support of an operator on a Hilbert space

Question

Let $x\colon\mathcal{H}\to\mathcal{H}$ be a self-adjoint operator,

the support $s(x)$ of $x$ is defined as the smallest projection $e\in B(\mathcal{H})$ such that $ex=xe=x$.

Let $x=\int\lambda \, dE_x$ be the spectral decomposition.

Then $s(x)=E_x(\mathbb{R}\setminus\{0\})$ und $E_x(\{0\})=\ker x$.

How can these properties be proven, or where can I find such elementary things with proofs?

Answer 1

Because $x$ commutes with $e$, then $e$ commutes with $E$ as well, including with $f=E_x(\mathbb{R}\setminus\{0\})$. Because $fx=xf=x$ and $fe=ef$, then $ef$ is also a projection such that $(ef)x=x(ef)=x$. Therefore $ef=fe=e$ follows from the minimality of $e$. Furthermore $x(f-e)=0$, which puts the range of $f-e$ in $\mathcal{N}(x)=\mathcal{R}(E_x\{0\})$; hence $0=f(f-e)=f^2-fe=f-e$.