 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/8e/a/i/solution

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

J_{d}

shifts each standard basis vector one place toward the first coordinate. Therefore

(J_{d}^{n})_{ij} = {1, 0, j - i = n, otherwise .

Thus

J_{d}^{0} = I

, for

1 \leq n < d

the matrix has ones precisely on its

n

th superdiagonal, and

J_{d}^{n} = 0 for n \geq d .

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