Proof of W as Subspace of Rn: Help & Explanation

[roam]

I saw this problem in a book, it asks if there are two subspaces of Rⁿ, say U & V and the following condition is true:
W={$$w \in R^n$$ : w=u+v for some $$u \in U$$ and $$v \in V$$}

Make a proof/show that W is a subspace of Rⁿ.

I think maybe we need to try to somehow prove that the set W is a subspace of Rⁿ by showing that it's non-empty and closed under addition/scalar multipication. Does anyone know how to show this? I'm not sure how we can do it, any explanation or links would be appreciated.

Thanks.

 

[HallsofIvy]

I saw this problem in a book, it asks if there are two subspaces of Rⁿ, say U & V and the following condition is true:
W={$$w \in R^n$$ : w=u+v for some $$u \in U$$ and $$v \in V$$}

Make a proof/show that W is a subspace of Rⁿ.

I think maybe we need to try to somehow prove that the set W is a subspace of Rⁿ by showing that it's non-empty and closed under addition/scalar multipication. Does anyone know how to show this? I'm not sure how we can do it, any explanation or links would be appreciated.

Thanks.

I don't see why just straightforward verification of the conditions you state wouldn't work.

W is not empty because U and V both contain the 0 vector. So what vector is in W?

Suppose w and w' are in W, we can write w= u+ v and w'= u'+ v' where u and u' are in U, v and v' are in V. what can you say about w+ w'? What about $\alpha w$ for $\alpha$ a number?

 

[roam]

Suppose w and w' are in W, we can write w= u+ v and w'= u'+ v' where u and u' are in U, v and v' are in V. what can you say about w+ w'? What about $\alpha w$ for $\alpha$ a number?

w+ w' = (u+v)+(u'+v'), therefore it's closed under addition. And for some $$\alpha \in R$$, $$\alpha w= \alpha u + \alpha v$$ & $$\alpha (w + w')= \alpha (u + v) + \alpha (u'+v')$$ so closed under scalar multipication. Is this right?

W is not empty because U and V both contain the 0 vector. So what vector is in W?

Could you please explain because I'm not sure. How do we write a proof to show that U and V contain the 0 vector?

 

[HallsofIvy]

Every subspace contains the 0 vector! Every subspace is non empty so either it contains the 0 vector or it contains some non-zero vector v. Since a subspace is closed under scalar multiplication, it contains (-1)v= -v. Since a subspace is closed under addition, it contains v+ (-v)= 0.

If you don't want to use 0 specifically, you can argue that U is a subspace so contains some vector u, V is a subspace so contains some vector v, therefore U+ V contains u+v and so is non-empty.

 

[de_brook]

w+ w' = (u+v)+(u'+v'), therefore it's closed under addition. And for some $$\alpha \in R$$, $$\alpha w= \alpha u + \alpha v$$ & $$\alpha (w + w')= \alpha (u + v) + \alpha (u'+v')$$ so closed under scalar multipication. Is this right?



Could you please explain because I'm not sure. How do we write a proof to show that U and V contain the 0 vector?

It is right. This is because, for any $$\alpha \in R$$, $$\alpha w= \alpha u + \alpha v\in {R^n}$$. why? recall that u and v are vectors respectively in subspaces U and V. and also for any w in W, w = u + v implies u is in U and v is in V.But
$$\alpha u$$ is in U and $$\alpha v$$ is V. Now think of the sum, their sum will surely be in W.

U and V contains the zero vector because of the fact that a linear space is an additive abelian group. You recall that a group must contain the identity element. Thus for a linear space, the identity element is the zero vector under the binary operation '+'. you recall that a subspace of a linear space is linear space in its own right. Thus, the subspaces U and V must contain the zero vector.