Every subspace of a vector space is itself a vector space
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In order to prove that every subspace of a vector space is itself a vector space, we need to show that it satisfies the three axioms of a vector space: closure, associativity, and distributivity.
Closure: If we take any two vectors from a subspace of a vector space, their sum will also be in the subspace. This is because a subspace is defined as a subset of a vector space that is closed under vector addition and scalar multiplication.
Associativity: The addition of vectors in a subspace is associative. This means that for any 3 vectors a, b and c in the subspace, (a+b)+c = a+(b+c). This is true in any vector space, and since the subspace is a subset of the vector space, the same property holds.
Distributivity: The distributive property of scalar multiplication over vector addition also holds in a subspace. This means that for any two vectors a and b in the subspace and any scalar c, we have c(a+b) = ca + cb. Once again, since the subspace is part of the larger vector space, this property still holds.
Therefore, we have shown that every subspace of a vector space is itself a vector space, since it satisfies all three axioms.
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