Theorem: $AB$ and $BA$ have the same eigenvalues, where $A_{n\times n}$ and  $B_{n\times n}$, and $\alpha\neq0$  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 $\alpha\neq0$. Q.E.D.
First of all, am I being crazy in thinking that if $\lambda$ is an eigenvalue of $AB$, where $A$ and $B$ are both $N \times N$ matrices (not necessarily invertible), then $\lambda$ is also an eigenvalue of $BA$?  If it's not true, then under what conditions is it true or not true?  If it is true, can anyone point me to a citation?  I couldn't find it in a quick perusal of Horn &amp; Johnson.  I have seen a couple proofs that the characteristic polynomial of $AB$ is equal to the characteristic polynomial of $BA$, but none with any citations.  A trivial proof would be OK, but a citation is better.
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