AB and BA have the same characteristic polynomial

Invertible $A$. If $A\in GL_n(\mathbb{R})$, then
$$A(BA)A^{-1}=ABAA^{-1}=AB,$$
so $AB$ and $BA$ are similar, and similar matrices have the same characteristic polynomial: $\chi_{AB}=\chi_{BA}$.

General $A$. Fix $B$ and consider $F(A)=\chi_{AB}-\chi_{BA}$, whose coefficients are polynomials in the entries of $A$ (the coefficients of $\chi_M$ are polynomials in the entries of $M$, and $AB$, $BA$ are linear in $A$). So $F$ is continuous. It vanishes on $GL_n(\mathbb{R})$, which is dense in $M_n(\mathbb{R})$ ($\det$ is a nonzero polynomial). Hence $F\equiv0$.

Check. For $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $B=\begin{pmatrix}0&1\\0&0\end{pmatrix}$: $AB=B$ and $BA=0$, which are *not* similar (different ranks), yet $\chi_{AB}=\chi_{BA}=X^2$.

Remark. Consequently $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ and $AB$, $BA$ have the same nonzero eigenvalues with the same algebraic multiplicities. For non-square $A\in M_{n,p}$, $B\in M_{p,n}$ one gets $X^p\chi_{AB}=X^n\chi_{BA}$ in the same way (pad with zeros). Lesson. When invertibility makes a problem trivial, density gives the rest.