Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Digital systems, linear algebra, introduction to formal proof / 6 / b

W_{1} \cap W_{2}

contains zero, so

(0, 0) \in U

. If

(u, - u), (v, - v) \in U

and

a, b

are scalars, then

a (u, - u) + b (v, - v) = (a u + b v, - (a u + b v)) \in U,

because

a u + b v \in W_{1} \cap W_{2}

by the vector subspace conditions. These pairs belong to

W_{1} \times W_{2}

, since both

u

and

- u

belong to the required factors. Thus $U$ is a vector subspace of $W_{1} \times W_{2}$ .

ii

f : W_{1} \cap W_{2} \to U

f (u) = (u, - u)

. It is well-defined by the definition of

U

, and

f (a u + b v) = (a u + b v, - a u - b v) = a f (u) + b f (v) .

Thus

f

f (u) = f (v)

, comparing first components gives

u = v

, proving injectivity. Every element of

U

has the form

(u, - u) = f (u)

, proving surjectivity. Hence $f$ is a linear isomorphism, with inverse

(u, - u) \mapsto u

, and

dim U = dim (W_{1} \cap W_{2})

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