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Proof of isomorphism of vector spaces

  • #1
Moved from a technical forum, so homework template missing
The theorem is as follows:
All finite dimensional vector spaces of the same dimension are isomorphic

Attempt:
If T is a linear map defined as :

T : V →W
: dim(V) = dim(W) = x < ∞
& V,W are vector spaces

It would be sufficient to prove T is a bijective linear map:
let W := {wi}ni
like wise let : let V:= {vi}ni
let ω ∈ W & ζ ∈ V
It can be shown:
ω = ∑iwiki
ζ = ∑ivioi
The above is a result of the definition of a vector, note ki and oi are of an arbitary vector field.
now:
T(ω) = ζ
T(∑iwiki) = ∑ivioi
iT(wiki) = ∑ivioi
thus:
i(T(wiki) - vioi) = 0
therefore:(a bit iffy)

T(wiki) - vioi=0
=>
T(wiki) = vioi
T(ωi)=ζi
Hence Surjective
a linear transform is by nature injective, therefore bijective

i have a hunch i am wrong.Any advice :)
 

Answers and Replies

  • #2
andrewkirk
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The usual way to do this would be to identify a basis B =(b1, b2, ...., bn) for V and a basis C = (c1, c2, ...., cn) for W, let f be the map from B to C that maps b1 to c1, b2 to c2 and so on. Then construct the map T from V to W that is the linear extension of f.

The map is linear since we constructed it to be so, and it is easily shown to be a bijection, using the fact that B and C are bases - hence spanning and linearly independent.
 
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  • #3
mathwonk
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forgive me, but this proof is trivial, assuming the concept of dimension. the hard one is the proof that dimension is well defined as the number of elments in any independent spanning set, i.e. the proof that any spanning set has at least as many elements as any independent set do you know the proof of that?
 
  • #4
forgive me, but this proof is trivial, assuming the concept of dimension. the hard one is the proof that dimension is well defined as the number of elments in any independent spanning set, i.e. the proof that any spanning set has at least as many elements as any independent set do you know the proof of that?
Well I did not find it trivial.
 
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  • #5
mathwonk
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sorry for the pejorative adjective, but my suggestion still applies, namely try to do the harder result. [the concept of dimension shows immediately that a space of dimension n over the field k is isomorphic to k^n, by the map taking v = a1v1+...anvn to (a1,...,an). hence two spaces both of dimension n are both isomorphic to k^n, hence to each other.] try to prove this and then try to see that it is straight forward.

to be briefly somewhat honest, I admit that "trivial" means that as a young man I struggled long and hard enough to finally understand this, and hence now as an old man I am tempted to torture younger people by saying it is easy! my apologies.
 
  • #6
sorry for the pejorative adjective, but my suggestion still applies, namely try to do the harder result. [the concept of dimension shows immediately that a space of dimension n over the field k is isomorphic to k^n, by the map taking v = a1v1+...anvn to (a1,...,an). hence two spaces both of dimension n are both isomorphic to k^n, hence to each other.] try to prove this and then try to see that it is straight forward.

to be briefly somewhat honest, I admit that "trivial" means that as a young man I struggled long and hard enough to finally understand this, and hence now as an old man I am tempted to torture younger people by saying it is easy! my apologies.
Alrighty , I have just begun really getting into linear algebra( at least proving generalise statements) I will attempt your suggestion.
 
  • #7
PeroK
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The theorem is as follows:
All finite dimensional vector spaces of the same dimension are isomorphic

Attempt:
If T is a linear map defined as :

T : V →W
: dim(V) = dim(W) = x < ∞
& V,W are vector spaces

It would be sufficient to prove T is a bijective linear map:
let W := {wi}ni
like wise let : let V:= {vi}ni
let ω ∈ W & ζ ∈ V
It can be shown:
ω = ∑iwiki
ζ = ∑ivioi
The above is a result of the definition of a vector, note ki and oi are of an arbitary vector field.
now:
T(ω) = ζ
T(∑iwiki) = ∑ivioi
iT(wiki) = ∑ivioi
thus:
i(T(wiki) - vioi) = 0
therefore:(a bit iffy)

T(wiki) - vioi=0
=>
T(wiki) = vioi
T(ωi)=ζi
Hence Surjective
a linear transform is by nature injective, therefore bijective

i have a hunch i am wrong.Any advice :)
I'm not sure this proof works. You have tried to show that an arbitrary linear map is a bijection. This can't be true. At the end you assume all linear maps are injective, which is not true.

Instead the proof requires you to construct a bijection between the spaces.
 
  • #8
WWGD
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Maybe this will work. If V,W ate vector spaces of the same dimension n,over the same field ##\mathbb F##, then each is isomorphic to ##\mathbb F^n## by the map that takes v=c1v1+....cnvn -->(c1,c2,....,cn), same for W. Now just compose the two maps ( or their inverses) the right way to get the isomorphism between V,W. EDIT: It comes down to: V homeo to Z, W homeo to Z ==> V homeo to W.
 
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  • #9
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forgive me, but this proof is trivial, assuming the concept of dimension. the hard one is the proof that dimension is well defined as the number of elments in any independent spanning set, i.e. the proof that any spanning set has at least as many elements as any independent set do you know the proof of that?
By trivial do you mean that proof is easily obtained and the two major ideas are the following theorems:

From Axler, page 54/ Theorem 3.5:
If A is a basis for V and w1,w2,.., wn are elements of W. Then there exist a unique linear map T from V to W such that Tvj = wj for each j=1,...,n

Also, the Fundamental Theorem Of Linear Maps: dimension V = dimension of null T + dimension of range T?

SO from this point, the other facts fall into place?


I do not want to write out the details of my proof, since this is the OP question and I do not want to hijack his thread. But is this what you mean Mathwonk?
 
  • #10
mathwonk
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i guess i meant that the definition of the isomorphism is very easy and the verification that it is injective, surjective, and linear, are all straightforward, no creativity is needed, just knowledge of the definitions.

i.e. if v1,...vn is the basis, then the map taking (a1,...,an) to a1v1+...+anvn is easily shown to be linear, and is injective because a basis is independent, and surjective because a basis is a spanning set. nonetheless to check these things one does need to know their definitions, but that's all. still learning those definitions, and being able to use them, is the most significant first step for a beginner.

anyway this shows every vector space of dimension n over k is isomorphic to k^n. i suppose then one should also show that the composition of two isomorphisms is an isomorphism, and this is again straightforward checking of definitions.

on the other hand showing that any two bases of the same vector space have the same number of elements, and hence that dimension is well defined, is quite hard.
 
  • #11
vela
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on the other hand showing that any two bases of the same vector space have the same number of elements, and hence that dimension is well defined, is quite hard.
I'd expect this result was already established in the OP's class since the problem refers to vector spaces of finite dimension. It wouldn't really make sense to ask a homework question referring to a concept that hasn't been defined.
 
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  • #12
WWGD
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I'd expect this result was already established in the OP's class since the problem refers to vector spaces of finite dimension. It wouldn't really make sense to ask a homework question referring to a concept that hasn't been defined.
Nice post. Happy to give you your 1,000nd upvote. Here is to (at least) 1,000 more soon! ;).
 
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  • #13
mathwonk
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well some courses skip the proof of that result and even if they give it, I still suggest thinking about how to prove it, since most students cannot, even those whose course has treated it. the idea, i recommend, is to try to understand and master the material. of course it is probably a good idea to try the present :"trivial" proof first, of that is a challenge, since one wants to build his/her strength up gradually, hence doing easier stuff first.
 
  • #14
Hey guys , just want to add this is not a course i am attending.Rather i am attempting to strengthen my understanding of tensors, linear algebra to QM is what meat is to a sandwich.
 
  • #15
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I guess that you mean surjective "onto" when you use the symbol "##\rightarrow##" in ##T: V \rightarrow W##. That is not the meaning that I would normally assume. But without that, the statement is clearly false.

Maybe I am missing something. It looks like you are trying to prove that T is surjective. That will not work unless you define T more specifically in a way that makes it surjective, which you have not. Suppose I say that ##Tv = 0 \forall v \in V##. How does that violate your (nonexistant) definition of T?
 
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