# A quite verbal proof that if V is finite dimensional then S is also...

Homework Statement:
If a linear space V is finite dimensional then S, a subspace of V, is also finite-dimensional and dim S is less than or equal to dim V.
Relevant Equations:
Let A = {u_1, u_2, ... u_n} be a basis for V.
If a linear space ##V## is finite dimensional then ##S##, a subspace of ##V##, is also finite-dimensional and ##dim ~S \leq dim~V##.

Proof: Let's assume that ##A = \{u_1, u_2, \cdots u_n\}## be a basis for ##V##. Well, then any element ##x## of ##V## can be represented as
$$x = \sum_{i=1}^{n} c_i u_i$$
As ## S \subset V## therefore, all the elements of ##S## can also be represented by linear combinations of ##u_i##s. As, ##u_i## s are finite, this implies that ##S## is finite dimensional.

Let's say ##dim~S \gt dim~V##. Then, a basis for ##S## would look like
$$A' = \{ v_1, v_2, \cdots v_m\}$$
where ##m \gt n##. But that would imply that ##A'## is independent thus contradicting that "any set of (n+1) elements in V would be dependent if ##L(A) = V## and number of elements in A is n". Hence, ##dim ~S \leq dim~V##.

Thats ends my proof.

It seems to me that my proof of the first part is quite verbal and needs to be a little more rigorous but the question is: Was the reasoning correct? And how to make it rigorous?

PeroK
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Homework Statement:: If a linear space V is finite dimensional then S, a subspace of V, is also finite-dimensional and dim S is less than or equal to dim V.
Relevant Equations:: Let A = {u_1, u_2, ... u_n} be a basis for V.

If a linear space ##V## is finite dimensional then ##S##, a subspace of ##V##, is also finite-dimensional and ##dim ~S \leq dim~V##.

Proof: Let's assume that ##A = \{u_1, u_2, \cdots u_n\}## be a basis for ##V##. Well, then any element ##x## of ##V## can be represented as
$$x = \sum_{i=1}^{n} c_i u_i$$
As ## S \subset V## therefore, all the elements of ##S## can also be represented by linear combinations of ##u_i##s. As, ##u_i## s are finite, this implies that ##S## is finite dimensional.
I agree with this, but why does this imply ##S## is finite dimensional? What's the definition of finite dimensional?

Let's say ##dim~S \gt dim~V##. Then, a basis for ##S## would look like
$$A' = \{ v_1, v_2, \cdots v_m\}$$
where ##m \gt n##. But that would imply that ##A'## is independent thus contradicting that "any set of (n+1) elements in V would be dependent if ##L(A) = V## and number of elements in A is n". Hence, ##dim ~S \leq dim~V##.

This is okay, but how do you know that "independent in ##S##" implies "independent in ##V##".

In fact, if you show this statement, then the result follows immediately. And, you'll have a very short proof.

I agree with this, but why does this imply S is finite dimensional?
Because it can be spanned by a a finite set which is independent. But, yes, the problem is those ##u_i## s' linear combinations surpass S.
What's the definition of finite dimensional?
Having a basis set which is finite (have finite number of elements). Or to put it more elaborately: if there exists a finite set which is indpendent and spans S, then S is said to be finite dimensional.

PeroK
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Because it can be spanned by a a finite set which is independent. But, yes, the problem is those ##u_i## s' linear combinations surpass S.
Okay, but one problem is that the vectors in the basis for ##V## may not be in the subspace ##S## at all. Maybe starting with a basis for ##V## is not the right approach?

This is okay, but how do you know that "independent in S" implies "independent in V".
Yes, the implication will not be true always.

PeroK
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Yes, the implication will not be true always.
It is true. That is the heart of the proof.

It is true. That is the heart of the proof.
Well, I mean a set is independent completely on its own accord, no matter whose subset it is. If a set is independent, then none of its elements can be represented as linear combination of its other elements.

PeroK
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Well, I mean a set is independent completely on its own accord, no matter whose subset it is. If a set is independent, then none of its elements can be represented as linear combination of its other elements.
Exactly, a linearly independent set in ##S## cannot be linearly dependent in ##V##. Can you construct a short proof from that?

Exactly, a linearly independent set in ##S## cannot be linearly dependent in ##V##. Can you construct a short proof from that?
Take an independent set in S that spans S, then this indpendent set would be a subset of basis for V. Hence, the basis of V is larger than that of S.

PeroK
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Take an independent set in S that spans S, then this indpendent set would be a subset of basis for V. Hence, the basis of V is larger than that of S.
That feels a bit loose to me. It's not wrong, but let me show you the alternative:

Suppose the dimension of ##V## is ##n##. We cannot find more than ##n## linearly independent vectors in ##V##. Now, any linearly independent set in ##S## is linearly independent in ##V##. Therefore, we cannot find more than ##n## linearly independent vectors in ##S##. Hence, ##S## is finite dimensional and ##dim \ S \le n##.

Hall
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Alternatively, you could do it by contradiction:

If we have a set of ##n + 1## linearly independent vectors in ##S##, then we have a set of ##n + 1## linearly independent vectors in ##V##. Which contradicts ##dim \ V = n##.

Therefore, ##S## is finite dimensional and ##dim \ S \le dim \ V##.

That feels a bit loose to me. It's not wrong, but let me show you the alternative:

Suppose the dimension of ##V## is ##n##. We cannot find more than ##n## linearly independent vectors in ##V##. Now, any linearly independent set in ##S## is linearly independent in ##V##. Therefore, we cannot find more than ##n## linearly independent vectors in ##S##. Hence, ##S## is finite dimensional and ##dim \ S \le n##.
That compelled me to yell out "That's elegant".

PeroK
@PeroK Can you please do something for the first part of question?

PeroK
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@PeroK Can you please do something for the first part of question?
What first part?

What first part?
Proving that S is finite-dimensional.

PeroK
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Proving that S is finite-dimensional.
We already have. I made no assumption that S was finite dimensional. I simply looked for ##n + 1## independent vectors in ##S##. I avoided the potential tangle created by looking for a basis for ##S## or a set that spans ##S##. That was part of the elegant simplicity!

Therefore, we cannot find more than n linearly independent vectors in S. Hence, S is finite dimensional and dim S≤n.
Yes, we cannot find more than n indpendent vectors in S, but how do we ensure that a set of independent vectors in S would span S? We can find indpendent vectors but assuming that they would span S is assuming that S is finite-dimensional.

PeroK
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Yes, we cannot find more than n indpendent vectors in S, but how do we ensure that a set of independent vectors in S would span S? We can find indpendent vectors but assuming that they would span S is assuming that S is finite-dimensional.
If you really wanted to, you could:

Assume ##S## is infinite dimensional. Therefore, we can find ##n + 1## linearly independent vectors in ##S##. Contradiction. Therefore, ##S## is finite dimensional.

But, it's not necessary to do that as a separate step.

The second point is really asking how we know that ##dim \ S## is well-defined? Perhaps ##S## doesn't have a basis? That's not something we need to prove here.

Hall
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Although, if you want a bit more practice, you could prove this:

Let ##V## be a finite-dimensional vector space and ##S## a subspace of ##V##. Prove that ##S## has a basis.

V be a finite-dimensional vector space and S a subspace of V. Prove that S has a basis
Suppose ##S## doesn't have a basis, that implies that there doesn't exist any independent set in ##S## that spans it. However, any independent set in ##S## is also indpendent in ##V## and therefore, forms a subset of a basis of ##V##. Now, this subset of a basis must span a part of ##V##, that is the span of indpendent elements of ##S## forms a subspace of ##V##.

Take any independent set A (number of elements in A being less than or equal to n) in S, then ##L(A) \subset S##. As no indpendent set in S can span S, ...

I have to think harder for that.

PeroK
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Suppose ##S## doesn't have a basis, that implies that there doesn't exist any independent set in ##S## that spans it. However, any independent set in ##S## is also indpendent in ##V## and therefore, forms a subset of a basis of ##V##. Now, this subset of a basis must span a part of ##V##, that is the span of indpendent elements of ##S## forms a subspace of ##V##.

Take any independent set A (number of elements in A being less than or equal to n) in S, then ##L(A) \subset S##. As no indpendent set in S can span S, ...

I have to think harder for that.
I wouldn't worry about it for this problem.

I wouldn't worry about it for this problem.
But it's a good and brain-demanding exercise. Should I create a new thread? I have no problem in continuing within this one.

Let's start with this one: How a subspace would look like if it doesn't have a basis?

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But it's a good and brain-demanding exercise. Should I create a new thread? I have no problem in continuing within this one.

Let's start with this one: How a subspace would look like if it doesn't have a basis?
It's more or less the same argument. You pick ##u_1 \in S##. If ##span \{u_1 \} = S## we are done. If not, we pick ##u_2 \in S - span \{u_1 \}## and ##u_3 \in S - span \{u_1, u_2 \}## etc.

This generates a set of at most ##n## linearly independent vectors that must span ##S##. I.e. we must be able to generate a finite basis for ##S##.

generates a set of at most n linearly independent vectors that must span S. I.e. we must be able to generate a finite basis for
I was thinking in the same line, coz if any set of those n indpendent elements doesn't span S then S won't be a subspace (however, I myself have some doubt regarding "then S won't be a subspace".)

PeroK