Why my Alternative Proof for eigenvalues of AB and BA Fails

Question

My book offers a different proof for the following question and after some consideration I reckon my attempt is incorrect, but I cannot pinpoint why.

Let A,B be square matrices of the same size. Show that the eigenvalues of AB and BA are equal.

I wrote $det(AB - \lambda I) = det(AB) - det(\lambda I) = det(BA) - det(\lambda I) = det(BA - \lambda I)$ ergo the characteristic polynomials are equal.

I ask: flaw in reasoning is where? Thanks.

Answer 1

The flaw is while applying determinants,see that $det(A+B)\neq det(A) + det(B)$ in general. Hope this helps!

Answer 2

In general, $\det(X+Y)\ne\det(X)+\det(Y)$, but in your case it could be true.

However, if the identity you use were true, you'd get $$ \det(AB-\lambda I)=\det(AB)-\det(\lambda I)=\det(AB)-\lambda^n $$ which is clearly absurd: take $B=I$ and $A$ any matrix with eigenvalues $1$ and $2$. (The identity is valid for $n=1$, actually, but it's not so useful in that case.)