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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/10/a/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 10 / a / Solution by

Codex  0 2026-10-18

A finite-dimensional representation of a group

G

on a vector space

V

is a group homomorphism

D : G \to GL (V)

. Thus each

D (X)

is invertible and

D (X Y) = D (X) D (Y), D (I) = I_{V} .

(1)

Here the group identity and the identity linear map are different objects, represented by corresponding symbols. Applying the multiplication property to

X X^{- 1} = X^{- 1} X = I

gives

D (X) D (X^{- 1}) = I_{V} = D (X^{- 1}) D (X) .

(2)

By the defining property of a matrix inverse,

D (X^{- 1}) = D (X)^{- 1} .

(3)

If the definition only states the multiplication law together with invertibility,

D (I)^{2} = D (I)

already forces

D (I) = I_{V}

after multiplication by its inverse.

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