Suppose that $A$, $B$ and $AB$ are Hermitian matrices. show that every eigenvalue of $AB$ is the product of an eigenvalue of $A$ and eigenvalue of $B$.
I'm stuck on this problem. I could show, any eigenvector of $A$ is an eigenvector of $B$ and $AB$ as well, but I don't know why any eigenvalue of $AB$ is the product of an eigenvalue of $A$ and eigenvalue of $B$. Here is a link, which I was told that's a answer for my question, but the assumptions seems slightly different to me and honestly I didn't get it. I appreciate any hint to start this problem.
First $(AB)^H=B^HA^H=BA=AB$, which implies $A$ and $B$ commute.
Let $\lambda \in spec(B)$ and let $v$ be an eigenvecotr of $B$ associated with $\lambda$ so $Bv=\lambda v$. Hence,
$$Bv=\lambda v$$ $$B(Av)=A(Bv)=\lambda Av,$$ so $Av\in W_\lambda$, the eigenspace of $B$ associated with $\lambda$, Since, both $v$ and $Av$ are in $W_\lambda$, there exists a scalar $k$ satisfying $Av=kv$, this implies $k\in spec(A)$ and it's associated with eigenvector $v$ of $A.$ Then, $AB(v)=\lambda kv.$
Let $k\in Spec(A)$, then there exists an eigenvector $v$ of $A$ associated with $k$, it's easy to show $v$ is an eigenvector of $A$, so there exists a $\lambda\in Spec(B)$ associated with $v$. Thus $v$ is an eigenvector of $AB$ associated with the eigenvalue $\lambda k\in spec(AB).$