Question on subspaces and spans of vector spaces

  • #1
Hi, I have read my notes and understand the theory, but I am having trouble understanding the following questions which are already solved (I am giving the answers as well).

The first question says:

Let [itex]U_{1}[/itex] and [itex]U_{2}[/itex] be subspaces of a vector space V. Give an example (say in [itex]V=\Re^{2}[/itex]) to show that the union [itex]U_{1}\bigcup U_{2}[/itex] need not be a subspace of V.

And the answer is:

Take [itex]U_{1}=\{(x_{1}, x_{2})\in\Re^{2}:x_{2}=0\}[/itex] and [itex]U_{2}=\{(x_{1}, x_{2})\in\Re^{2}:x_{1}=0\}[/itex]

So I really don't understand this answer... I would really appreciate it if someone could explain it to me.


And the second question says:

Let S be the set of all vectors [itex](x_{1}, x_{2})[/itex] in [itex]\Re^{2}[/itex] such that [itex]x_{1}=1[/itex]. What is the span of S?

And the answer is:

span S = [itex]\Re^{2}[/itex] because [itex](x_{1}, x_{2})=x_{1}(1, x^{-1}_{1}x_{2})[/itex] when [itex]x_{1}\neq 0[/itex] and [itex](x_{1}, x_{2})=(1, 0)-(1, -x_{2})[/itex] when [itex]x_{1}=0[/itex].

So I don't understand this explanation, mainly because I thought that [itex]x_{1}[/itex] is supposed to be 1... :/

Sorry if the questions sound silly, and thanks for any help you can give me! :)
 

Answers and Replies

  • #2
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3,292
For the first one: what don't you understand?? The solution claims three things:

1) U1 is a subspace of [itex]\mathbb{R}^2[/itex]
2) U2 is a subspace of [itex]\mathbb{R}^2[/itex]
3) [itex]U_1\cup U_2[/itex] is not a subspace of [itex]\mathbb{R}^2[/itex]

Which of these three points is bothering you??

For the second one: A typical vector in S is (1,a) for a arbitrary. We are interesting in whether the span of S is [itex]\mathbb{R}^2[/itex].
The span is all linear combinations of elements of S. Thus if we claim that the span of S is [itex]\mathbb{R}^2[/itex], then we actually claim that each vector in [itex]\mathbb{R}^2[/itex] is a linear combination of elements in S. Can you find such a linear combination?
 
  • #3
Fredrik
Staff Emeritus
Science Advisor
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It looks like you haven't really studied the definitions of the terms you're using, in particular "span". The span of S is the set of all linear combinations of members of S. So it's almost obvious that span S=ℝ2. To prove it, it's sufficient to show that an arbitrary member of ℝ2 can be written as ax+by, where a and b are real numbers, and x and y are members of S. (Actually, this proves that ℝ2 is a subset of span S, but since it's obvious that span S is a subset of ℝ2, this is what we need to conclude that span S=ℝ2).

The first question makes me wonder if you understand the terms "union" and "subspace". It should be pretty obvious that the union isn't closed under the vector space operations (addition and scalar multiplication).

(I actually wrote this before micromass wrote his reply above, but my brother who is visiting me this week pulled me away from the computer for a while).
 
  • #4
Thanks a lot, I feel like a retard now... I got very confused with the concept of subspace. I do understand it now though and it seems easy, so all the other problems make sense :)
 

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