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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?

Hanno
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User8976
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    What does it mean for a linear operator to be positive? – Jacob Wakem Nov 06 '16 at 04:27
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    Its an operator with a good attitude. Or, $\langle x, L x \rangle > 0$ for all $x \neq 0$. – copper.hat Nov 06 '16 at 04:30
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    With @copper.hat's definition, the problem is probably wrong. We need to require $L$ to be Hermitian (aka self-adjoint). – darij grinberg Nov 06 '16 at 05:26
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    Also, the real case with the dot product is solved in http://math.stackexchange.com/questions/1365079/the-product-of-two-positive-definite-matrices-has-real-and-positive-eigenvalues . I think the generalization is pretty straightforward. – darij grinberg Nov 06 '16 at 05:28

2 Answers2

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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: $$\begin{align} T \: & =\: \sqrt{I+E} \;=\;\sqrt{(I-E)+2E}\\[1ex] & =\: (I-E) + \sqrt2\,E \;=\; I + \big(\sqrt 2 -1\big) E \end{align}$$

Hanno
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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.

glS
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