Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3 / 19G / b / Solution

For a normal operator $\varphi$, the adjoint operator identity and ${\varphi }^{\ast }\varphi =\varphi {\varphi }^{\ast }$ givePositive definiteness implies $\ker \varphi =\ker {\varphi }^{\ast }$. Also $(\mathrm{im}\varphi {)}^{\perp }=\ker {\varphi }^{\ast }$, since orthogonality to every $\varphi v$ means $⟨{\varphi }^{\ast }u,v⟩=0$ for every $v$. In finite dimension this gives $\mathrm{im}\varphi =(\ker \varphi {)}^{\perp }$.

The idempotent linear map decomposes $U=\ker \varphi \oplus \mathrm{im}\varphi$ as in part (a), now orthogonally, and acts as zero on its kernel and identity on its image. Writing $u=k+w$ and $v=k^{\prime }+w^{\prime }$ in these orthogonal summands,Thus $\varphi$ is self-adjoint, and together with idempotence this proves $\varphi$ is an orthogonal projection.