Let and be subspaces of a vector space .
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To show something is a subspace we have to check the three properties: containing the zero vector, closure under addition, and closure under scalar multiplication.
as and are subspaces so contain .
Let . is a subspace, so closed under addition, so . For the same reason . Therefore .
Let be a scalar and . is a subspace, so closed under scalar multiplication, so . For the same reason . Therefore .
is in and as they are subspaces, so is in .
Any two elements of have the form and , where and .
by associativity and commutativity. But as is a subspace and as is a subspace, so this is in which is therefore closed under addition.
Let be a scalar.
as is a subspace so closed under scalar multiplication, for the same reason, so their sum is in which is therefore closed under scalar multiplication. ∎