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?
Best Answer
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.