Is it a property of vector spaces?

Question

Given vector space $V$, if $\rm\ v\in V\ $ then $\rm\ w\in V\ \iff\ w+c\:v\in V\:.\ $

Could anyone tell me why? Is it an axiom of vector space, that it's closed under addition and multiplication? According to Wikipedia, a subspace has to be closed under addition and multiplication, but it doesn't say a word about whether the same applies to vector spaces.

What is $c$ here? And for the case that $w\notin V$, how is $w+cv$ even defined? — Hagen von Eitzen, Mar 05 '16 at 16:46
@HagenvonEitzen http://math.stackexchange.com/a/20210/290354 — user5539357, Mar 05 '16 at 16:52
Here is a very narrow-minded answer: if you know something is true for subspaces of any vector space, then it must also be true for any vector space, because $V$ is a subspace of $V$. — Eric Stucky, Mar 06 '16 at 18:46

Workaholic · Answer 1 · 2016-03-05T16:43:58.103

2

Subspaces are vector spaces. So you can definitely state that a vector space is closed under addition and scalar multiplication. (Actually this last fact follows directly from vector spaces' axioms.)

edited Mar 05 '16 at 16:43

answered Mar 05 '16 at 16:35

Workaholic

6,763

score 0 · Answer 2 · answered Mar 05 '16 at 16:36

0

did you check the definition part of the wiki page?

vector spaces are closed under addition and scalar multiplication (i'm assuming your c is a scalar)

answered Mar 05 '16 at 16:36

homedoggieo

164

score 0 · Answer 3 · answered Mar 05 '16 at 16:40

Of course it is, a subspace is in particular a vector space so if it characterizes a subspace that is a vector space it is also a property for vector spaces. Also, under addition it is a group and under multiplication it is a semi group so it is closed under linear combinations of elements.

score 0 · Answer 4 · answered Mar 06 '16 at 18:39

cv is a member of V by closure under scaler multiplication. Then, w + cv is a member of V by closure under vector addition (since cv has been shown to be a vector in V).

Going the other way, we can still deduce that cv is a member of V, and similarly so is -cv. So, by closure under vector addition, (w + cv) + (-cv) = w, w is also a vector in V.

Is it a property of vector spaces?

4 Answers4