 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-1/8g/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Suppose that

Q

were not surjective. Its image would be a proper subspace of

F^{n}

, so there would be a nonzero functional

ℓ (y_{1}, \dots, y_{n}) = \sum_{j = 1}^{n} c_{j} y_{j}

vanishing on

im Q

. Then

0 = ℓ (Q x) = \sum_{j = 1}^{n} c_{j} q_{j} (x)

for every

x \in V

, contradicting the linear independence of the

q_{j}

Q : V \to F^{n} is surjective .

Let

K = ⋂_{j} ker q_{j} = ker Q

and suppose

K \subseteq ker f

g : F^{n} \to F, g (Q x) = f (x) .

This is well-defined: if

Q x = Q y

, then

x - y \in K

, so

f (x) = f (y)

. It is linear, and surjectivity of

Q

means it is defined on all of

F^{n}

. Thus there are scalars

a_{1}, \dots, a_{n}

such that

g (y) = \sum_{j} a_{j} y_{j}

. Therefore

f (x) = g (Q x) = \sum_{j = 1}^{n} a_{j} q_{j} (x),

and hence

f \in span {q_{1}, \dots, q_{n}} .

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