There is a base such $A$ and $B$ are both upper triangular on these base, and if $A$ and $B$ are diagonalizable, then $A$ and $B$ are diagonalizable simultaneously.
For the first I have no idea. To the second one I am trying to show something like: Iv $v$ is an eigenvector for $A$ then $Bv$ is an eigenvector for $A$.
To the second I proceeded as:
$$v \in \ker (A - \lambda * Id)$$ for same $\lambda \neq 0$, then
$$Bv = B\frac{1}{\lambda}Av = \frac{1}{\lambda}BAv = \frac{1}{\lambda}ABv$$
Then
$$A(Bv) = \lambda Bv$$ and then $Bv$ is an eigenvector for $A$ also with eigenvalue $\lambda$.
But I am stuck here. I mean, I don't know what to do with this.
I do appreciate any hint or answer.
