Linear Algebra - Vector Subspaces

In summary, Homework statement gives an example of a nonempty subset U of R2 that is closed under scalar multiplication, but not vector addition. However, this subset is not a subspace of R2 because 0 is not in it.
  • #1
steelphantom
159
0

Homework Statement


Give an example of a nonempty subset U of R2 such that U is closed under scalar multiplication, but U is not a subspace of R2.

Homework Equations





The Attempt at a Solution


I think I have it, but I just want to make sure it's right:

Let U = {(x, x + 2)} | x is in R}. Let u be in U. Then au = a(x, x + 2) = (ax, a(x + 2)), which is still in U (i.e. still on the same line). But this is not a subspace of R2 because 0 is not in this subset.

Thanks!
 
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  • #2
steelphantom said:

The Attempt at a Solution


I think I have it, but I just want to make sure it's right:

Let U = {(x, x + 2)} | x is in R}. Let u be in U. Then au = a(x, x + 2) = (ax, a(x + 2)), which is still in U (i.e. still on the same line). But this is not a subspace of R2 because 0 is not in this subset.

You get [itex]<0,0>[/itex] if you multiply by [itex]a=0[/itex].
(Hint: What are the properties of subspaces?)
 
  • #3
NateTG said:
You get [itex]<0,0>[/itex] if you multiply by [itex]a=0[/itex].
(Hint: What are the properties of subspaces?)

1) Additive Identity: 0 is in U.

2) Closed Under Addition: u, v in U implies u + v is in U.

3) Closed Under Scalar Multiplication: a is in R (or C) and u is in U implies au is in U.

So I need to come up with a subset that fails for one (or both) of the first two properties, but not for property 3.

How about checking for property 2? If u = (x, x + 2) and v = (y, y + 2), where u, v are in U, then u + v = (x + y, (x + y) + 4), which is not in U, so U is not a subspace of R2.
 
  • #4
steelphantom said:
How about checking for property 2? If u = (x, x + 2) and v = (y, y + 2), where u, v are in U, then u + v = (x + y, (x + y) + 4), which is not in U, so U is not a subspace of R2.

...better, but:
[tex]2<3,(3+2)>=<6,10> \notin U[/tex]
so you don't have closure under scalar multiplication.
 
  • #5
NateTG said:
...better, but:
[tex]2<3,(3+2)>=<6,10> \notin U[/tex]
so you don't have closure under scalar multiplication.

Wow, I guess I don't. I really don't know where to go from here. Any hints on a a type of subset that would work in this situation?
 
  • #6
I was searching the forums and found pretty much the exact same question. Here's the thread if anyone is interested: https://www.physicsforums.com/showthread.php?t=41884"

A set that works is the following: U = {(x, y) | xy = 0}. This set is closed under scalar multiplication, but not vector addition.
 
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  • #7
steelphantom said:

Homework Statement


Give an example of a nonempty subset U of R2 such that U is closed under scalar multiplication, but U is not a subspace of R2.

Homework Equations





The Attempt at a Solution


I think I have it, but I just want to make sure it's right:

Let U = {(x, x + 2)} | x is in R}. Let u be in U. Then au = a(x, x + 2) = (ax, a(x + 2)), which is still in U (i.e. still on the same line). But this is not a subspace of R2 because 0 is not in this subset.

Thanks!


You are absolutely wrong.U={(x,x+2)| x is in R} is not closure under multiplication.
Ex:put x=0,a=2 (0,2) is in U but (0,4) is not in U.
 
  • #8
studguy said:
You are absolutely wrong.U={(x,x+2)| x is in R} is not closure under multiplication.
Ex:put x=0,a=2 (0,2) is in U but (0,4) is not in U.
Yes. he was told that almost two years ago!
 

Related to Linear Algebra - Vector Subspaces

1. What is a vector subspace?

A vector subspace is a subset of a vector space that is closed under addition and scalar multiplication. This means that any two vectors in the subspace can be added together and the result will still be in the subspace, and any vector in the subspace can be multiplied by a scalar and the result will still be in the subspace.

2. How do you determine if a set of vectors is a subspace?

To determine if a set of vectors is a subspace, you can use the three subspace tests: closure under addition, closure under scalar multiplication, and containing the zero vector. If all three tests are satisfied, then the set is a subspace.

3. What is the difference between a vector space and a subspace?

A vector space is a set of vectors that is closed under addition and scalar multiplication, while a subspace is a subset of a vector space that is also closed under these operations. In other words, a subspace is a smaller vector space contained within a larger one.

4. Can a subspace have an infinite number of vectors?

Yes, a subspace can have an infinite number of vectors. As long as the subspace satisfies the three subspace tests, it can contain any number of vectors, including an infinite number.

5. What is the relationship between linear independence and vector subspaces?

A set of linearly independent vectors can form a basis for a subspace. This means that any vector in the subspace can be written as a linear combination of the basis vectors. Additionally, if a set of vectors is linearly dependent, then they cannot form a basis for a subspace.

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