2

enter image description here

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 $P_\mu$.

Shirly Geffen
  • 2,571

1 Answers1

1

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 $T=\lambda_1\,E$.

Martin Argerami
  • 205,756
  • Very simple. Thank you! – Shirly Geffen Apr 22 '16 at 17:09
  • My pleasure. Davidson's book is a good place to learn how to write proofs that are concrete and direct. – Martin Argerami Apr 22 '16 at 17:41
  • I shall be grateful if you could clarify what is meant by minimal projection? I am reading this chapter on my own . – Noob mathematician Aug 26 '21 at 12:20
  • 1
    It means that it does not have a proper subprojection. In a context with lots of projections, like the algebra of compact operators or for instance a von Neumann algebra, it equivalently means that $E\mathfrak A E=\mathbb CE$, as the Lemma says. – Martin Argerami Aug 26 '21 at 15:20
  • This minimal projection might not be unique right? – Noob mathematician Aug 26 '21 at 16:43
  • And should it always be of rank 1. – Noob mathematician Aug 26 '21 at 17:03
  • 1
    No, a rank-one projection is minimal, but a minimal projection needs not to have rank 1, as minimal is a relative notion. For instance, in $M_3(\mathbb C)$ consider the subalgebra $$A={a,E_{11}+b,(E_{22}+E_{33}):\ a,b\in\mathbb C}.$$ The projection $E_{11}$ is minimal in $A$, but so is $E_{22}+E_{33}$. – Martin Argerami Aug 26 '21 at 18:20
  • Thank you sir for your help . This means a lot to me . – Noob mathematician Aug 26 '21 at 19:46
  • There is another question that I wanted to ask . For the second part of the proof we taking $T\in E\mathfrak A E$ self adjoint that is not a scalar multiple of $E$. Shouldn't we take $T\in E\mathfrak A E$ ( not necessarily self adjoint ) ? I couldn't understand why it suffices to show for self adjoints only . – Noob mathematician Aug 26 '21 at 21:09
  • 1
    Every element is a linear combination of selfadjoints. If you show that all selfadjoints are scalar multiples of the identity, it follows immediately that all elements are. – Martin Argerami Aug 27 '21 at 00:47
  • @MartinArgerami, could you please clarify how does the second line of the proof follow from continuous functional calculus (Theorem I.3.3)…. – sigma Aug 17 '22 at 05:07
  • By “spectral projection” corresponding to a non zero eigenvalue, say, $\lambda$ I assume the projection onto the $\ker(A-\lambda)$…. But then why it belongs to the C*-algebra generated by A?? How to write $P$ as $f(A)$?? – sigma Aug 17 '22 at 05:09
  • The nonzero elements of the spectrum of a compact operator are isolated eigenvalues. Take $f$ to be any continuous function with $f(\lambda)=1$ and zero on all other eigenvalues. – Martin Argerami Aug 17 '22 at 05:12