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?

Alternative proof #1:

If $n\times n$ matrices $X$ and $Y$ are such that $\mathrm{tr}(X^k)=\mathrm{tr}(Y^k)$ for $k=1,\ldots,n$, then $X$ and $Y$ have the same eigenvalues.

See, e.g., this question.
Using $\mathrm{tr}(UV)=\mathrm{tr}(VU)$, it is easy to see that 
$$
\mathrm{tr}[(AB)^k]=\mathrm{tr}(\underbrace{ABAB\cdots AB}_{\text{$k$-times}})
=\mathrm{tr}(\underbrace{BABA\cdots BA}_{\text{$k$-times}})=\mathrm{tr}[(BA)^k].
$$
Now use the above with $X=AB$ and $Y=BA$.
Alternative proof #2:
$$
\begin{bmatrix}
I & A \\ 0 & I
\end{bmatrix}^{-1}
\color{red}{\begin{bmatrix}
AB & 0 \\ B & 0
\end{bmatrix}}
\begin{bmatrix}
I & A \\ 0 & I
\end{bmatrix}
=
\color{blue}{\begin{bmatrix}
0 & 0 \\ B & BA
\end{bmatrix}}.
$$
Since the $\color{blue}{\text{red matrix}}$ and the $\color{red}{\text{blue matrix}}$ are similar, they have the same eigenvalues. Since both are block triangular, their eigenvalues are the eigenvalues of the diagonal blocks.
