I am stuck with this seemingly easy problem but I am having trouble showing this:

Let $\mathcal{A}\subseteq\mathcal{B}(\mathcal{H})$ be a von Neumann algebra realized inside a subalgebra of the bounded operators on a Hilbert space $\mathcal{H}$ and $p\in\mathcal{A}$ an orthogonal projection on a (non-zero) subspace of $\mathcal{H}$. Let furthermore $A\in\mathcal{A}$ be positive and invertible.

Show that $pAp$ is positive & invertible and the inverse is given by $$(pAp)^{-1} = p A^{-1} p.$$

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    $\begingroup$ For $p=0$ we have an obviuously false statement. If things should have a sense, than the text is longer. Maybe $\mathcal A$ is realized inside some $B(H)$, algebra of bounded operators on the Hilbert space $H$, then $p$ is an orthogonal projector on some subspace $H_1=[p]=pH$ of $H$, then we may consider "some restriction" of $A$... $\endgroup$
    – dan_fulea
    Feb 4 '19 at 12:00
  • $\begingroup$ Thank you very much! Yes, indeed! I will adjust the text accordingly $\endgroup$
    – Alvo
    Feb 4 '19 at 12:04
  • $\begingroup$ $pAp$ is positive with respect to what? The cone of $\mathcal{H}$ is not guaranteed to lie in the image of $p$. For example let $\mathcal{H}=\mathbb{R}^{2}$ with positive cone $C=\{(x,y)\in\mathbb{R}^{2}:x,y\geq0\}$, $A=I$ the identity operator and $p$ the projection on span$((1,-1))$. Then $pAp((2,1))=(1,-1)\not\in C$. $\endgroup$ Feb 4 '19 at 12:26
  • $\begingroup$ By "positive & invertible" I mean that the operator is self-adjoint and its spectrum lies in $\mathbb{R}^+.$ $\endgroup$
    – Alvo
    Feb 4 '19 at 12:29

If $A$ is positive and invertible, let $\lambda=\min\sigma(A)$. Then $A\geq\lambda I$. Thus $$ pAp\geq \lambda p, $$ and so $\sigma(pAp)\subset [\lambda,\infty)$ and $pAp$ is invertible in $pB(H)p$.

But the equality $(pAp)^{-1}=pA^{-1}p$ does not usually hold. For instance take $H=\mathbb C^2$, and $$ A=\begin{bmatrix} 2&1\\1&1\end{bmatrix},\ \ p=\begin{bmatrix} 1&0\\0&0\end{bmatrix}. $$ Then, as $A^{-1}=\begin{bmatrix} 1&-1\\-1&2\end{bmatrix}$, we have $$ pAp=2p\ne p=pA^{-1}p. $$

  • $\begingroup$ I see. Thank you very much! I thought I needed this to show that for a finite and faithful trace $\phi$ I have that $\phi(pAppA^{-1}p)=\phi(p).$ So I guess the above idea was too strong... $\endgroup$
    – Alvo
    Feb 4 '19 at 17:00
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    $\begingroup$ But that's not true. Take my example above and the usual trace: you are trying to show that $\phi(2p)=\phi(p)$. $\endgroup$ Feb 4 '19 at 17:04
  • $\begingroup$ Of course! Now I get it! Thanks so much!! :) $\endgroup$
    – Alvo
    Feb 4 '19 at 17:08

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