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

For any idempotent linear map $\tau$, each vector decomposes asIf $w\in \ker \tau \cap \mathrm{im}\tau$, write $w=\tau v$; then $0=\tau w={\tau }^{2}v=\tau v=w$. Thus the kernel of a linear map and image of a linear map give a direct sum.

For the self-adjoint map in the question, $k\in \ker \tau$ and $w=\tau v$ satisfy $⟨k,w⟩=⟨k,\tau v⟩=⟨\tau k,v⟩=0$. Hence the direct sum is an orthogonal decomposition: