Find the spectral projection of a positive operator explicitly

Question

Let $a$ be an element in $\mathcal B(\mathcal H)$. How can one show that $\chi_{(0,\infty)}(aa^∗)=[a\mathcal H]$? Here $[a\mathcal H]$ is the projection corresponding to the closed subspace $\overline{a\mathcal H}$.

Answer 1

You have $$ (a\mathcal H)^\perp=\ker a^*=\ker aa^*=(aa^*\mathcal H)^\perp. $$ So $[a\mathcal H]=[aa^*\mathcal H]$.

Now, from $t\,\chi_{(0,\infty)}(t)=t$ on $[0,\infty)$, you get $$ \chi_{(0,\infty)}(aa^*)\,aa^*=aa^*\,\chi_{(0,\infty)}(aa^*)=aa^*. $$ So $$ \chi_{(0,\infty)}(aa^*)\,\overline{a\mathcal H}=\chi_{(0,\infty)}(aa^*)\overline{aa^*\mathcal H}=\overline{aa^*\mathcal H}=\overline{a\mathcal H}. $$ This shows that $\chi_{(0,\infty)}(aa^*)\geq[a\mathcal H]$. Conversely, $\chi_{(0,\infty)}(aa^*)$ can be written as a limit of polynomials $p_n(aa^*)$ with $p_n(0)=0$, so $$[a\mathcal H]\,\chi_{(0,\infty)}(aa^*)=\chi_{(0,\infty)}(aa^*),$$' showing that $\chi_{(0,\infty)}(aa^*)\leq[a\mathcal H]$.

Answer 2

As @Martin Argerami suggests, we need to prove that $\chi_{(0,\infty)}(a^*a) = [aa^*\mathcal{H}]$.

Since $\sigma(aa^*) \subseteq [0, \infty\rangle$ we have $$\chi_{(0,\infty)} = 1 - \chi_{\{0\}}$$ In general we have $\chi_{\{0\}}(aa^*) = [\ker aa^*]$ so $$\chi_{(0,\infty)}(aa^*) = 1(aa^*) - \chi_{\{0\}}(aa^*) = I-[\ker aa^*]= [aa^*\mathcal{H}]$$

The last equality is clear from $\mathcal{H} = \ker aa^* \oplus \overline{aa^*\mathcal{H}}$.