V^{'} = Hom_{F} (V, F),

(ϕ + ψ) (v) = ϕ (v) + ψ (v), (a ϕ) (v) = a ϕ (v) .

Given the basis

e_{1}, \dots, e_{n}

, its dual basis consists of the coordinate linear functionals

f_{i} (j = 1 \sum n a_{j} e_{j}) = a_{i}, f_{i} (e_{j}) = δ_{ij} .

In particular,

dim V^{'} = dim V = n

ii

Solution

T^{'} : W^{'} ⟶ V^{'}, T^{'} (ϕ) = ϕ \circ T .

The composition is a linear functional on

V

, and pointwise operations give

T^{'} (a ϕ + b ψ) = a T^{'} ϕ + b T^{'} ψ

. Thus this really is a linear transformation; its direction reverses that of

T

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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