Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Digital systems, linear algebra, introduction to formal proof / 5 / f

For a sum of mutually orthogonal projectors, orthonormality gives $q_i^T{q}_j=0$ for $i\ne j$ and one for $i=j$. Therefore $P_i{P}_j=q_i(q_i^T{q}_j)q_j^T$ equals zero for different indices and $P_i$ for equal indices. Expanding the finite sum yieldsThus every stated partial sum is a projector. Its image of a linear map is the span of the first $k$ orthonormal set of vectors; arbitrary sums of idempotent matrices would not have this conclusion.

The image of a linear map of $P_1$ is exactly the one-dimensional span of $q_1$: its formula always returns a multiple of $q_1$, and $P_1{q}_1=q_1\ne 0$. Hence $\mathrm{rank}(P_1)=1$.

The rank-nullity theorem gives $\dim \mathcal{N}(P_1)=n-1$. Since $\mathcal{C}(I_n-P_1)=\mathcal{N}(P_1)$ by (e)(i), $\mathrm{rank}(I_n-P_1)=n-1$. This includes the case $n=1$, where the complementary projector is zero.