 Solution

ID: past-exam-of-the-linear-algebra-course-of-the-university-of-oxford/2017/2/a/iii/solution

For

ϕ \in V^{'}

and

v \in V

(I^{'} ϕ) (v) = ϕ (I v) = ϕ (v),

I^{'}

S : U \to V

and

T : V \to W

, then, for

ψ \in W^{'}

and

u \in U

((TS)^{'} ψ) (u) = ψ (T (S u)) = (S^{'} (T^{'} ψ)) (u) .

Equality on all vectors proves

(TS)^{'} = S^{'} T^{'} .

Apply this to

R R^{- 1} = R^{- 1} R = I

. It gives both

R^{'} (R^{- 1})^{'} = I^{'}

and

(R^{- 1})^{'} R^{'} = I^{'}

. Consequently the dual map is invertible and

(R^{'})^{- 1} = (R^{- 1})^{'} .

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