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
No. This is a subset of $\displaystyle \mathbb{R}^5$ 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 $\displaystyle \bold{w}_1,\bold{w}_2\in W$ then $\displaystyle \bold{w}_1+\bold{w}_2\in W$ and that if $\displaystyle k\in \mathbb{R}$ then $\displaystyle k\bold{w}\in W$ for any $\displaystyle \bold{w}\in W$.
I am curious: The empty set {} is also a subset of $\displaystyle \mathbb{R}^5$, 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.
Indeed.
Because each element of W is a 5-tuple of reals $\displaystyle (a_1, a_2, a_3, a_4, a_5), a_i \in \mathbb{R}$.Ok but how did you know that it is a subset of R^5?
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 $\displaystyle \forall \vec{x} \in W, 0\vec{x} = \vec{0} \in W$.(Corrected 30/3/09)I also have to prove that 0 belongs to W no?which is evident
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.
For closure under vector addition, you must show that for any $\displaystyle \bold{x}, \bold{y} \in W$, that $\displaystyle \bold{x} + \bold{y} \in W$.Anyway can you help me prove the other ones?
for example for the first one how do i begin with it
thank you
E.g. Let $\displaystyle t_1, t_2, s_1, s_2 \in \mathbb{R}$ and $\displaystyle \bold{x}, \bold{y}$ be elements of W such that,
$\displaystyle \bold{x} = (t_1 + s_1, t_1 - s_1, 2t_1 + 3s_1, t_1, s_1)$ and
$\displaystyle \bold{y} = (t_2 + s_2, t_2 - s_2, 2t_2 + 3s_2, t_2, s_2)$.
Then,
$\displaystyle \bold{x} + \bold{y} = ((t_1 + t_2) + (s_1 + s_2), (t_1 + t_2) - (s_1 + s_2), 2(t_1 + t_2) + 3(s_1 + s_2), t_1 + t_2, s_1 + s_2)$ 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.
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
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?
thank yu for your cooperation