 Solution

ID: 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/i/solution

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 $W_{1} \times W_{2}$ .

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