Proving Vector Spaces: W & U in F

In summary, the conversation discusses the proof that V is a vector space over F, the proof that W is a subspace of V, and the proof that U is a subspace of V and is contained in W. The conversation also touches on the idea of showing that sums and scalar multiples of elements in U remain in U, and the importance of showing that the sum of every sequence squared is finite in order to prove this.
  • #1
fk378
367
0

Homework Statement


Let F be the field of all real numbers and let V be the set of all sequences (a1,a2,...a_n,...), a_i in F, where equality, addition, and scalar multiplication are defined component-wise.

(a) Prove that V is a vector space over F
(b) Let W={(a1, a2,...,a_n,...) in V | lim a_n = 0 as n-->inf}. Prove that W is a subspace of V.
(c) Let U={(a1,...,a_n,...) in V | summation of (a_i)^2 is finite, i evaluated from 1 to inf}. Prove that U is a subspace of V and is contained in W.

The Attempt at a Solution


I know that in order for W to be a subspace of V, W must form a vector space over F under the operations of V. I've already proved (a). Do I need to know the limit of a_n to prove (b) or is that just for (c)? It seems like proving (b) is pretty similar to (a), right? Any tips on proving (c)?
 
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  • #2
In part b, the lim a_n = 0 as n --> inf just means that only those a_n that are contained in V that meet the criteria lim a_n = 0 as n --> inf are contained in the set.

So to prove part b, you just need to show that addition and scalar multiplication are closed in the subspace.

Sorry, I'm not too sure about part C (I don't want to guess and tell you wrong either).
 
  • #3
Suppose {a_i} is a sequence in U such that the a_i's do not tend to 0 as i increases without bound. Consider the infinite series Sum[(a_i)^2] = Sum[b_i] where each b_i is positive and does not tend to 0. Suppose the sum converges to L, which means for each r > 0, there exists a number N so that | L - Sum[b_i] | < r where i ranges from 0 to n for all n > N. If the terms b_i are never negative and never tend to 0, can this condition be satisfied? (Note that the condition that the b_i's tend to 0 is that for all d > 0, there is some N so that |b_i| < d for all i > N.)
 
  • #4
I don't really understand, but I would say that the condition cannot be satisfied because if the terms b_i never tend to 0 then the summation must diverge.
 
  • #5
fk378 said:
I don't really understand, but I would say that the condition cannot be satisfied because if the terms b_i never tend to 0 then the summation must diverge.

If you can show that rigorously, then you have shown that each element of U is necessarily an element of W, and is thus contained in W. The only thing left is to show that sums of elements in U remain in U, and so do scalar multiples, which is the easy part.
 
  • #6
How would I show that the sum of any multiples of elements of U is still in U? Wouldn't I have to show that the sum of every sequence^2 is finite?
 
  • #7
fk378 said:
How would I show that the sum of any multiples of elements of U is still in U? Wouldn't I have to show that the sum of every sequence^2 is finite?

As an example, suppose {a_i} is an element of U. Then the series (a_1)^2 + (a_2)^2 + ... converges. Consider the element s*{a_i} defined to be {s*a_i}, so we now consider the series (s*a_1)^2 + (s*a_2)^2 + ... = s^2*(a_1)^2 + s^2*(a_2)^2 + ... = s^2*((a_1)^2 + (a_2)^2 + ...) = s^2*A where A is the number that the original series converges to, showing constructively that the component-wise defined multiple also converges. Now you just have to consider what happens to a component-wise sum.
As for the latter question, you would already have shown that by showing that U is contained in W (by contradiction). (Need more of a hint?)
 

Related to Proving Vector Spaces: W & U in F

1. What is a vector space?

A vector space is a mathematical structure that consists of a set of vectors and two operations, vector addition and scalar multiplication, that satisfy certain properties. These properties include closure, associativity, commutativity, existence of a zero vector, existence of additive inverse, and distributivity.

2. How do you prove that a set is a vector space?

To prove that a set is a vector space, you need to show that it satisfies all the properties of a vector space. This includes showing that the set is closed under vector addition and scalar multiplication, that a zero vector exists, and that each vector has an additive inverse. You also need to verify that the operations of vector addition and scalar multiplication follow the properties of associativity, commutativity, and distributivity.

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

A subspace is a subset of a larger vector space that also satisfies the properties of a vector space. This means that a subspace is itself a vector space. A vector space, on the other hand, is a standalone mathematical structure that does not necessarily need to be a subset of another vector space.

4. How do you prove that a set is a subspace of a given vector space?

To prove that a set is a subspace of a given vector space, you need to show that it satisfies all the properties of a vector space. This includes showing that the set is closed under vector addition and scalar multiplication, that a zero vector exists, and that each vector has an additive inverse. You also need to verify that the operations of vector addition and scalar multiplication follow the properties of associativity, commutativity, and distributivity. Additionally, you need to show that the set is a subset of the given vector space.

5. What is the significance of W & U in F when proving vector spaces?

W & U in F are used to represent two sets within a given vector space. W represents a subspace of the vector space, while U represents a set of vectors that span the vector space. Using these two sets, we can prove that the given vector space is a direct sum of W and U. This is important because it allows us to understand the structure of the vector space and how it can be broken down into smaller subspaces.

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