= Solution
For a <normal operator> $\phi$, the <adjoint operator> identity and $\phi^*\phi=\phi\phi^*$ give
$$
\|\phi u\|^2=\langle\phi^*\phi u,u\rangle=\langle\phi\phi^*u,u\rangle=\|\phi^*u\|^2.
$$
Positive definiteness implies $\ker\phi=\ker\phi^*$. Also $(\operatorname{im}\phi)^\perp=\ker\phi^*$, since orthogonality to every $\phi v$ means $\langle\phi^*u,v\rangle=0$ for every $v$. In finite dimension this gives $\operatorname{im}\phi=(\ker\phi)^{\perp}$.
The <idempotent linear map> decomposes $U=\ker\phi\oplus\operatorname{im}\phi$ as in part (a), now orthogonally, and acts as zero on its <kernel> and identity on its <image>. Writing $u=k+w$ and $v=k'+w'$ in these orthogonal summands,
$$
\langle\phi u,v\rangle=\langle w,w'\rangle=\langle u,\phi v\rangle.
$$
Thus $\phi$ is <self-adjoint>, and together with idempotence this proves \b[$\phi$ is an orthogonal projection].
Back to article page
