I know you need closure under addition and closure under scalar multiplication. But can you also prove something a subspace by claiming it spans the set and is linearly independent?
2026-04-03 03:30:47.1775187047
Proving subspace?
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No. The set $\{1\}$ is linearly independent and spans the one dimensional vector space $\mathbb{R}$ but it isn't a subspace.
In general, what you have described is a basis. A basis is never a subspace since (at the very least) a basis can't contain the $0$ vector and a subspace must.