I am reading Curtis - Abstract Linear Algebra to pump up my knowledge a little bit and I found exercise I.F.7 (page 41), where I am asked to prove the following:
If $V$ is a vector space over a field $\mathbb{F}$ and $A, B \in End(V)$, then $AB$ and $BA$ have the same eigenvalues.
To begin with... Is this statement true?
This statement is indeed false and that a corrected version could be the following:
The proof of the first assertion should go as follows.
If $\lambda$ is a non-zero eigenvalue of $AB$, then $$ AB v = \lambda v $$ for some non-zero $v \in V$ and since $\lambda \neq 0$, $Bv$ cannot be zero. So we can apply $B$ to both sides and get $$ BA (Bv) = \lambda (Bv) $$ which means that $\lambda$ is an eigenvalue of $BA$
The second assertion is in general false in an infinite-dimensional space.
For example take $V = \mathbb{R}^{\omega}$, $A(v_1, v_2, \dots) = (0, v_1, v_2, \dots)$ and $B(v_1, v_2, \dots) = (v_2, \dots)$ Then $0$ is an eigenvalue of $AB$ (because $AB(v) = (0, v_2, v_3, \dots)$ has clearly a non-trivial kernel) but is definitely not an eigenvalue of $BA$ since $BA = I$.
To prove the second assertion, we could reason like the following.
In general, if $AB$ is injective (resp., surjective) then $B$ is injective (resp., $A$ is surjective), which implies that if $AB$ is invertible then $B$ is injective and $A$ is surjective. If $V$ is finite dimensional, we can make this result stronger and say that invertibility of $AB$ implies invertibility of both $A$ and $B$, thus of $BA$. So in the finite-dimensional case, if $0$ is an eigenvalue of $AB$, then $AB$ is not injective, i.e. not invertible, then (by the contrapositive of the result above) $BA$ is not invertible, i.e. not injective, which is to say that $0$ is also an eigenvalue of $BA$.