On proving real vector spaces (subspaces)

In summary, the conversation was about how to show that a subset W of a vector space V is a subspace if and only if given u and v as vectors in W and a and b as scalars, then au + bv is in W. The definition of a subspace and how to prove it using the given statement were discussed. The set W={(x,y)€R^2;|x|=|y|} was also mentioned as an example of a set that is not a subspace due to a non-satisfying property.
  • #1
franz32
133
0
I hope someone can help me (guide) in this theorem.

How can I show that a "subset W of a vector space V is indeed
a subspace of V if and only if given u and v as vectors in W and
a and b are said to be scalars, then au + bv is in W."?

Can I assume a vector with my desired number of elements?
Also I am sure that W is a subset of V because it's given.
 
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  • #2
What do you mean by "number of elements"? The vector space is not stated to have a particular dimension (and in fact, the statement is true for infinite dimensional vectors spaces as well) so I wouldn't recommend trying to do this by looking at components.

What is the definition of "subspace". What you need to do is show that if "given u and v as vectors in W and
a and b are said to be scalars, then au + bv is in W", then all of the properties of the definition of subspace are satisfied.
(Hint: many of them, such as commutativity of addition, follow from the fact that u, v are in the V and those are true for V.)

You will also need to show(since this is "if and only if" that, IF W is a subspace of V, THEN "given u and v as vectors in W and a and b are said to be scalars, then au + bv is in W" but that's much easier.
 
  • #3
Hallsof Ivy

Hello there.

Yeah, what I mean about the "number of elements" is that the number of components in a vector. But anyway, ... it may not be the concern.

So au + bv is in W... does it mean that au + bv = bv + au or
au + (bv + cw) = (au + bv) + cw for w as a vector and c as a constant?

How about the converse of that statement? WHat is the approach?
 
  • #4
Originally posted by Franz32So au + bv is in W... does it mean that au + bv = bv + au or
au + (bv + cw) = (au + bv) + cw for w as a vector and c as a constant?

Yes, of course. au+bv is in W so it is in V. au+ bv= bv+ au is true because V is a vector space and so addition is commutative.
au, bv, and cw are in W so they are in V. au + (bv + cw) = (au + bv) + cw because addition is associative in V.

(By the way, a, b, and c are "scalars" or "numbers". They are not necessarily "constant".)
 
  • #5
Thank you very much

Hello HallsofIvy!

Now, I get it... =) thank you. PArdon me for the thread on

the resistances. I didn't take a glance at it and I just

simply placed it there without knowing that it was easy.
 
  • #6
i want to similar question...
the question is shown as below...
Explain why the set w={(x,y)€R^2;|x|=|y|}, is not a real subspace.
anyone can help me??
thanks a lot...
 
  • #7
xiaobai5883 said:
i want to similar question...
the question is shown as below...
Explain why the set w={(x,y)€R^2;|x|=|y|}, is not a real subspace.
anyone can help me??
thanks a lot...

Consider vectors
[tex] (2,2), (-1,1) \in W [/tex]
But
[tex] (2,2)+ (-1,1) = (1,3) [/tex]
is not in W. Hence not a vector space.
 

1. What is a real vector space?

A real vector space is a mathematical structure that consists of a set of elements, called vectors, and two operations: vector addition and scalar multiplication. The elements of a real vector space can be any real numbers, and the operations must follow certain properties such as closure, associativity, and distributivity.

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

To prove that a set is a subspace of a real vector space, you must show that it satisfies three conditions: it contains the zero vector, it is closed under vector addition, and it is closed under scalar multiplication. These conditions ensure that the set is a valid subset of the vector space and follows the properties of a vector space.

3. What is the significance of proving a set is a subspace?

Proving that a set is a subspace is important because it allows us to use all the properties and theorems of vector spaces on that set. This makes it easier to solve problems and make conclusions about the elements in the set. Additionally, subspace proofs are often used in more complex mathematical proofs and theories.

4. Can a set be a subspace of more than one real vector space?

No, a set can only be a subspace of one real vector space. This is because the operations of vector addition and scalar multiplication must follow the properties of a vector space, which are specific to each vector space. If a set satisfies the properties of a vector space for one real vector space, it may not necessarily satisfy them for a different real vector space.

5. Are all real vector spaces finite in size?

No, not all real vector spaces are finite in size. Some real vector spaces may have infinite elements, such as the set of all real numbers. However, there are also real vector spaces that are finite in size, such as the set of 2D or 3D vectors with real number components.

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