How to show that $det(AB-xI)=det(BA-xI)$ ,for any $x\in \mathbb F$. There is a problem in Hoffmann Kunze:
Show that $AB$ and $BA$ have the same characteristic polynomials,where $A,B$ are both $n\times n$ matrices in $\mathbb F$.
If $A$ or $B$ is invertible it can be easy done:
$det(AB-xI).det(A)=det(ABA-xA)=det(A).det(BA-xI)$ and if $detA\neq 0$,we can cancel it to get our desired result but what about the case when both $detA $ and $detB$ are $0$.How to proceed in that case?
By the way the case $n=2$ is easy as $p_A(x)=x^2-tr(A)x+det(A)$ and so characteristic polynomial of $AB$ and $BA$ must be same as they have the same trace and same determinant.But this idea cannot be applied for higher order matrices.
 A: A bit tricky way, but which it even can be applied when $A$ and $B$ are not square matrices, so $A$ is $n\times m$ and $B$ is $m\times n$, is the following: 
Consider the identity of the product of matrices with dimensions $(n+m)\times (n+m)$: $$\begin{pmatrix} I_n & -A \\ B & I_m \end{pmatrix}= \begin{pmatrix} I_n & 0 \\ B & I_m \end{pmatrix} \begin{pmatrix} I_n & -A \\ 0 & I_m+BA \end{pmatrix}= \begin{pmatrix} I_n+AB & -A \\ 0 & I_m \end{pmatrix}\begin{pmatrix} I_n & 0 \\ B & I_m \end{pmatrix}.$$
Now, using the following result on the determinant of "block" triangular matrices 
$$\text{det}\begin{pmatrix} A & 0 \\ B & C\end{pmatrix}=\text{det}(A)\text{det}(C)$$
we get that 
$$\text{det}(I_n+AB)=\text{det}(I_m+BA).$$
Finally, one gets from this that
$$(-x)^{m}\text{det}(AB-xI_n)=(-x)^{n} \text{det}(BA-xI_m).$$
which, in your case $n=m$, gives the result. 
A: How about this? introduce a new indeterminate $t$, and consider
$(A+tI)B-xI$ and $B(A+tI)-xI$. Over the extension field $\Bbb F(t)$ the matrix
$(A+tI)$ is invertible, so
$$\det((A+tI)B-xI)=\det(B(A+tI)-xI).$$
Both sides of this equation are elements of the polynomial ring $\Bbb F[x,t]$.
Now we can replace $t$ by $0$ to get
$$\det(AB-xI)=\det(BA-xI).$$
