Prove that if \(E\) is the projection on \(R\) along \(N\), then \((I - E)\) is the projection on \(N\) along \(R\).
Solution
Because \(E\) is the projection onto \(R\) along \(N\), one has \[ V=R\oplus N, \] and every \(v\in V\) has a unique decomposition \[ v=r+n, \qquad r\in R,\ n\in N, \] with \[ E(v)=r \] [@HK71]. Therefore \[ (I-E)(v)=v-E(v)=(r+n)-r=n. \] Thus \(I-E\) fixes every vector of \(N\) and kills every vector of \(R\). Consequently \[ \operatorname{range}(I-E)=N, \qquad \ker(I-E)=R. \] Moreover, \[ (I-E)^2=I-2E+E^2=I-E \] since \(E^2=E\). Hence \(I-E\) is precisely the projection onto \(N\) along \(R\).