On proving real vector spaces (subspaces)

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Discussion Overview

The discussion revolves around the conditions under which a subset W of a vector space V qualifies as a subspace. Participants explore the theorem stating that W is a subspace if and only if for any vectors u and v in W and scalars a and b, the linear combination au + bv is also in W. The conversation includes definitions, properties of vector spaces, and specific examples.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant seeks guidance on proving that a subset W is a subspace based on the linear combination condition.
  • Another participant emphasizes the importance of the definition of subspace and suggests showing that all properties of a subspace are satisfied if the linear combination condition holds.
  • A participant questions the meaning of "number of elements" in the context of vectors and clarifies that the dimension of the vector space is not fixed.
  • There is a discussion about the properties of vector addition, specifically commutativity and associativity, and how they relate to the elements being in W and V.
  • One participant expresses gratitude for the clarification received regarding the properties of vector addition.
  • Another participant introduces a related question about why a specific set W defined by |x|=|y| in R^2 is not a subspace, providing an example to illustrate their point.

Areas of Agreement / Disagreement

Participants generally agree on the definitions and properties of vector spaces and subspaces, but there are multiple views on how to approach the proof and the implications of the linear combination condition. The discussion about the specific set W in R^2 introduces a new question that remains unresolved.

Contextual Notes

Some participants express uncertainty regarding the implications of the linear combination condition and the definitions involved. The discussion does not resolve the question about the specific set W in R^2, leaving it open for further exploration.

Who May Find This Useful

This discussion may be useful for students studying linear algebra, particularly those interested in understanding the properties of vector spaces and subspaces, as well as those tackling related homework problems.

franz32
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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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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.
 
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?
 
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".)
 
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.
 
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...
 
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.
 

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