1
$\begingroup$

Let $T$ be a bounded self-adjoint operator. I already showed that the operator $\mathbb{1}_{U} (T)$ is an orthogonal projection, where $\mathbb{1}_{U}$ is the characteristic function of the Borel set $U \subset \mathbb{R}$. (To make sense of this, one can think of $\mathbb{1}_U(T) $ as restricted to the spectrum $\sigma(T)$.)

However, now I wish to show that $\mathbb{1}_{\left\{\lambda \right\}} (T)$ is non-zero if and only if $\lambda$ is an eigenvalue of $T$.

My attempt is this: I think we have the identity $$ (x- \lambda) \mathbb{1}_{\left\{ \lambda \right\}}(x) = 0 $$ for all $x \in U$. Then applying the star homomorphism $\Phi$ to this, we get $$ (T - \lambda) \mathbb{1}_{ \left\{ \lambda \right\} } (T) = 0. $$ Does this prove that $\lambda$ is an eigenvalue of $T$?

$\endgroup$
1
$\begingroup$

Yes. If $1_{\{\lambda\}}(T)\ne0$, there exists nonzero $v\in H$ with $1_{\{\lambda\}}(T)v=v$. Then $$ Tv=T1_{\{\lambda\}}(T)v=\lambda 1_{\{\lambda\}}(T)v=\lambda v, $$ so $\lambda $ is an eigenvalue. Conversely, if $\lambda$ is an eigenvalue then $Tv=\lambda v$ for some nonzero $v$, and so you can deduce that $f(T)v=f(\lambda)v$ for all bounded Borel functions $f$ (start with monomials, then polynomials, then take limits). So $$ 1_{\{\lambda\}}(T)v=1_{\{\lambda\}}(\lambda)v=v. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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