Proving U+V is a Subspace of R^n

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SUMMARY

The discussion centers on proving that the sum of two subspaces U and V of R^n, denoted as U+V, is also a subspace of R^n. The proof is established by demonstrating that U+V contains the zero vector, is closed under addition, and is closed under scalar multiplication. The user successfully shows that since U and V both contain the zero vector, their sum will also contain it. Additionally, the properties of closure under addition and scalar multiplication are verified, confirming that U+V is indeed a subspace of R^n.

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  • Knowledge of linear algebra concepts, particularly subspace criteria
  • Basic proficiency in mathematical proofs and logic
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Students and educators in linear algebra, particularly those seeking to understand the properties of vector spaces and subspaces. This discussion is beneficial for anyone looking to solidify their grasp of fundamental concepts in linear algebra.

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Homework Statement


If U and V are subsets of R^n, then the set U+V is
defined by

U+V={x:x=u+v,u in U, and v in V} prove that U and V are subspaces of R^n
then the set U+V is a subspace of R^n.
I am just having trouble proving U+V is a subspace.

Homework Equations



To be a sub-space...
1. it needs to contain the zero vector
2. x+y is in W whenever x and y are in W.
3. ax is in W whenever x is in W and a is any scalar.

The Attempt at a Solution



1. U and V both contain the zero vector, so their sum will also contain the zero vector.
2. any u1 plus u2 should be in U because U is a subspace, and any v1+v2 should be in V becuse V is a subspace. So (u1+v1)+(u2+v2)=(u1+u2)+(v1+v2)=u+v.
3. below is just the matrix u+v times a
a(v1+u1)=av1+au1
(v2+u2)=av2+au2
(v3+u3)=av3+au3

Because u is in U, and v is in V then au must be in U, and av must b in V,
and u+v is in U+V. Therefore a(U+V)must be in U+V.

Is this sufficient?
 
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Yes, it's sufficient. Though I'm having a little trouble understanding why you felt you needed u1, u2, u3 and v1, v2, v3 in part 3. a(u+v)=au+av. Isn't that enough?
 
I did that because I am a linear algebra newb lol.

Thanks for the help.
 
EV33 said:
I did that because I am a linear algebra newb lol.

Thanks for the help.

That's a good reason! :)
 

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