TITLE: A result about commuting matrices in $ M(n, \mathbb{C} ) $
QUESTION [5 upvotes]: Let $ A $ be a matrix in $ M(n, \mathbb{C} ) $ and let $ A^{*} $ be its Hermitian adjoint. Suppose that the matrices $ A $ and $ AA^{*}-A^{*}A $ commute. Show that $ AA^{*} = A^{*}A $.
Here is a sketch of my solution : 
Lemma : Let $ F $ be a field of characteristic $ 0 $ and $ S,T \in M(n, F) $. Suppose that $ ST-TS $ commutes with $ S $. Then, $ ST-TS $ is nilpotent.
We have that $ A $ and $ B = AA^{*} - A^{*}A $ commute, so by the lemma, $ B $ is nilpotent. Also, $ B^{*} = AA^{*} - A^{*}A = B $, so $ B $ is Hermitian. But, $ B $ being both, nilpotent and Hermitian $ \implies $ $ B=0 $ $ \implies $ $ AA^{*} = A^{*}A $
I was wondering whether there is a simpler solution as mine uses some machinery. Any hints please?

REPLY [2 votes]: Your solution is a good one. Another one is:
Since a hermitian matrix is diagonalizable in an orthonormal basis, we may assume that $AA^*-A^*A=diag(\lambda_1I_{n_1},\cdots,\lambda_kI_{n_k})$ where $n_1+\cdots+n_k=n$. Thus $A$ has the form $A=diag(U_1,\cdots,U_k)$ and $AA^*-A^*A=diag(U_1U_1^*-U_1^*U_1,\cdots,U_kU_k^*-U_k^*U_k)$. Finally $\lambda_iI_{n_i}=U_iU_i^*-U_i^*U_i$, that implies $\lambda_i=0$ (cf. the trace) and we are done.