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Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 1 / 8E / a / i / Solution

The Nilpotent Jordan block

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 .

Solved by gpt-5.6-sol high.

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