Let $\mathbf{A},\mathbf{B}\in\mathrm{M}_n(\mathbb{R})$ such that $\mathbf{A}$ invertible and diagonalisable, and $\mathbf{AB}=\lambda \mathbf{BA}$ for some $\lambda >1$. I want to show that $\mathbf{B}$ is nilpotent i.e. $\mathbf{B}^k=0$ for some $k$. What I have so far:
The condition implies $\mathbf{AB}^m=\lambda^m \mathbf{B}^m\mathbf{A}$. Taking a basis in which $\mathbf{A}$ is diagonal, and writing $b_{ij}^{(m)}$ as the $ij$th entry in $\mathbf{B}^m$, we have: $$a_ib_{ij}^{(m)}= \lambda^m a_j b_{ij}^{(m)}\implies b_{ij}^{(m)}=\frac{1}{\lambda^m a_i\big/a_j-1}\overset{m\to\infty}\longrightarrow 0$$ where $a_j\neq 0$ since $\mathbf{A}$ invertible. So $\mathbf{B}^m\to 0$ but it's not clear to me why the sequence must actually hit $0$? Would appreciate some help.
Let $\alpha$ be a (possibly complex) eigenvalue of $B$, with eigenvector $v$. Then $$BA v=\frac{1}{\lambda}ABv=\frac{\alpha}{\lambda}Av,$$ and $Av\neq 0 $ since $A$ is invertible, so $\frac{\alpha}{\lambda}$ is also an eigenvalue of $B$. You can iterate this argument to show that $\frac{\alpha}{\lambda^m}$ is an eigenvalue for all $m\in \mathbb{N}$. But $B$ cannot have infinitely many eigenvalues; thus $\alpha=0$.
Since $B$ only has $0$ as an eigenvalue, its characteristic polymial is $$\det(tI-B)=t^n,$$ and by the Cayley-Hamilton Theorem $B^n=0$.