 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/a/solution

For nonzero

x

, let

ϕ

be its angle to the unit vector

q_{1}

. The scalar projection on the oriented

q_{1}

axis is

∥ x ∥ cos ϕ

. Since

∥ q_{1} ∥ = 1

, the dot product identity gives

q_{1}^{T} x = ∥ x ∥ cos ϕ

. Multiplying that signed scalar by

q_{1}

places the foot at $p = (q_{1}^{T} x) q_{1}$ .

The residual is perpendicular to the axis, because

q_{1}^{T} (x - p) = q_{1}^{T} x - (q_{1}^{T} x) (q_{1}^{T} q_{1}) = 0

. An obtuse angle gives a negative scalar and a foot on the negative side of the line;

x = 0

has projection zero without needing an angle.

The first panel is a constructed geometric illustration. The second illustrates the directional stretch used later in Question 6.

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