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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?
It suffices to show that $AB$ and $BA$ have the same characteristic polynomial. First assume that $A$ is invertible then
$$\chi_{AB}(x)=\det(AB-xI)=\det A\det(B-xA^{-1})\\=\det(B-xA^{-1})\det A=\det(BA-xI)=\chi_{BA}(x)$$ Now since $\operatorname{GL}_n(K)$ is dense in $\operatorname{M}_n(K)$ then there's a sequence of invertible matrices $(A_n)$ convergent to $A$ and by the continuity of the $\det$ function we have $$\chi_{AB}(x)=\det(AB-xI)=\lim_{n\to\infty}\det(A_nB-xI)=\lim_{n\to\infty}\det(BA_n-xI)\\=\det(BA-xI)=\chi_{BA}(x).$$
Here is a proof similar to what the OP has tried:
Let $\lambda$ be any eigenvalue of $AB$ with corresponding eigenvector $x$. Then
$$ABx = \lambda x \Rightarrow \\ BABx = B\lambda x \Rightarrow\\ BA(Bx) = \lambda (Bx) $$
which implies that $\lambda$ is an eigenvalue of $BA$ with a corresponding eigenvector $Bx$, provided $Bx$ is non-zero. If $Bx = 0$, then $ABx = 0$ implies that $\lambda = 0$.
Thus, $AB$ and $BA$ have the same non-zero eigenvalues.
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.
If $A$ is invertible, user63181 already showed that
$$\det(AB-xI)=\det(BA-xI)$$
We now prove this equality in general.
Fix $x$
Let $$P_x(y):=\det[(A-yI)B-xI]-\det[B(A-yI)-xI] \,.$$
Then $P_x(y)$ is a polynomial of degree at most $n$ in $y$. Whenever $y$ is not an eigenvalue of $A$ the matrix $A-yI$ is invertible, thus by the first pat $P_x(y)=0$. Hence $P_x$ has infinitely many roots, and hence $P_x \equiv 0$.
This proves that $P_x(0)=0$ which is exactly what you need to prove.
Here is a more "algebraic" approach from the other answers by user63181 and N. S. which, as far as I can see generalizes to other fields (where the continuity argument might fail) although I thought the argument by continuity was cool!
First we use the following characterization of non-zero eigenvalues: $\lambda$ is an eigenvalue for $AB$ iff $I - \lambda AB$ is not invertible.
I claim the following: $I - \lambda AB$ is (not) invertible iff $I - \lambda BA$ is (not) invertible.
Proof: Suppose $I - \lambda AB$ is invertible, then let $U := 1 + \lambda B (I - \lambda AB)^{-1}A$. Now show that $U$ is an inverse to $I - \lambda BA$ (by multiplying out and using distributivity). The converse direction follows by letting $V := 1 + \lambda A(1-\lambda BA)^{-1}B$.
From this it follows that $AB$ and $BA$ have the same non-zero eigenvalues.
