Prove Field F Has Only 2 Subspaces: {0} & F Itself

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In summary: So, in other words, U is a subspace of F because it is a subset of F that is also a vector space. Then, since scalar multiplication is defined for vector spaces, it follows that U is closed under scalar multiplication.
  • #1
Anonymous217
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Homework Statement


Prove whether or not a field F has just two subspaces: {0} and F itself.2. The attempt at a solution
I'm not exactly sure where to start. I believe it's true but I don't know how to prove that F has only two subspaces. I tried doing proof by contradiction (ie. supposing there was a third subspace for F), but I don't know where I can go with that.
 
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  • #2
Not sure on this one, but... Can you show that the zero subspace, and the field itself has: (a) the "zero object" of F exists in both subspaces. (b) Each subspace is closed under "object" addition. (c) Each subspace is closed under scalar multiplication. ? Just a thought.
 
  • #3
If F is a vector space over itself, what dimension is it?

Then remember any proper subspace has to have a smaller dimension
 
  • #4
Samuelb88 said:
Not sure on this one, but... Can you show that the zero subspace, and the field itself has: (a) the "zero object" of F exists in both subspaces. (b) Each subspace is closed under "object" addition. (c) Each subspace is closed under scalar multiplication. ? Just a thought.
Well, yes you can, but that would just prove that the empty set and the field itself are subspaces. It wouldn't prove that those are the only two subspaces of F.

Office_Shredder said:
If F is a vector space over itself, what dimension is it?

Then remember any proper subspace has to have a smaller dimension
We haven't gotten over dimensions in my class so I don't know that to be honest. Do you know a link that can direct me to know more about this, or do you have the time to explain it? It would seem pretty easy once I know this piece of information.
 
  • #5
Bumppp.
 
  • #6
Suppose that U is a non-zero subspace of F. Then U is closed under scalar multiplication. Let x be any non-zero element of U and let y be any element of F. Can you find a scalar multiple of x that equals y? What does that tell you?

Petek
 
  • #7
Do you mean subfield rather than subspace? I don't know what it means to say that a field has a subspace.
 
  • #8
Petek said:
Suppose that U is a non-zero subspace of F. Then U is closed under scalar multiplication. Let x be any non-zero element of U and let y be any element of F. Can you find a scalar multiple of x that equals y? What does that tell you?

Petek

Is it supposed to show that any non-zero subspace of F must be F itself? I can understand the rest of the proof from then on, but how do you show that U must be = F? What scalar multiple of x equals y?
 
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  • #9
Anonymous217 said:
Is it supposed to show that any non-zero subspace of F must be F itself?
Yes.
I can understand the rest of the proof from then on, but how do you show that U must be = F? What scalar multiple of x equals y?

A subspace of a vector space is, by definition, closed under scalar multiplication. I'm trying to get you to show that every element of F is a scalar multiple of a non-zero element of U. Thus, F [itex]\subseteq[/itex] U and so F = U. I'm sure that with some thought you can figure out what scalar (that is, what element of F) times x equals y.

Petek
 
  • #10
Anonymous217 said:
Is it supposed to show that any non-zero subspace of F must be F itself? I can understand the rest of the proof from then on, but how do you show that U must be = F? What scalar multiple of x equals y?

HallsofIvy said:
Do you mean subfield rather than subspace? I don't know what it means to say that a field has a subspace.

I'm taking this to mean that the field is being regarded as a vector space over itself and then subspace means a subset of F that also is a vector space over F.
 

1. What is a subspace?

A subspace is a subset of a vector space that satisfies the properties of a vector space. This means that it is closed under vector addition and scalar multiplication, and contains the zero vector.

2. How do you prove that a field has only two subspaces?

To prove that a field has only two subspaces, we need to show that the only subspaces that exist are the trivial subspace {0} and the entire field itself. This can be done by showing that any other subset of the field fails to satisfy the properties of a vector space.

3. Why is it important to prove that a field has only two subspaces?

Proving that a field has only two subspaces is important because it helps us understand the structure of the field. It also allows us to simplify and generalize certain theorems and proofs in linear algebra and other fields of mathematics.

4. Can a field have more than two subspaces?

No, a field can only have two subspaces: the trivial subspace {0} and the entire field itself. This is because any other subset of the field will either fail to satisfy the properties of a vector space or will be equal to the entire field.

5. How does proving the number of subspaces of a field relate to linear independence?

The number of subspaces of a field is related to linear independence because the number of subspaces corresponds to the number of linearly independent vectors in the field. Since a field can only have two subspaces, this means that there can only be two linearly independent vectors in the field. Any additional vectors will be linearly dependent on these two vectors.

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