If a subspace of a finite linear space contains all the elements of that linear space, can the set be still considered a subspace.
What would be the 0 vector of such a subspace and how would one go about showing that. It has to exist right? Because the subspace is all the elements of the linear space and by definition of linear space the 0 element has to exist?
Additive and scalar multiplication would still be closed under by the same reasoning in (2). Would this be correct reasoning?
2026-04-02 18:38:42.1775155122
Short subspace and linear spaces questions
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I stopped for a moment to wonder what a "short subspace" would be!
We can think of the entire space as a "subspace" of itself in the same way we think of a set being a "subset" of itself. Yes, all theorems about subspace allow the entire space as a subspace.
The 0 vector of every subspace of a vector space is the original vector space. One of the requirements that U be a subspace of V is that the 0 vector in V is also in U.
Yes, the definition of subspace is that it be closed under addition of vectors and scalar multiplication.