 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/41e/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

A_{k} = Q_{k} R_{k}

gives

A_{k + 1} = R_{k} Q_{k} = Q_{k}^{T} A_{k} Q_{k} .

A_{k + 1} = \overline{Q}_{k}^{T} A \overline{Q}_{k}, \overline{Q}_{k} = Q_{0} Q_{1} \dots Q_{k} .

We next prove the corresponding factorization of a matrix power. The case

k = 0

is just

A = Q_{0} R_{0}

. If

A^{k + 1} = \overline{Q}_{k} \overline{R}_{k}, \overline{R}_{k} = R_{k} R_{k - 1} \dots R_{0},

then the similarity formula above implies

A \overline{Q}_{k} = \overline{Q}_{k} A_{k + 1} .

Consequently,

A^{k + 2} = A \overline{Q}_{k} \overline{R}_{k} = \overline{Q}_{k} A_{k + 1} \overline{R}_{k} = \overline{Q}_{k} Q_{k + 1} R_{k + 1} \overline{R}_{k} = \overline{Q}_{k + 1} \overline{R}_{k + 1} .

The product

\overline{Q}_{k}

is orthogonal, and the product

\overline{R}_{k}

A^{k + 1} = \overline{Q}_{k} \overline{R}_{k}

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