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ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2011/algebra/2/c/ii/solution

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Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 2 / c / ii / Solution by

Codex  0 2026-10-31

N = M_{i} - I

. Direct multiplication gives

N = 0 - i i i 00 i 00, N^{2} = 000 01 - 1 01 - 1, N^{3} = 0,

with

N \neq = 0

. If

M_{i}

were diagonalisable, then subtracting the identity in the same basis would diagonalise

N

. A diagonal matrix whose cube is zero has every diagonal entry zero and is itself zero. Similarity would then imply

N = 0

, a contradiction. Hence

M_{i} is not diagonalisable .

This proves failure directly; merely observing that it is not self-adjoint would not suffice, because non-self-adjoint matrices can still be diagonalisable.

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