Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 4 / 1E / ii

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- Solution ii

Solution

For

X = (a c b d),

direct multiplication gives

ϕ_{A} (X) = A X - X A = (c 0 d - a - c) .

With respect to the ordered standard basis

(E_{11}, E_{12}, E_{21}, E_{22})

, its matrix is therefore

[ϕ_{A}] = 0 - 1 00 0000 100 - 1 0100 .

This nilpotent operator satisfies

ϕ_{A}^{3} = 0

, while

ϕ_{A}^{2} \neq = 0

. Moreover,

rank ϕ_{A} = 2, rank ϕ_{A}^{2} = 1.

These ranks determine one nilpotent block of size three and one of size one. Hence its Jordan normal form is

J_{3} (0) \oplus J_{1} (0) .

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