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ID: past-exam-of-the-numerical-analysis-course-of-the-university-of-oxford/2015/3/d/ii/solution

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Past exam of the numerical analysis course of the University of Oxford / 2015 / 3 / d / ii / Solution by

Codex  0 2026-10-31

Put

C = Q_{1} A

and write its second column as

v

. For

n \geq 3

, take

Q_{2} = J (2, n) J (2, n - 1) \dots J (2, 3),

(1)

where each angle is computed from the current second column, in the right-to-left application order. These Givens rotations eliminate its entries in rows

3, \dots, n

by accumulating their Euclidean norm in row two.

Each rotation acts only on rows with indices at least two. The first column of

C

∥ a ∥_{2} e_{1}

, whose entries in all those rows are zero. Mixing two zeros leaves them zero, and row one is untouched. Thus the first column is unchanged. The resulting matrix has the form

Q_{2} Q_{1} A = ∥ a ∥_{2} 00 ⋮ 0 v_{1} v_{2}^{2} + \dots + v_{n}^{2} 0 ⋮ 0 (n \geq 3) .

(2)

This verifies both required properties, including the case of zero second-column tail. When

n = 2

there are no rows to eliminate: take

Q_{2} = I_{2}

, the empty product. No sign change of

v_{2}

is required in that case. Each Givens row-pair update preserves previously eliminated columns, which is the central invariant of QR decomposition by Givens rotations.

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