Do $AB$ and $BA$ have the same eigenvalues?

A more general result:

If $A$ and $B$ are $n \times n$ matrices, then we have for a scalar $\lambda \ne 0$:

$ \lambda$ is an eigenvalue of $AB$ iff $ \lambda$ is an eigenvalue of $BA$ .

Proof: if $ \lambda$ is an eigenvalue of $AB$ , then there is $x \ne 0$ such that

$(*)$ $ABx= \lambda x$.

Let $y:=Bx$. Then $y \ne 0$ (otherwise we would get from $(*)$ that $ \lambda =0$ or $x=0$).

Now we have

$$ BAy=BABx=B(ABx)=B( \lambda x)=\lambda Bx = \lambda y.$$

It follows that $\lambda$ is an eigenvalue of $BA$.

Yes, this is correct. They have the same characteristic polynomial. Look at Wikipedia.