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