Proof involving subspaces of finite-dimensional vector spaces

In summary: S is finite-dimensional.If S does not form a basis for V, then add a new vector to S such that S is still linearly independent and S' will then form a basis for V. However, since all vectors in V are also in S, then S' contains at most n vectors. Therefore, S does not span V and W cannot be a subspace of V.
  • #1
Essnov
21
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This is an exercise in a linear algebra textbook that I initially thought was going to be easy, but it took me a while to make the proof convincing.

Prove: Any subspace of a finite-dimensional vector space is finite-dimensional.

Here's my attempt. I am not sure about some details and I'm hoping you guys can tell me if it makes sense or suggest improvements, give some feedback, suggest a better method, etc.

Let V be a finite-dimensional vector space such that dim(V) = n.

Now let W be a subspace of V and let S be a linearly independent set of vectors {w1, w2, ... , wr} in W.

If S spans W, then S is a basis for W and W has dimension r. If S does not span W, then add a new vector from W to S such that S is still linearly independent. If W is infinite-dimensional, then we should be able to add as many vectors as we like and S will still not span W.

However, since all vectors in W are also in V, then S is also a set of vectors in V. Therefore, S can contain at most n vectors while still being linearly independent. And if S contains n vectors, then S spans V and thus W as well.

This implies that W has dimension of at most n, and thus is finite-dimensional.

Thanks in advance for any comments.
 
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  • #2
That's rather awkwardly stated but I think it is basically correct. One difficulty I see is that you haven't stated explictely what your definition of "finite dimensional" is. Not knowing that, I can't say whether your proof covers it or not. In most books, the definition of "finite dimensional" is NOT that it has a basis containing only a finite number of vectors, because you need finiteness to show that a vector space has a basis (without going into "transfinite induction" or "Zorn's lemma" that is used to prove the existence of a basis for infinite dimensional spaces). Most books say a vector space is "finite dimensional" if and only if it can be spanned by a finite set of vectors. You then show that there exist a basis which is a subset of that spanning set. So I would not worry about a basis. Since V is finite dimensional, it can be spanned by a finite set of vectors. Since W is as subspace of V it can be spanned by that same set and so is finite dimensional.
 
  • #3
Thank you for your reply.

The textbook gives only the following definition:

"A nonzero vector space V is called finite-dimensional if it contains a finite set of vectors {v1, v2, ... vn} that forms a basis. If no such set exists, V is called infinite-dimensional. "

It makes the proof a little more difficult than the definition you stated, where if a vector space can be spanned by a finite set of vectors, then it is finite-dimensional. Using the definition you stated, it seems like the proof immediately follows from the fact that a basis for V spans all of its subspaces as well, so all of its subspaces are finite-dimensional -- like you said, basically.
 
  • #4
Sure, but you can prove if a space is spanned by a finite number of vectors, then it has a basis with a finite number of vectors, right?
 
  • #5
It makes sense that it would be true. Let's see.

"If a vector space is spanned by a finite number of vectors, then it contains a finite set of vectors which form a basis."

Let V be a vector space and let S be a linearly independent set of n vectors which spans V.

While the vectors in S are not necessarily in V, all linearly independent sets of vectors in V can be constructed from linear combinations of the vectors in S.

This implies that the largest number of vectors that a linearly independent set in V can contain is n. Then, construct from S a set of linearly independent vectors in V to which no other vector from V can be added without making it linearly dependent.

This new set S' has the following properties:
a) All vectors in V can be expressed as a linear combination of the vectors in S', otherwise those vectors could be added to S' without making it linearly dependent. Therefore, S' spans V.
b) It is linearly independent.
c) It is in V.
d) It contains at most n vectors.
And so S' contains a finite number of vectors which form a basis for V.

Therefore, all vector spaces spanned by a finite number of vectors contain a finite set of vectors which form a basis.

Now using this result...

"Any subspace of a finite-dimensional vector space is finite-dimensional."

If S forms a basis for a finite-dimensional vector space, then it spans its subspaces as well. Then, all subspaces of finite-dimensional vector spaces contain finite sets of vectors which form bases for those subspaces. Therefore, all subspaces of finite-dimensional vector spaces are finite-dimensional.

How does that look? :)
 
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  • #6
That's a bit confusing. I'd be a little more explicit about how you get a basis for W. Pick a vector in W, say w1. If {w1} spans W, then you are done. If not, you can find a vector w2 in W that is linearly independent of {w1}, say w2. If {w1,w2} spans W then it's a basis and you are done. If not you can find a third linearly independent vector in W, say w3. You know this has to stop at some point because the maximum number of linearly independent vectors in V is n. Then the w's form a basis for W. Something like that.
 

1. What is a subspace of a finite-dimensional vector space?

A subspace of a finite-dimensional vector space is a subset of the vector space that also satisfies all the properties of a vector space. This means that it is closed under addition and scalar multiplication, contains the zero vector, and is closed under linear combinations.

2. How can I prove that a subset is a subspace of a finite-dimensional vector space?

To prove that a subset is a subspace of a finite-dimensional vector space, you must show that it satisfies all the properties of a vector space. This can be done by showing closure under addition and scalar multiplication, as well as containing the zero vector. Additionally, you must show that it is closed under linear combinations.

3. What is the dimension of a subspace of a finite-dimensional vector space?

The dimension of a subspace of a finite-dimensional vector space is the number of linearly independent vectors that span the subspace. This is also known as the number of basis vectors for the subspace.

4. How can I show that two subspaces of a finite-dimensional vector space are equal?

To show that two subspaces of a finite-dimensional vector space are equal, you must first show that they are subsets of each other. Then, you must show that they have the same dimension, which means that they have the same number of linearly independent vectors that span the subspace. This can be done by showing that any basis for one subspace can be extended to form a basis for the other subspace.

5. Can a subspace of a finite-dimensional vector space be infinite-dimensional?

No, a subspace of a finite-dimensional vector space must also be finite-dimensional. This is because it is a subset of the vector space, which by definition is finite-dimensional. If a subspace were to be infinite-dimensional, it would not satisfy the properties of a finite-dimensional vector space.

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