How many vectors in span({v})

  • Thread starter brru25
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In summary: So (1,1)*x does indeed yield (1,1,0), (1,1,1), (1,1,2)."a" I believe can be any number in F. So (1,1)*x does indeed yield (1,1,0), (1,1,1), (1,1,2).
  • #1
brru25
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Homework Statement



Let v be an element of F2p \ {(0,0)}. How many vectors does Span({v}) have? How many 1-dimensional vector subspaces does F2p have?

F2p is the two-dimensional field (a,b) where each a, b are elements of Fp, where p is a prime number.

The Attempt at a Solution



I know the total number of elements in the vector space F2p \ {(0,0)} is p2 - 1. I also thought that this was the number of vectors in Span({v}) but I've told by a couple people that is not so. I started with the definition of span but I just couldn't see the rest unfold.

Thank you ahead of time for your help.
 
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  • #2
An element in span(v) must be of the form av for some a in the field. How many choices of a do you have? Do all of these choices yield a different vector, or can av=bv if a=/=b?
 
  • #3
Take an example. Let F=Z_3, the field with three elements {0,1,2}. Yes, there are 8 elements in (Z_3)^2-{0,0}. How many elements are in span((1,1))? That's (1,1)*x for all x in Z_3. Does that help you to see things unfold?
 
  • #4
Dick said:
Take an example. Let F=Z_3, the field with three elements {0,1,2}. Yes, there are 8 elements in (Z_3)^2-{0,0}. How many elements are in span((1,1))? That's (1,1)*x for all x in Z_3. Does that help you to see things unfold?

Would (1,1)*x have 3 elements? (1,1,0), (1,1,1), (1,1,2)? Or am I reading this incorrectly?
 
  • #5
brru25 said:
Would (1,1)*x have 3 elements? (1,1,0), (1,1,1), (1,1,2)? Or am I reading this incorrectly?

Well, yeah. Would have three elements. But they would be (0,0)=(1,1)*0, (1,1)=(1,1)*1 and (2,2)=(1,1)*2.
 
  • #6
Dick said:
Well, yeah. Would have three elements. But they would be (0,0)=(1,1)*0, (1,1)=(1,1)*1 and (2,2)=(1,1)*2.

Okay now I see what you meant sorry about that.
 
  • #7
Office_Shredder said:
An element in span(v) must be of the form av for some a in the field. How many choices of a do you have? Do all of these choices yield a different vector, or can av=bv if a=/=b?

"a" I believe can be any number in F.
 

1. How do you determine the number of vectors in span({v})?

To determine the number of vectors in span({v}), you need to find the linearly independent vectors that make up the span. The number of linearly independent vectors will give you the number of vectors in span({v}).

2. Can the number of vectors in span({v}) be more than the number of original vectors?

Yes, the number of vectors in span({v}) can be more than the number of original vectors. This is because the span of a set of vectors can contain linear combinations of those vectors, resulting in additional vectors in the span.

3. Is it possible for the number of vectors in span({v}) to be less than the number of original vectors?

No, the number of vectors in span({v}) cannot be less than the number of original vectors. This is because the span contains all possible linear combinations of the original vectors, so it will always have at least the same number of vectors as the original set.

4. How does the dimension of the vector space affect the number of vectors in span({v})?

The dimension of the vector space does not directly affect the number of vectors in span({v}). However, the dimension of the vector space will determine the maximum number of linearly independent vectors that can be in the span. If the dimension of the vector space is n, then the span can have at most n linearly independent vectors.

5. Can there be an infinite number of vectors in span({v})?

Yes, there can be an infinite number of vectors in span({v}). This is possible when the original set of vectors contains an infinite number of vectors or when the span includes all possible linear combinations of the original vectors, resulting in an infinite number of vectors.

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