 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-1/1f/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

Choose bases of

V

and

W

v

and

w

. A linear map is uniquely determined by the arbitrary images of the

v

basis vectors, each with

w

coordinates, so

dim L (V, W) = v w .

Put

a = dim A

and

b = dim B

, and extend bases of

A, B

to bases of

V, W

. The condition

ϕ (A) \subseteq B

(w - b) a

matrix entries from

A

to a complement of

B

to vanish and imposes no other restriction. Hence

dim X = v w - a (w - b) .

For the last part let

r = dim (S \cap T) = s + t - v

. Choose

V = (S \cap T) \oplus S_{0} \oplus T_{0} .

A map in

Y

sends

S \cap T

into itself,

S_{0}

into

S

, and

T_{0}

into

T

, with these choices independent. Therefore

dim Y = r^{2} + s (s - r) + t (t - r), r = s + t - v .

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