# Inverse of compression (von Neumann algebra)

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 $$(pAp)^{-1} = p A^{-1} p.$$

• 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 $A$... Feb 4 '19 at 12:00
• Thank you very much! Yes, indeed! I will adjust the text accordingly
– Alvo
Feb 4 '19 at 12:04
• $pAp$ 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 $pAp((2,1))=(1,-1)\not\in C$. Feb 4 '19 at 12:26
• By "positive & invertible" I mean that the operator is self-adjoint and its spectrum lies in $\mathbb{R}^+.$
– Alvo
Feb 4 '19 at 12:29

## 1 Answer

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 $$pAp=2p\ne p=pA^{-1}p.$$

• 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...
– Alvo
Feb 4 '19 at 17:00
• But that's not true. Take my example above and the usual trace: you are trying to show that $\phi(2p)=\phi(p)$. Feb 4 '19 at 17:04
• Of course! Now I get it! Thanks so much!! :)
– Alvo
Feb 4 '19 at 17:08