My preferred answer was given by Robert Israel, as it is a great opportunity to meet the Fuglede-Putnam-Rosenblum Theorem.
Here is an interpolation argument in the finite-dimensional case where you stand. I will first prove a lemma for matrices. Of course, the same holds for operators on a finite-dimensional complex inner-product space.
Lemma: a matrix $A\in M_n(\mathbb{C})$ is normal if and only if there exists a polynomial $P\in \mathbb{C}[X]$ such that $A^*=P(A)$.
Proof: the implication $\Leftarrow$ is obvious. So assume $A$ is normal. Equivalently, there exists a unitary matrix $U$ and a diagonal matrix $D=\mbox{Diag}\{\lambda_1,\ldots,\lambda_n\}$ such that $A=UDU^*$. Then $A^*=UD^*U^*$ where $D^*=\mbox{Diag}\{\overline{\lambda_1},\ldots,\overline{\lambda_n}\}$.
By Lagrange interpolation, we can find a polynomial $P\in\mathbb{C}[X]$ such that $P(\lambda_j)=\overline{\lambda_j}$ for $j=1,\ldots,n$. Hence $D^*=P(D)$ and it follows that $$P(A)=P(UDU^*)=UP(D)U^*=UD^*U^*=A^*.$$
QED.
Now assume that $T$ is normal and commutes with $S$. Then $S$ commutes with every power of $T$, hence with every polynomial in $T$. By the above, $S$ commutes with $T^*$.
The facts you want to prove follow easily.
Note: this remains true in $\mathbb{R}$. But note that we need to be a little bit careful as real normal matrices are not diagonalizable in $M_n(\mathbb{R})$ in general.
To be clear and correct in the following it is assumed that $\,W$ is positive$\,$
signifies that
- $W$ is self-adjoint, i. e. $W^*=W$, and
- $\langle W\alpha|\alpha\rangle > 0$ for all $\alpha \neq0\,$.
Ad $1)\:\:$ If $\,T,S\,$ are positive and they commute, then $\sqrt S\,$
( = the unique positive square root, being a power series in $S$) also commutes with $T$, that is $T\sqrt S = \sqrt S\,T$. Then
$$\langle TS\alpha|\alpha\rangle \:=\:
\langle T\sqrt S\,\alpha|\sqrt S\,\alpha\rangle> 0\,$$
for any $\alpha\neq0\,$.
Ad $2b)\:\:$ Decomposing the identity along the subspace $V$ as $\,I=(I-E)+E\,$, you may, thanks to orthogonality, take summand-wise the positive square root:
$$\begin{align}
T \: & =\: \sqrt{I+E} \;=\;\sqrt{(I-E)+2E}\\[1ex]
& =\: (I-E) + \sqrt2\,E \;=\; I + \big(\sqrt 2 -1\big) E
\end{align}$$
Best Answer
You can use Fuglede's theorem: http://en.wikipedia.org/wiki/Fuglede%27s_theorem which ensures, in your case that $ST^*=T^*S$ and $S^*T=ST^*$.
In view of the facts that $(TS)^*=S^*T^*$, $SS^*=S^*S$ and $TT^*=T^*T$, the conclusion follows by easy computations: $$ TS(TS)^*=T(SS^*)T^*=(TS^*)ST^*=S^*T(ST^*)=S^*(TT^*)S=(S^*T^*)TS=(TS)^*TS $$
The property is not longer true in the unbounded case. You can follow the paper "M. H. Mortad, On the closedness, the self-adjointness and the normality of the product of two unbounded operators, Demonstratio Math., 45 (2012), 161-167" for an example. It is also stated there that the normality of $TS$ is valid when, in addition, $S$ (or $T$) is unitary.