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?
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.