I am not sure that I explained to myself the missing details in the proof right, so please check my explanations. (The proof is taken from "$C^*$ -Algebras by Example"-Davidson)
First, I don't know how we use the fact that A is a positive operator. We can assume it's self adjoint just to have non-zero point in it's spectrum.
For the second part of the proof: Let $T \in E \mathscr{A}E$ self-adjoint non-scalar multiple of E.
$T$ is compact self-adjoint, thus the spectrum of $T$ is finite or a sequence converges to zero+zero itself, and each element in the spectrum, besides zero, is an eigenvalue with finite dimensional eigenspace. If we denote by $P_{\lambda i}$ the projections onto the eigenspaces $W_{\lambda i}$, then $T=\sum_{i=0}^\infty \lambda_i P_{\lambda i}$.
Now, As $Im(P_{\lambda i})=W_{\lambda i} \subseteq Im(E)$ $\forall i$ we conclude $P_{\lambda i} \le E \; \forall i$ . But $E$ is minimal, so $\forall i$ $P_{\lambda i} = E$.
Thus, we can write $T=\sum_{i=0}^\infty \lambda_i E$ and we got a contradiction, as we assumed $T$ is not a scalar multilpe of $E$.
I am not only worried that there are mistakes in my proof, but also I have a feeling I overcomplicate it.
Thank you for your time.
EDIT: Ok, maybe another approach:
- If the spectrum contains exactly one non-zero point $\lambda$ which is an eigenvalue with projection $P_\lambda$ , from the arguments above, we get $T=\lambda P_\lambda =\lambda E$, a contradiction.
-If the spectrum contains at least two non-zero eigenvalues $\lambda$ and $\mu$ then we conclude from minimality of $E$, as above, $P_\lambda=P_\mu=E$ but this is not a possibility because $P_\lambda$ is orthogonal to $P_\mu$.
I think you make it unncessarily complicated. By the minimality of $E$, you have $P_{\lambda_i}=P_{\lambda_j}$ for all $i,j$. But each $P_{\lambda_i}$ is the particular projection corresponding to $\lambda_i$; so $\lambda_i=\lambda_j$ for all $i,j$ (projections corresponding to different eigenvalues are orthogonal to each other). Thus $T=\lambda_1\,E$.