Show that if $A$ and $B$ are commuting $n*n$ real matrices, then each eigenvalue of $AB$ is a product of some eigenvalue of $A$ with some eigenvalue of $B$.
I know the fact that since $AB=BA$ , $A$ and $B$ have a common eigenvector but I could not use it.
Two matrices which commute are simultaneously triangularizable, i.e., there exists an invertible matrix $P$ such that both $A'=PAP^{-1}$ and $B'=PBP^{-1}$ are upper triangular. Now we can show: $$ AB = P^{-1}A'PP^{-1}B'P=P^{-1}A'B'P $$ $A'B'$ is still upper triangular. The diagonal entries of an upper triangular matrix are just its eigenvalues which are unaffected by a similarity transformation (that is, taking $PA'B'P^{-1}$), so the eigenvalues of $AB$ are products of eigenvalues of $A$ and $B$.