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ID: 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

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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 by

Codex  0 2026-10-25

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.

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