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A subspace of a vector space $V$ is a subset $H$ of $V$ that has three properties:

a) The zero vector of $V$ is in $H$.

b) $H$ is closed under vector addition. That is for each $u$ and $v$ in $H$, the sum $u+v$ is in $H$.

c) $H$ is closed under multiplication by scalars. That is, for each $u$ in $H$ and each scalar $c$, the vector $cu$ is in $H$.

It would be great if someone could "dumb" this down. It already seems extremely simply, but i'm having a very difficult time applying these.

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I'm not sure what you're asking...This means $V$ is stable by linear combination. In particular, whenever you take two vectors in $V$ their span (which can be $\{0\}$, a line, or a plane) is always contained in $V$. – 1015 Mar 28 '13 at 0:25
This is just saying that $H$ is a vector space in it's own right with respect to the operations on the big space $V$. – Wintermute Mar 28 '13 at 0:25
you can add vectors, and you can multiply them by scalars. also, we need the zero vector. the significance is that it is both easy to show that many spaces are vector spaces, and, sort of going the other way, many results are true in general for vector spaces. you will see these properties used over and over again in proofs of vector space theorems. :) – Brady Trainor Mar 28 '13 at 0:27

A subspace of a vector space is simply a subset, that is itself a vector space under the same operations.

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oh right, related to my comment above, we can now include that many facts that are true for vector spaces, are now true also in these subspaces. – Brady Trainor Mar 28 '13 at 0:29
and it is usually not hard to show that a subspace, is a subspace. not hard properties to check. and look forward to it. – Brady Trainor Mar 28 '13 at 0:30

If your original vector space was $V=\mathbb R^3$, then the possible subspaces are:

  • The whole space
  • Any plane that passes through $0$
  • Any line through $0$
  • The singleton set, $\{0\}$

One reading for the definition is that $H$ is a subspace of $V$ if it is a sub-set of $V$ and it is also a vector space under the same operations as in $V$.

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and it is not hard to show that they are isomorphic to one of $\mathbb R^n$, $n\in\{0,1,2,3\}$. – Brady Trainor Mar 28 '13 at 0:35
Yes, I just didn't want to get into isomorphisms. I wanted to give an idea of the geometric nature of subspaces. – Thomas Andrews Mar 28 '13 at 0:36
totally, just "adding", pun intended – Brady Trainor Mar 28 '13 at 0:37

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