$AB$ and $BA$ have the same eigenvalues when $\alpha=0$, $\alpha\neq0$, where $A_{n\times n}$ and $B_{n\times n}$

Question

I've had a couple of stabs at this and I really appreciate the help people provide. I think it's correct, but would appreciate any comments otherwise or confirming this. Thanks.

Theorem: $AB$ and $BA$ have the same eigenvalues when $\alpha=0$, $\alpha\neq0$, where $A_{n\times n}$ and $B_{n\times n}$

Proof. Step 1: Let $v$ be an eigenvector corresponding to the eigenvalue $\alpha\neq0$ of $AB$. $ABv=\alpha v$ and by definition $v\neq0$. We are looking for a vector $w$ such that $BAw=\alpha w$

2. If we apply $B$ to both sides of $ABv=\alpha v$, we have $BABv=B\alpha v=\alpha Bv$. Then, $BA(Bv)=\alpha (Bv)$ and $w=Bv$. If we can show that $Bv\neq0$ then $w$ is an eigenvector, $\alpha\neq0$ is an eigenvalue of $BA$, and $AB$ and $BA$ have the same eigenvalues when $\alpha\neq0$.

3. $\alpha\neq0$ and take $Bv=0$. If $Bv=0$, $ABv=0=\alpha v$. As $v\neq0$, $\alpha=0$, however this is a contradiction and therefore $Bv\neq0$ and $w$ is an eigenvector. As $w$ is an eigenvector, $\alpha\neq0$ is an eigenvalue of $BA$, and $AB$ and $BA$ have the same eigenvalues when $\alpha\neq0$.

4. Now we must prove $AB$ and $BA$ have the same eigenvalue when $\alpha=0$. If $\alpha=0$ then det$(AB-0I)$ is the characteristic polynomial of $AB$, and det$(AB-0I)$=det$(AB)$=det$(A)$det$(B)$ by the theorem det$(AB)$$=$det$(A)$det$(B)$.

5. As real numbers are commutative under multiplication, det$(A)$det$(B)$$=$det$(B)$det$(A)$ and det$(B)$det$(A)$$=$det$(BA)$$=$det$(BA-0I)$. det$(BA-0I)$ is the characteristic polynomial of $BA$ and $0$ is an eigenvalue of $BA$ when $0$ is an eigenvalue of $AB$.

6. When $\alpha\neq0$, $\alpha=0$ $AB$, $BA$ have the same eigenvalues. Therefore $AB$, $BA$ have the same eigenvalues.

Q.E.D.

Answer 1

If $\alpha = 0$ then $ABv=0$ for all $v$ in the eigenspace corresponding to zero. Then $\det AB$ is the characteristic polynomial of $AB$, and $$\det AB=\det A\det B.$$ Since real numbers are a field and hence commutative under multiplication, $\det A\det B = \det B\det A$ and $$\det B\det A = \det BA$$ is the characteristic polynomial of $BA$, hence zero is an eigenvalue of $BA$.

If $\alpha\ne 0$, let $v$ be an eigenvector associated to $\alpha$ of $AB$. Then $ABv =\alpha v$ and $v\ne 0$. Applying $B$ to both sides of $ABv = \alpha v$, we have $BABv = \alpha Bv$, then $Ba(Bv) = \alpha Bv$ and $w=Bv$.If $B_v=0$ then $ABv=0 =\alpha v$. As $v\ne 0$, $\alpha=0$, but this is a contradiction since we assumed that $\alpha\ne0$, and hence $Bv\ne 0$ so that $w$ is an eigenvector of $B$. It follows that $\alpha\ne0$ is an eigenvalue of $BA$, so that $AB$ and $BA$ have the same eigenvalues.