 Solution

ID: past-exam-of-the-linear-algebra-course-of-the-university-of-oxford/2023/1/b/ii/solution

U \subseteq V

U^{0} = {φ \in V^{'} : φ (u) = 0 for every u \in U} .

Let

dim U = k

and

dim V = n

. Extend a basis

u_{1}, \dots, u_{k}

U

u_{1}, \dots, u_{n}

V

, and let

ε^{1}, \dots, ε^{n}

be the dual basis. Every linear functional has the unique expression

φ = \sum_{j = 1}^{n} φ (u_{j}) ε^{j}

. It belongs to

U^{0}

precisely when its first

k

coefficients vanish. Consequently

U^{0} = span {ε^{k + 1}, \dots, ε^{n}}, dim U + dim U^{0} = dim V .

The listed dual basis elements are linearly independent, so counting them proves the formula, including

U = 0

and

U = V

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