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Bases of linearly isomorphic vector spaces

  1. Nov 22, 2011 #1
    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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  3. Nov 22, 2011 #2

    Office_Shredder

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    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)?
     
  4. Nov 22, 2011 #3
    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?
     
  5. Nov 22, 2011 #4

    Fredrik

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    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.
     
  6. Nov 22, 2011 #5
    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?
     
  7. Nov 22, 2011 #6

    Fredrik

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    Good definition.

    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?
     
  8. Nov 22, 2011 #7
    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?
     
  9. Nov 22, 2011 #8

    Fredrik

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    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.

    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.

    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.
     
  10. Nov 22, 2011 #9
    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?
     
  11. Nov 22, 2011 #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).
     
  12. Nov 23, 2011 #11
    Any idea on how to set that up? I tried to show it by contradiction.
     
  13. Nov 23, 2011 #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.
     
  14. Nov 23, 2011 #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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