Solution
= Solution
The <intersection of vector subspaces> $W_1\cap W_2$ contains zero, so $(0,0)\in U$. If $(u,-u),(v,-v)\in U$ and $a,b$ are <scalars>, then
$$
a(u,-u)+b(v,-v)=(au+bv,-(au+bv))\in U,
$$
because $au+bv\in W_1\cap W_2$ 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 \b[$U$ is a <vector subspace> of $W_1\times W_2$].