# Eigenvalues of $AB$ vs eigenvalues of $BA$ (incl. 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.

I am seeking a confirmation that this result as stated 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 $$BA$$.

• Yes, that's all fine. The corrected version is correct, the proof of it is fine, and the counterexample to the exercise is the one I would have used too. – Theo Bendit Aug 4 '19 at 11:27
• @TheoBendit Thanks a lot. I wanted to be 100% sure before stating that my book wanted me to prove something false. I m not sure how it works in these cases.. if you post it as an answer I can accept it. – Tom Aug 4 '19 at 19:19
• One way it would is to answer the question yourself. Answer the question, "Is this exercise wrong?", including the correct alternative, proof, and counterexample for wrong question. Then I'll give you a +1. – Theo Bendit Aug 5 '19 at 0:50
• @TheoBendit Thank you.. actually the one who takes credit should be you.. :):) in any case I ve done as you suggested here math.stackexchange.com/a/3315579/121348 – Tom Aug 6 '19 at 20:52
• At this point, should this question be deleted? – Tom Aug 6 '19 at 20:53