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 / b / Solution

Associativity of matrix multiplication identifies $(q_i^{T}x)q_i=q_i(q_i^{T}x)=(q_i{q}_i^T)x$. Thus each matrix $P_i=q_i{q}_i^T$ has the required action and $x={\sum }_i{P}_{i}x$.

Consequently $({\sum }_i{P}_i-I_n)x=0$ for every $x$. Taking $x$ successively as the standard basis vectors shows that every column of this difference is zero. Hence ${\sum }_{i=1}^n{P}_i=I_n$. Equality on one selected vector would not suffice; here it holds on all vectors.