1. vector space

Hello

How can I prove that the set

W={(t+s, t-s, 2t+3s, t, s);t and s belong to R} is a vector space under the standard operations?

do i have to prove all the axioms ?

thank you

2. Originally Posted by qwerty321
do i have to prove all the axioms ?
No. This is a subset of $\mathbb{R}^5$ which is a vector space. So immediately most of the definitions for a vector space are immediately verified. All you need to do is show if that if $\bold{w}_1,\bold{w}_2\in W$ then $\bold{w}_1+\bold{w}_2\in W$ and that if $k\in \mathbb{R}$ then $k\bold{w}\in W$ for any $\bold{w}\in W$.

Ok but how did you know that it is a subset of R^5?
I also have to prove that 0 belongs to W no?which is evident

Anyway can you help me prove the other ones?
for example for the first one how do i begin with it
thank you

4. Originally Posted by ThePerfectHacker
No. This is a subset of $\mathbb{R}^5$ which is a vector space. So immediately most of the definitions for a vector space are immediately verified.
I am curious: The empty set {} is also a subset of $\mathbb{R}^5$, but it is not a vector space. If S is a subset of vector space V, does that really imply that S is itself a vector space?

Ah, nevermind, I see that your statement is true for any non-empty subset.

5. this is a theorem scorpion:
if V is a subset of W(which is a vector space) and you can prove that V is a subspace of W, then V is a subset itself.Now can you help me prove the properties?

6. Originally Posted by qwerty321
this is a theorem scorpion:
if V is a subset of W(which is a vector space) and you can prove that V is a subspace of W, then V is a subset itself.Now can you help me prove the properties?
Indeed.

Ok but how did you know that it is a subset of R^5?
Because each element of W is a 5-tuple of reals $(a_1, a_2, a_3, a_4, a_5), a_i \in \mathbb{R}$.

I also have to prove that 0 belongs to W no?which is evident
No, because if you can prove that W is closed under scalar multiplication, then the fact that is has a 0 element is implied, by $\forall \vec{x} \in W, 0\vec{x} = \vec{0} \in W$.(Corrected 30/3/09)

So as long as you can show that it is not empty, and is closed under scalar multiplication and vector addition, that will suffice to prove that it is a subspace of R^5.

Anyway can you help me prove the other ones?
for example for the first one how do i begin with it
thank you
For closure under vector addition, you must show that for any $\bold{x}, \bold{y} \in W$, that $\bold{x} + \bold{y} \in W$.

E.g. Let $t_1, t_2, s_1, s_2 \in \mathbb{R}$ and $\bold{x}, \bold{y}$ be elements of W such that,

$\bold{x} = (t_1 + s_1, t_1 - s_1, 2t_1 + 3s_1, t_1, s_1)$ and
$\bold{y} = (t_2 + s_2, t_2 - s_2, 2t_2 + 3s_2, t_2, s_2)$.

Then,

$\bold{x} + \bold{y} = ((t_1 + t_2) + (s_1 + s_2), (t_1 + t_2) - (s_1 + s_2), 2(t_1 + t_2) + 3(s_1 + s_2), t_1 + t_2, s_1 + s_2)$ which is also in W. (I grouped the terms in the same step, but you can show it separately first.)

Hence, W is closed under vector addition.

Scalar multiplication can be shown similarly.

7. ok that's what i did and it worked
now suppose i have the same set except that instead of 2t+3s, i have 2t+3s+1..will it be a vector space?
no right?
because w1+w2 is not in W(because of the 1 we will have 2)

8. you said:
No, because any non-empty subset of a vector space is itself a vector space, and so most of the axioms are inherited and subsequently verified, as ThePerfectHacker pointed out.

but when we want to prove that w is a vector subspace of v we have to prove 3 properties:

the scalar multiplication, the addition and the fact that 0 belongs to W, that is, that w is a non empty set

9. I'm a bit confused
For example, when we want to prove that thet set W={3x+2y=0, for (x,y)belong to R^2}, we prove that (x1,y1)+(x2,y2) belong to W.

here, u proved that w1+w2 belong to W.Don't we have to pove that for al s and t, (t1,s1)+(t2,s2) belong to W?

or because it is a set we can't say that?