conjugated matrices 
due to $BA = A^{-1}(AB)A$. In particular they have the same trace. Equivalently, they share the same eigenvalues (counted according to their
algebraic multiplicity) hence they share the sum of such eigenvalues. On the other hand, if $A$ is a singular matrix then 
$A_\varepsilon\stackrel{\text{def}}{=} A+\varepsilon I$ is an invertible matrix for any $\varepsilon\neq 0$ small enough. It follows that 
$\operatorname{Tr}(A_\varepsilon B) = \operatorname{Tr}(B A_\varepsilon)$, and since $\operatorname{Tr}$ is a continuous operator, 
by considering the limits of both sides as $\varepsilon\to 0$ we get $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$ just as well.