We know that a subspace (of a vector space $V$) is a vector space that follows the same addition and multiplication rules as $V$, but is a vector space a subspace of itself?

Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10, which are the ten axioms of vector spaces. So for example in $\Bbb R^2$, we have that $\vec{x} + \vec{y} = \vec{y} + \vec {x}$, etc..

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    $\begingroup$ How do you define a subspace of a vector space? $\endgroup$ – Brian Apr 9 '19 at 2:02
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    $\begingroup$ Is a set a subset of itself?? What’s V1-V10? $\endgroup$ – J. W. Tanner Apr 9 '19 at 2:16
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    $\begingroup$ The term "proper" subspace is often used to denote a subspace space that is not the entire vector space. $\endgroup$ – Theo Bendit Apr 9 '19 at 2:19
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    $\begingroup$ As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too. $\endgroup$ – Xander Henderson Apr 9 '19 at 3:39
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    $\begingroup$ @CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace. $\endgroup$ – Theo Bendit Apr 9 '19 at 8:57

I'm guessing that V1 - V10 are the axioms for proving vector spaces.

To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.

There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.

Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.

I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.

You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.

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    $\begingroup$ Ohh okay I think I understand. So when we prove that something is a subspace of say $\Bbb R^2$, we don't have to prove that $c(\vec{x} + \vec{y}) = c\vec{x} + c\vec{y}$ because we already know that's true inside of $\Bbb R^2$ and we know this subspace is a vector space INSIDE of $\Bbb R^2$ following all its rules. $\endgroup$ – ming Apr 9 '19 at 16:37
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    $\begingroup$ But then for questions like "A sequence is a infinite list of real numbers. For example 1, 2, 3, 4, 5 is a sequence, and so is 1, -1, 2, -2, 4, -4. We define addition and scalar multiplication so that...." We actually don't have any known information/facts about this vector space so then we need to prove all ten axioms? $\endgroup$ – ming Apr 9 '19 at 16:39
  • $\begingroup$ @ming Exactly. Couldn't put it better myself. $\endgroup$ – Theo Bendit Apr 9 '19 at 22:34

Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.


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