 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2015/3/a/i/solution

The dual basis is the collection of linear functionals characterized by

e_{i}^{'} (e_{j}) = δ_{ij} .

Explicitly,

e_{i}^{'} (\sum_{j} α_{j} e_{j}) = α_{i}

ℓ \in V^{'}

satisfies

ℓ = \sum_{i} ℓ (e_{i}) e_{i}^{'}

, proving that the displayed functionals span the dual space. Evaluating a zero linear combination at each

e_{j}

makes all its coefficients zero, so they are also linearly independent. Thus they form a basis of

V^{'}

 New to topics? Read the docs here!