Subspaces of Functions- Linear Algebra

In summary, the conversation discusses whether certain sets, namely U, V, S, and T, are subspaces of F[R]. It is determined that U and V are not subspaces due to a lack of closure under addition. The process of proving whether something is a subspace or not is discussed, including the need to verify all 10 axioms of vector spaces, with emphasis on the importance of proving closure.
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Homework Statement


Which of the following are subspaces of F[R] = {f |f:R-->R}?
a) U = {f e F[R]|f(-1)f(1)=0
b) V = " |f(1)+f(2)=0
c) S = " |f(x)=f(-x)
d) T = " |f(1)<= 0


Homework Equations





The Attempt at a Solution



I got S and V or c) and b), is that correct?

I thought U or a) would be too, but just found out that it is not closed under addition:
Lets say if: f(1)=2 f(-1)=0 whose product is 0, however another possibility is f(1)=0 f(-1)=a number (non-zero), so if you add them it would not be in V (2+0, 0 + a number) whose product would not be zero. Am I correct?

I somehow know whether it is in the subspace or not, but no idea how to write a proof. How do I show it? Do I do as I typed above, like with actual number examples? or do I have to do it with symbols? ex. (f+g)(x) = f(x) + g(x) there closed under addition...like that?

Thanks
 
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  • #2
To show its NOT a vector space, what you've done suffices (although I'm not following your notations of those 4 sets). You only need one counterexample: so showing its not closed under addition suffices.

To show that something IS a vector space (subspace in this case) however, is much harder. You must verify the 10 (if my memory serves me) axioms for vector space. But since you're trying to show its a subspace, many of these follow directly from U,V,S, and T inheriting their operations from F[r].
As you mentioned, closure does not follow trivially (like associativity does). So go over the axioms of vector fields to see which ones don't follow trivially, and you must write a proof for each of these.

For instance, if you think something IS a subspace, one of the things you must show is closure (as you've mentioned). So let 2 arbitrary elements be in your (supposed) subspace, and show that their 'sum' is in your subspace. Then take another arbitrary element in your subspace and show that its 'scalar product' is in your subspace. Then conclude that since the elements were arbitrarily chosen, closure holds for ALL elements of the subspace.

You must do this for all of the axioms!

Good luck! Hope this helps!
 

Related to Subspaces of Functions- Linear Algebra

1. What is a subspace?

A subspace is a subset of a vector space that maintains the same operations and properties as the larger vector space. In other words, it is a smaller space that is contained within a larger space.

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

To determine if a set of functions is a subspace, you must check three criteria: closure under addition, closure under scalar multiplication, and containing the zero function. If all three criteria are met, then the set of functions is a subspace.

3. What is the significance of subspaces in linear algebra?

Subspaces are important in linear algebra because they allow us to study certain properties and operations in a smaller, more manageable space. They also help us understand the structure and relationships within a larger vector space.

4. Can a subspace be infinite-dimensional?

Yes, a subspace can be infinite-dimensional. This means that the subspace contains an infinite number of linearly independent vectors.

5. How are subspaces related to linear independence?

Subspaces and linear independence are closely related. A set of vectors is linearly independent if and only if the subspace they span is the entire vector space. In other words, if the vectors in a subspace are linearly independent, then they span the entire vector space.

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