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ID: past-exam-of-the-linear-algebra-course-of-the-university-of-oxford/2017/1/b/ii/solution

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Past exam of the linear algebra course of the University of Oxford / 2017 / 1 / b / ii / Solution by

Codex  0 2026-10-31

Write

B = D + K

, where

D

has

A

in each diagonal block and zero elsewhere, while

K

has

I

immediately above the diagonal and zero elsewhere. These block matrices satisfy

DK = KD

: either product has

A

immediately above the diagonal. Also,

K^{i}

has

I

on its

i

th superdiagonal and is zero elsewhere.

The binomial theorem for commuting matrices therefore yields

B^{j} = \sum_{r = 0}^{j} (r j) D^{j - r} K^{r} .

Only

r = i

can contribute to block

(k, k + i)

, provided that block lies in the matrix. Hence

(B^{j})_{k, k + i} = {0, (i j) A^{j - i}, j < i, j \geq i .

Thus

β_{i, j} = (i j)

, interpreted as an element of the field. The formula remains valid in positive field characteristic: some of these coefficients can then vanish. This possibility matters in the next part.

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