# What is the proof of this theorem in Vector Spaces ?

• Maths Lover
In summary: Theorem to allow us to say that there cannot be such a set.In summary, the theorem states that if a set S of linear independent vectors spans the space V, then n is bigger than or equal to m. This theorem is important because it allows us to talk about the "dimension" of a vector space.
Maths Lover
Theorem :
if S ={ v1 , ... , vn} spans the V.Space V
, L={w1 , ... , wm} is set of linear independent vectors in V
then , n is bigger than or equal to m

How can we prove this ?

_____________

I read this theorem as a important note but the proof was ommited

Hey Maths Lover.

What can you say about the span of the space and a set of linear independent vectors with regard to the dimensionality of the space?

(Hint: How can you relate span to dimensionality and linear independence to said dimensionality of your vector space)?

Maths Lover said:
Theorem :
if S ={ v1 , ... , vn} spans the V.Space V
, L={w1 , ... , wm} is set of linear independent vectors in V
then , n is bigger than or equal to mHow can we prove this ?

_____________

I read this theorem as a important note but the proof was ommited
Chiro, I think the results you suggest to be used are consequences of this theorem.

I know of two different proofs. One is based upon the thereom which says that a linear homogeneous equation system with more unknowns than equations has nontrivial solutions.

Another proof goes like this:

For each k (0<=k<=m), let Sk = {w1, w2, ... , wk, v1, v2, ... , vn}
Each Sk spans W, since S does.

Now, for each k (0<=k<=m), let Tk be the result when we remove from Sk all vectors which are linear combinations of the previous vectors in Sk. Then, each Tk is linearly independent. Since L is linearly independent, no wi will ever be removed when we form Tk, so w1, w2, ... wk in Tk for all k (0<=k<=m). But when we form T(k+1) (0<=k<m), we must remove at least all vectors from S(k+1) which we remove when we form Tk from Sk. But this is not enough, for if we only remove these vectors from S(k+1) to form T(k+1), w(k+1) would still be a linear combination of the other vectors in T(k+1), i.e. of the vectors in Tk, which contradicts that T(k+1) is linearly independent. Thus, we must remove at least one vector more to obtain T(k+1).

It follows that for each k (1<=k<=m), we must remove at least k vectors from Sk to form Tk, but none of the wi:s will be removed. In particular Tm contains all the vectors w1, w2, ... wm, and still it contains at least m vectors fewer than Sm, which contains m+n vectors. It follows that m<=(m+n)-m=m, that is: m<=n,

"... if we only remove these vectors from S(k+1) to form T(k+1), w(k+1) would still be a linear combination of the other vectors in T(k+1), i.e. of the vectors in Tk,..."

it might help to state why this claim is true, i.e. Tk is not only independent but also spans.

(this proof is due to riemann, in the special case of proving invariance of rank of homology groups of a surface, but is usually attributed to steinitz.)

You can get lost iin a sea of notation here. If we are talking about finite dimension vector spaces (whcih seems a reasonable assumption, from the notation of "m" and "n" vectors) then:

If the vectors in S are linearly independent, the dimension of V is n, and there can not be more than n independsnt vectors in L.

If the vectors in S are not linearly dependent, remove vectors from S one at a time until you are left with a set of k < n independent vectors that span V. Repeating the previous argument, there can not be more than k indepedent vectors in L.

AlephZero said:
If the vectors in S are linearly independent, the dimension of V is n, and there can not be more than n independsnt vectors in L.
But you cannot prove this without the theorem in the OP. It is this theorem which makes it possible to talk about "the dimension" of a vector space. Without it, we wouldn't know that all bases have the same number of elements.

The definition of "dimension" that I use is that dim V=n if V contains a linearly independent set with cardinality n, but no linearly independent set with cardinality n+1. With this definition, we don't need a theorem to make it possible to talk about "the dimension" of a vector space.

Fredrik said:
The definition of "dimension" that I use is that dim V=n if V contains a linearly independent set with cardinality n, but no linearly independent set with cardinality n+1. With this definition, we don't need a theorem to make it possible to talk about "the dimension" of a vector space.
OK, we can have such a definition. But it does not á priori exclude that there exists a linearly independent set with k<n elements which cannot be extended to a linearly independent set with k+1 elements. We need a theorem like the one above, with a complex proof like the one above, to ensure this.

Erland said:
OK, we can have such a definition. But it does not á priori exclude that there exists a linearly independent set with k<n elements which cannot be extended to a linearly independent set with k+1 elements.

If a linearly independent set of vectors S spans a vector space, then every vector in the space has a unique representation as a linear combination of the vectors in S. (If there were two different representations, you could subtract them and get a linear dependency between the vectors in S).

You can use that fact to make a proof of the OP's theorem, by translating the argument about linear homegenous equations (which you mentioned) into the language of vector algebra.

People may have different opinions about whether that is a "simpler" proof than your (or Riemann's) proof, of course.

You can try with this proof (apagoge).

Suppose m > n.

Each vector of L is a linear combination of vectors v1, ... , vn, then:

w1 = a1(v1) + ... + an(vn)

At least one a_i coefficient is not zero (wlog, suppose a1 ≠ 0). Let's multiply both sides of equation by (a1)^(-1):

a1(v1) = w1 - a2(v2) - ... - an(vn)

[(a1)^(-1)](a1)(v1) = [(a1)^(-1)][w1 - a2(v2) - ... - an(vn)]

v1 = [(a1)^(-1)][w1 - a2(v2) - ... - an(vn)] (*)

Each vector of V is a linear combination of vectors v1, ... , vn, then:

v = b1(v1) + ... + bn(vn)

Recall the last result about v1:

v = b1(v1) + ... + bn(vn) = (b1)[(a1)^(-1)][w1 - a2(v2) - ... - an(vn)] + ... + bn(vn)

that is a linear combination of L1 = {w1, v2, ... vn}. This is another spanning set of V.

Let's express w2 as linear combination of L1 vectors:

w2 = c1(w1) + c2(v2) + ... + cn(vn)

There is at least one nonzero coefficient among c2...cn (otherwise w1 and w2 would not be linearly independent vectors). We can reiterate the procedure used to obtain (*), then L2 = {w1, w2, v3, ... vn} is another spanning set of V.

Replay this procedure n times, then Ln = L = {w1, ... , wn} is a spanning set of V.
Then, each vector of V is a linear combination of vectors of L, then

v = h1(w1) + ... + hn(wn)

In this case, it is true that

w_(n+1) = k1(w1) + ... + kn(wn)

but we supposed L to be a set of linearly independent vectors, then the last result is impossible and we proved the theorem.

the problem with Fredrik's definition is it is not clear that R^n has dimension n, without this theorem.

## 1. What is a vector space?

A vector space is a mathematical structure that consists of a set of objects (called vectors) and operations (such as addition and scalar multiplication) that can be performed on those objects. These operations must follow certain rules in order for the set to be considered a vector space.

## 2. What is a theorem in vector spaces?

A theorem in vector spaces is a statement that has been proven to be true within the context of vector spaces. It is based on the axioms and definitions of vector spaces and can be used to make conclusions about the properties and behaviors of vectors and operations within a vector space.

## 3. How is a theorem proven in vector spaces?

A theorem in vector spaces is typically proven using logical reasoning and mathematical techniques such as algebra, calculus, or linear algebra. The proof must be rigorous and follow the rules and definitions of vector spaces in order to be considered valid.

## 4. Why is it important to prove theorems in vector spaces?

Proving theorems in vector spaces is important because it allows us to understand and make predictions about the behavior of vectors and operations within a vector space. It also provides a solid foundation for further mathematical concepts and applications.

## 5. Can a theorem in vector spaces be proven in different ways?

Yes, a theorem in vector spaces can be proven in different ways as long as the proof follows the rules and definitions of vector spaces. Some theorems may have multiple proofs, each using different techniques and approaches.

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