Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $H$ be a complex Hilbert space, $T\in H'$ and $T=T^*$. Here is where I need help: If $\sigma(T)\subset\{0,1\}$ then $T=T^2$.

Using the spectral theorem I know that $\{0,1\} \supset q(\sigma(P)) = \sigma(q(T))$ for $q:x\mapsto x^2$, thus $\sigma(T^2)=\sigma(T)$. How can I continue?

share|improve this question
    
Let $E$ be the spectral measure of $T$, then we have, as $\lambda^2 = \lambda$ for $\lambda \in \sigma(T)$, that $T^2 = \int_{\sigma(T)} \lambda^2\, dE_\lambda = \int_{\sigma(T)}\lambda\, dE_\lambda = T$. –  martini Dec 1 '12 at 10:37

1 Answer 1

Or more simple: Consider the polynomial $p(x) = x^2 - x$ and let $S := p(T)$. By the spectral theorem we have $\sigma(S) = p\bigl(\sigma(T)\bigr) = \{0\}$. Also $S$ is selfadjoint since $S^* = p(T)^* = p(T^*) = p(T)$. Hence $$\|S\| = \rho(S) = \max_{\lambda\in \sigma(S)} |\lambda| = 0 $$ So $0 = S = T^2 - T$.

share|improve this answer
    
the last equation should be $T^2 - T$ not $T^2 = T$. –  mland Dec 1 '12 at 12:22
1  
@mland Thx. Corrected it. –  martini Dec 1 '12 at 12:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.