# Product of two matrices sharing the same eigenvector

I have never learned how to prove things, so I don't know how or when a proof is finished. If you prove this, it may help me understand a bit how things are done:

The vector $e$ is an eigenvector of each of the $n\times n$ matrices $A$ and $B$, with the corresponding eigenvalues $\lambda$ and $\mu$, respectively. Prove that $e$ is an eigenvector of the matrix $AB$ with the eigenvalue $\lambda\mu$.

What you know is $Ae=\lambda e$ and $Be=\mu e$. This is what your are allowed to use.
Can you compute $ABe$ now just using the above?
• Is it: $ABe = (A(Be) = A(µe) = (Ae)µ = µλe$ ? – George May 28 '14 at 11:36
• Yes, this is right. Does this prove that $e$ is an eigenvector of $AB$? – daw May 28 '14 at 11:39
• Yes because $ABe = µλe$. Thanks guys by the way, for helping me solve. – George May 28 '14 at 11:43