 Solution

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/f/ii/solution

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} \neq = 0

. Hence $rank (P_{1}) = 1$ .

dim N (P_{1}) = n - 1

. Since

C (I_{n} - P_{1}) = N (P_{1})

by (e)(i), $rank (I_{n} - P_{1}) = n - 1$ . This includes the case

n = 1

, where the complementary projector is zero.

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