Eigenvalues of $AB$ and $BA$ where $A$ and $B$ are square matrices 
Show that if $A,B \in M_{n \times  n}(K)$, where $K=\mathbb{R}, \mathbb{C}$, then the matrices $AB$ and $BA$ have same eigenvalues.

I do that like this:
let $\lambda$ be the eigenvalue of $B$ and $v\neq 0$
$ABv=A\lambda v=\lambda Av=BAv$
the third equation is valid, because $Av$ is the eigenvector of $B$. Am I doing it right?

Observe that there is some $v$ such that $ABv=\lambda v$ iff there are $v,w$ that solve the linear system:
$$\begin{cases}Aw=\lambda v, \\Bv=w.\end{cases}\tag1$$
On the other hand, from (1), if $\lambda\neq0$, one also immediately gets $BAw=\lambda w$.
It follows that $ABv=\lambda v$ for some $\lambda\neq0$ iff $BAw=\lambda w$, where $w=Bv$ (and thus also $v=\frac1\lambda Aw$).
Example
For example, consider
$$A\equiv \begin{pmatrix}1&0\\0&0\end{pmatrix},
\qquad B\equiv \frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}.$$
Then
$$
AB = \frac12\begin{pmatrix}1&1\\0&0\end{pmatrix},
\qquad BA = \frac12\begin{pmatrix}1&0\\1&0\end{pmatrix}.
$$
Thus $AB e_1=\frac12 e_1$ and $AB(e_1-e_2)=0$, while $BA(e_1+e_2)=\frac12(e_1+e_2)$ and $BA e_2=0$.
And we can notice how the nonzero eigenvectors of $AB$ are related as spelled out before: $Be_1= \frac12(e_1+e_2)$, and $A(e_1+e_2)=e_1$.
On the other hand, this same example shows how the result fails for zero eigenvalues. It remains true that both matrices have kers of the same dimensions, but the correponding eigenvectors are not related as in the nonzero eigenvalues case. Here the ker of $AB$ is spanned by $e_1-e_2$, but $B(e_1-e_2)=0$, which thus clearly does not give the ker of $BA$, which is spanned by $e_2$.
I'd also observe here that the reason we don't get the stated relation between eigenvectors of $AB$ and $BA$ is that we can have situations where $ABv=0$ because $Bv=0$ (which is also what happens in the example). If however we have $ABv=0$ but $Bv\neq0$, then we can see that we recover (at least partially) the result, because $(BA)(Bv)=0$.
