Why do $S^{1/2}$ and $T^{1/2}$ commute This question is actualy related to my old question Product and sum of positive operators is positive
If $S,T \in B(H)$ are bounded, linear and normal operators on a Hilberspace $H$, i.e. $SS^*=S^*S$ and  $TT^*=T^*T $
Assume that $S,T\geq 0$ and $ST=TS$. My goal was to show $ST\geq 0$.
Since $S,T\geq 0$ both have a unique positive square root, and are self-adjoint ($S=S^*, T=T^*$). I got 2 ways of doing this, but in both cases I need something I cant prove. I either need that $S^{1/2}$ and $T^{1/2}$ commute and $S^{1/2}T^{1/2}$ is selfadjoint. Or I need that $S^{1/2}$ commutes with $T$, and that $S^{1/2}$ is self-adjoint. (I know now that $S^{1/2}$ is self-adjoint since it is positive)
Then we can see that $(STx,x) = (S^{1/2}T^{1/2}S^{1/2}T^{1/2}x,x) = (S^{1/2}T^{1/2}x, S^{1/2}T^{1/2}x) \geq0$ 
Or in the second case $(STx,x) = (S^{1/2}TS^{1/2}x, x)  = (TS^{1/2}x,S^{1/2}x) \geq 0$
How can I show this?
 A: The positive square roots are defined by the continuous functional calculus. For a normal operator $T$, a continuous function $f(T)$ is the norm-limit of polynomials in $T$ and $T^*$. (You need the identity operator $I$ for $f$'s that don't vanish at $0$ but that doesn't apply here.) 
But by your assumption, any polynomials in $T$ and $S$ commute. So passing to the limit gives $f(T)g(S) = g(S)f(T)$ for any $f$ and $g$ that are continuous on their respective spectra. 
A: Here's an alternate approach to the problem: once you've shown that $ST$ is self-adjoint (using $ST = TS$), it is enough to show that $(STx,x)>0$ for all $x$ for which $(x,x)=1$.  Note that for any such $x$, we have
$$
\begin{align}
0 &< (\inf_{(x,x)=1} \left\{(Sx,x)\right\})( \inf_{(x,x)=1} \left\{(Tx,x)\right\})\\
& \leq (Sx,x)(Tx,x)\\
&= (x^*Sx)(x^*Tx) = x^* S(x^*x)Tx =x^*STx
\end{align}
$$

The following works in the finite-dimensional case: since $A,B$ form a commuting family of matrices, they are simultaneously triangularizable with respect to some orthonormal basis.  Since $A$ and $B$ are normal, this means that they are simultaneously diagonalizable.
Thus, the question is reduced to computing the product of two positive definite and diagonal operators.
