I know that given a self-adjoint operator $A$ and $F \subset \mathbb{R}$ a Borel set, one can define the bounded operator $\mathbb{1}_F(A)$. Now, what I don't understand is why if $\lambda$ is an eigenvalue of $A$, then $\mathbb{1}_{\{\lambda\}}(A)$ is the orthogonal projection on $Ker(A - \lambda I)$? What I get is that $\mathbb{1}_F$ is an orthogonal projection for all $F$, but I don't know how to recover the space on which we're projecting...
Thanks for any help!
You have the equality (as functions) $$ t\,1_{\{\lambda\}}(t)=\lambda\,1_{\{\lambda\}}(t). $$ So for any $\xi\in 1_{\{\lambda\}}(A)H$ you have $$ A\xi=A\,1_{\{\lambda\}}(A)\,\xi=\lambda\,1_{\{\lambda\}}(A)\xi=\lambda\,\xi. $$ And if $\xi\in \ker (A-\lambda I)$, then $A\xi=\lambda\,\xi$. If $\lambda\ne0$, then $\xi\in AH$, so $\xi\in \overline{1_{\{\lambda\}}(A)H}=1_{\{\lambda\}}(A)H$. If $\lambda=0$, then $A\xi=0$, so for any polynomial (and thus any continuous function) $f$ we have $f(A)\xi=f(0)\xi$; the function $f=1_{\{0\}}$ satisfies $f(0)=1$, so we may approximate it with continuous functions that also take the value $1$ at $0$; thus $1_{\{0\}}(A)\xi=\xi$.