Prove that the product of two positive linear operators is positive if and only if they commute. Having problem in the following problems on positive forms:
$1)$ Prove that the product of two positive linear operators is positive if and 
only if they commute. 
I am able to do one direction that if the product of two positive linear operators is positive then they commute. But unable to do the opposite direction.
Let $T,S$ be two positive linear operators and they commute , i.e. $ST = TS$. To show the product of two positive linear operators is positive we have to show that $\langle TS\alpha,\alpha\rangle > 0$ for any $\alpha \neq0$ and $(TS)^* = TS$. I have shown the part  $(TS)^* = TS$.
I need help to show that $\langle TS\alpha,\alpha\rangle > 0$ for any $\alpha \neq0$.
$2)$ Let $V$ be a finite-dimensional inner product space and $Ε$ the orthogonal 
projection of $V$ onto some subspace.
$(a)$ Prove that, for any positive number $c$, the operator $cI + Ε$ is positive.
$(b)$ Express in terms of $Ε$ a self-adjoint linear operator $Τ$ such that $T^2 = I + E$. 
In this I am able to do part $(a)$ but unable to the second part.
Can anyone give me any lead to the problems?
 A: 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:
$$\begin{align}
T \: & =\: \sqrt{I+E} \;=\;\sqrt{(I-E)+2E}\\[1ex]
 & =\: (I-E) + \sqrt2\,E \;=\; I + \big(\sqrt 2 -1\big) E
\end{align}$$
A: Suppose $T,S,TS\ge0$. Then
$$ST= S^\dagger T^\dagger = (TS)^\dagger = TS.$$
For the other direction, suppose $T,S\ge0$ and $[T,S]=0$. Consider the unique positive square root of $S$: the operator $\sqrt S\ge0$ such that $(\sqrt S)^2=S$.
Then $[T,\sqrt S]=0$, and therefore for any vector $x$ we have
$$\langle TS x,x\rangle = \langle T\sqrt S x,\sqrt S x\rangle \ge 0$$
(this is the argument already provided in the other answer).
If the underlying space if finite-dimensional, and thus $T,S$ are matrices, we can also prove the latter direction observing that if $[T,S]=0$ then $T$ and $S$ are simultaneously diagonalisable. The eigenvalues of $TS$ must then be products of pairs of eigenvalues of $T$ and $S$, and therefore $TS$ must be positive.
