 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/40a/b/iv/solution

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

Let

q_{1}

and

q_{n}

be the first and last columns of

Q_{k}

. Since

R_{k}

is upper triangular,

A^{k} e_{1} = (R_{k})_{11} q_{1} .

Thus

q_{1}

is the normalized

k

th power method iterate starting from

e_{1}

Taking the transpose of

A^{k} = Q_{k} R_{k}

and using symmetry gives

A^{k} = R_{k}^{T} Q_{k}^{T}

. Therefore

A^{k} q_{n} = (R_{k})_{nn} e_{n}, q_{n} = (R_{k})_{nn} A^{- k} e_{n} .

So the last column is the normalized

k

th inverse iteration iterate with shift zero and starting vector

e_{n}

 New to topics? Read the docs here!