 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2011/1/b/i/solution

Let

ε_{i}

denote the coordinate linear functional dual to the

i

x = c_{1} v_{1} + c_{2} v_{2} + c_{3} v_{3} + c_{4} v_{4} = (c_{1} + c_{2} + c_{3} + c_{4}, c_{2} + c_{3} + c_{4}, c_{3} + c_{4}, c_{4})^{T} .

Successive differences recover the coefficients. Hence the dual basis is

v_{1}^{'} = ε_{1} - ε_{2}, v_{2}^{'} = ε_{2} - ε_{3}, v_{3}^{'} = ε_{3} - ε_{4}, v_{4}^{'} = ε_{4} .

These formulas directly give

v_{i}^{'} (v_{j}) = δ_{ij}

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