Schur decomposition and upper triangular matrix

Question

Can someone please help me with this problem. If $A\in \mathbb C^{n\times n}$ has distinct eigenvalues. How do I show that if $Q^*AQ=T$ is the Shur decomposition and $AB=BA$, then $Q^*BQ$ is upper triangular?

Answer 1

The crucial argument is as follows.

Consider any eigenvector $\lambda$ of $A$ and associated eigenvector $v$. Then $$A(Bv)=(AB)v=(BA)v=B(Av)=B(\lambda v)=\lambda(Bv)$$

Therefore $Bv$ is also an eigenvector of $A$ associated with $\lambda$ and so $Bv$ is a scalar multiple of $v$. Thus $v$ is also an eigenvector of $B$.

If a basis is used with $v$ as its first element, the linear transformations represented by $A$ and $B$ will therefore have all first row entries zero apart (possibly) from the first entry which will be the respective eigenvalues.

An inductive argument can then be used to complete the proof.