TITLE: Proof of an equivalence with respect to self-adjoint operators
QUESTION [5 upvotes]: I am trying to solve this problem but I fail in the converse part...

Show  that  the  product of two  self-adjoint operators is  self-adjoint if and  only 
  if the  two  operators  commute.

($\Rightarrow$) If we suppose that $T,U,V$ are self-adjoint operators such that $T=UV$ and $U^*=U,V^*=V,T^*=T$ we have by Theorem that:
$$(UV)^*=V^*U^*=VU$$
but we know, by hypothesis, that: $$(UV)^*=T^*=T=UV$$.We already know that the adjoint operator is unique, so we hav that $UV=VU$ and we are done with this part.
So my troubles are this the converse, can anyone help please.

REPLY [2 votes]: Suppose $UV=VU$. Then, for any $x,y$ in the given space
$$ 0=\langle UVx,y\rangle  - \langle VUx,y\rangle=\langle x,V^*U^*y\rangle - \langle x,U^*V^*y\rangle = \langle x, (V^*U^* - U^*V^*)y\rangle .$$
By the Zero-Operator Lemma (i.e., $T=0$ iff $\langle Tx,  y\rangle =0$ for all $x,y$), $V^*U^* - U^*V^*=0$. Since $U^*=U$ and $V=V^*$, we have $0=V U  - U^*V^*= VU - (VU)^*.$