$AB$ and $BA$ have the same characteristic polynomial Can someone please help me with this proof. 
For $A,B$ ∈ $F^{n×n}$, show that $AB$ and $BA$ have the same characteristic polynomial.
 A: In the case $F=\mathbb{C}$ or $F=\mathbb{R}$:
First if $A\in\mathrm{GL}_n(F)$ then we have 
$$\chi_{AB}(x)=\det(xI-AB)=\det(A(xI-BA)A^{-1})=\det(xI-BA)=\chi_{BA}(x)$$
Now by the density of $\mathrm{GL}_n(F)$ in $\mathcal{M}_n(F)$ and the continuity of the $\det$ function we have the desired result.
A: Edit: There is the argument given by Sami Ben Romdhane which works when we have the density of invertible matrices. I had long known this approach only. But recently I came across this purely algebraic approach which works over any commutative unital ring $R$.
I'll start with a slightly more general result for rectangular matrices.
For every $n\times m$ matrix $A$ and every $m\times n$ matrix $B$, we have
$$
\left(\matrix{I&A\\B&tI}\right)\left(\matrix{tI&-A\\0&I}\right)=\left(\matrix{tI&0\\*&tI-BA}\right)
$$
and
$$
\left(\matrix{I&A\\B&tI}\right)\left(\matrix{tI&0\\-B&I}\right)=\left(\matrix{tI-AB&*\\0&tI}\right).
$$
Taking the determinant of all these yields 
$$
t^m\det(tI-AB)=t^n\det \left(\matrix{I&A\\B&tI}\right)=t^n\det(tI-BA).
$$
In the case of square $n\times n$ matrices, this yields, since $n=m$:
$$
t^n(\det(tI-AB)-\det(tI-BA))=0\quad\Rightarrow\quad \det(tI-AB)-\det(tI-BA)=0
$$
where the implication follows from the fact that $\det(tI-AB)-\det(tI-BA)$ is a polynomial in $t$.
Thus
$$
\chi_{AB}(t)=\det(tI-AB)=\det(tI-BA)=\chi_{BA}(t)\qquad\forall t\in R.
$$
