Bases of linearly isomorphic vector spaces

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Finite-dimensional [itex]V[/itex] and [itex]W[/itex] are linearly isomorphic vector spaces over a field. Prove that if [itex]\{v_{1},...,v_{n}\}[/itex] is a basis for [itex]V[/itex], [itex]\{T(v_{1}),...,T(v_{n})\}[/itex] is a basis for [itex]W[/itex].

My attempt at a proof:

Let [itex]T:V\rightarrow W[/itex] be an isomorphism and [itex]\{v_{1},...,v_{n}\}[/itex] be a basis for [itex]V[/itex]. Since [itex]T[/itex] is an isomorphism, [itex]\forall v\in V,\exists T(v)\in W[/itex]. Therefore, if [itex]\{v_{1},...,v_{n}\}[/itex] spans [itex]V[/itex], [itex]\{T(v_{1}),...,T(v_{n})\}[/itex] spans [itex]W[/itex]. Since [itex]dim V=dimW[/itex], [itex]\{v_{1},...,v_{n}\}[/itex] is linearly independent, and is a basis for [itex]W[/itex].

I'm pretty sure I went wrong somewhere/claimed something I needed to prove. Any ideas where?
 

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  • #2
Office_Shredder
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Therefore, if [itex]\{v_{1},...,v_{n}\}[/itex] spans [itex]V[/itex], [itex]\{T(v_{1}),...,T(v_{n})\}[/itex] spans [itex]W[/itex].
You just state this, but you never really proved it. In fact, you never used that T is an isomorphism.... that given [itex]v\in V[/itex] there exists [itex]T(v)\in W[/itex] is merely a consequence of T being a function. Given [itex]w\in W[/itex], how can you go about determining the coefficients of w in terms of the T(vi)?
 
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Do I use the fact that any [itex]w\in W [/itex] can be written as [itex]w=a_{1}T(v_{1})+...+a_{n}T(v_{n}) [/itex] or does that only follow after I show something else?
 
  • #4
Fredrik
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Do I use the fact that any [itex]w\in W [/itex] can be written as [itex]w=a_{1}T(v_{1})+...+a_{n}T(v_{n}) [/itex]
That's part of what you're trying to prove.

I think you should start by writing down the definition of "basis" that you want to use.
 
  • #5
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I think you should start by writing down the definition of "basis" that you want to use.
A basis is a linearly independent spanning set. So [itex]\{v_{1},...,v_{n}\}[/itex] spans [itex]V[/itex] and is linearly independent. Does that imply that the set of all [itex]T(v)[/itex] is linearly independent since T is an isomorphism?
 
  • #6
Fredrik
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A basis is a linearly independent spanning set.
Good definition.

So [itex]\{v_{1},...,v_{n}\}[/itex] spans [itex]V[/itex] and is linearly independent. Does that imply that the set of all [itex]T(v)[/itex] is linearly independent since T is an isomorphism?
It does, but the question is how. You need to verify that [itex]\{Tv_1,\dots,Tv_n\}[/itex] is a linearly independent set. Can you use the definition of linearly independent (and isomorphism) to do that?
 
  • #7
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Can you use the definition of linearly independent (and isomorphism) to do that?
Since [itex]T[/itex] is an isomorphism, [itex]\forall v\in V[/itex] there exists only one [itex]T(v)\in W[/itex] such that [itex]v=T(v)[/itex]. Since [itex]\{v_{1},...,v_{n}\}[/itex] is linearly independent, [itex]\alpha_{1}v_{1}+...+\alpha_{n}v_{n}=0 [/itex] iff [itex]\forall i,\alpha_{i}=0[/itex]. Thus [itex]T(\alpha_{1}v_{1}+...+\alpha_{n}v_{n})=T(0)[/itex] and [itex]\{T(v_{1}),...,T(v_{n})\}[/itex] is linearly independent ...am I on the right track?
 
  • #8
Fredrik
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Since [itex]T[/itex] is an isomorphism, [itex]\forall v\in V[/itex] there exists only one [itex]T(v)\in W[/itex] such that [itex]v=T(v)[/itex].
This doesn't quite make sense, but you probably meant something different from what you said. Note that v and Tv aren't even members of the same space. If you meant that for all [itex]v\in V[/itex], there's a unique [itex]w\in W[/itex] such that [itex]Tv=w[/itex], then this is just what it means for T to be a function from V into W.

Since [itex]\{v_{1},...,v_{n}\}[/itex] is linearly independent, [itex]\alpha_{1}v_{1}+...+\alpha_{n}v_{n}=0 [/itex] iff [itex]\forall i,\alpha_{i}=0[/itex].
This is correct, but I want your first step to be to apply the definition of "linearly independent" directly to [itex]\{Tv_1,\dots,Tv_n\}[/itex] to see if the definition is satisfied.

Thus [itex]T(\alpha_{1}v_{1}+...+\alpha_{n}v_{n})=T(0)[/itex]
You can't say something like that when you haven't even specified what the v's and a's are. It looks like you have assumed that they are such that [itex]a_1v_1+\cdots+a_nv_n=0[/itex]. So you appear to be trying to prove that [itex]\{T(v_{1}),...,T(v_{n})\}[/itex] is linearly independent by having T act only on 0. This can't work.
 
  • #9
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This is correct, but I want your first step to be to apply the definition of "linearly independent" directly to {Tv1,…,Tvn} to see if the definition is satisfied.
Suppose it is linearly dependent. Then, [itex]\alpha_{1}T(v_{1})+...+\alpha_{n}T(v_{n})=0 [/itex] for some nonzero [itex]\alpha_{i}[/itex]. Then there exists a [itex]j\in\mathbb{N}, 1\leq j\leq n [/itex] such that [itex]\alpha_{1}T(v_{j}) =-(\alpha_{1}T(v_{1}) +...+\alpha_{j-1}T(v_{j-1})) [/itex]. Is this on the right track?
 
  • #10
Fredrik
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That could work, but I haven't tried to do it that way. This is what I had in mind: Prove that for all [itex]a_1,\dots,a_n\in\mathbb R[/itex] such that [itex]\sum_i a_iTv_i=0[/itex], we have [itex]a_1=a_2=\cdots=a_n=0[/itex]. (Replace ℝ with ℂ if it's a vector space over ℂ).
 
  • #11
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Any idea on how to set that up? I tried to show it by contradiction.
 
  • #12
Fredrik
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Just start like this: Let [itex]a_1,\dots,a_n[/itex] be arbitrary real numbers such that [itex]\sum_i a_i Tv_i=0[/itex]. We have [tex]0=\sum_i a_i Tv_i=\cdots[/tex] Now think of something to replace the dots with, and figure out what this means.
 
  • #13
Deveno
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it's often easier to prove linear independence rather than "not linear dependence".

make a linear combination of the T(vj) and set it = 0.

T is linear. use this fact to get T "of something".

then use the fact that T is an isomorphism, and thus 1-1,

to show what the "something" must be.

THEN....you should have a statement involving only the vj, and these are linearly independent, so....this should say something about a certain linear combination of them, which will maybe, just maybe, tell us something about the T(vj).
 

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