Proving U+V is a Subspace of R^n

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In summary, the given conversation discusses the definition of the set U+V as a sum of two subsets U and V in R^n. The task at hand is to prove that U and V are subspaces of R^n and therefore, U+V is also a subspace. The proof is based on the three conditions for a subspace, which include containing the zero vector, being closed under addition, and being closed under scalar multiplication. By showing that U and V both satisfy these conditions, it is proven that the set U+V is indeed a subspace of R^n.
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EV33
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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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  • #2
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?
 
  • #3
I did that because I am a linear algebra newb lol.

Thanks for the help.
 
  • #4
EV33 said:
I did that because I am a linear algebra newb lol.

Thanks for the help.

That's a good reason! :)
 

FAQ: Proving U+V is a Subspace of R^n

1. What is a subspace?

A subspace is a subset of a vector space that satisfies all the properties of a vector space. This means that it is closed under addition and scalar multiplication, and contains the zero vector.

2. How do you prove that U+V is a subspace of R^n?

To prove that U+V is a subspace of R^n, you need to show that it satisfies the three properties of a vector space: closure under addition, closure under scalar multiplication, and containing the zero vector. This can be done by showing that for any two vectors in U+V, their sum and scalar multiplication are also in U+V, and that the zero vector is also in U+V.

3. What is the difference between U+V and U∩V?

U+V refers to the set of all vectors that can be written as the sum of a vector in U and a vector in V. U∩V, on the other hand, refers to the set of all vectors that are common to both U and V. In other words, U+V contains all possible combinations of vectors from U and V, while U∩V only contains vectors that are in both U and V.

4. Can U+V still be a subspace if U and V are not subspaces of R^n?

No, U+V can only be a subspace if both U and V are subspaces of R^n. This is because in order for U+V to satisfy the properties of a vector space, it must contain the zero vector and be closed under addition and scalar multiplication, which can only be ensured if U and V themselves are subspaces.

5. How does proving U+V is a subspace relate to linear independence?

If U+V is a subspace of R^n, then it means that all combinations of vectors from U and V are also in R^n. This implies that the vectors in U and V are linearly independent, as they are able to span the entire vector space of R^n. However, proving U+V is a subspace does not necessarily guarantee that the vectors in U and V are linearly independent, as there may be other combinations of vectors that are not in U+V but can still span R^n.

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