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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 8 / b / Solution 2026-10-20

Since

M^{T} = I - a b^{T} = I - N

, nilpotence already gives

(I + N) (I - N) = I

. Explicitly, in index notation,

(L M^{T})_{ij} = \sum_{k} (δ_{ik} + a_{i} b_{k}) (δ_{jk} - b_{j} a_{k}) = δ_{ij} - b_{j} a_{i} + a_{i} b_{j} - a_{i} b_{j} \sum_{k} a_{k} b_{k} = δ_{ij} .

Thus

L M^{T} = I

, or equivalently

M^{T} = L^{- 1}

. Here

δ_{ij}

is the Kronecker delta.

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