# Eigenvalues of $AB$ vs eigenvalues of $BA$ (finite and infinite-dimensional case)

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:

$$AB$$ and $$BA$$ have the same non-zero eigenvalues. If $$V$$ is finite dimensional, then $$AB$$ and $$BA$$ have the same eigenvalues.

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 $$BA$$.