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L_A is isomorphism

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data

    if [itex]L_A: ℝ^n -> ℝ^n : X-> A.X [/itex] is a linear transformation, and A is an orthogonal matrix, show that L_A is an isomorphism.
    also given is that [itex] (ℝ,ℝ^n,+,[.,.]) [/itex] , the standard Euclidian space which has inproduct [X,Y]= X^T.Y

    2. Relevant equations

    ortogonal matrix, so [itex]A^T=A^{-1}[/itex]
    isomorphism = bijective and linear (so what is left to show is bijective)


    3. The attempt at a solution

    don't know where to start
     
  2. jcsd
  3. Sep 23, 2012 #2

    jbunniii

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    Well, what's the definition of bijective?
     
  4. Sep 23, 2012 #3
    surjective and injective ofcourse:
    injective:
    I must show that if [itex]X_1 != X_2[/itex] then [itex]A.X_1 ≠A.X_2.[/itex]
    So choose [itex]X_1,X_2 \in ℝ^n[/itex]. because [itex] [AX_1,AX_2]=X_1^T.A^T.A.X_2 = X_1^T.X_2=[X_1,X_2] [/itex], length, distance and orthogonality will be preserved. so [itex]A.X_1 ≠A.X_2 [/itex].
    surjective:
    I must show that for every [itex]Y \in ℝ^n : A.X=Y [/itex]for some [itex]X \in ℝ^n [/itex]
    Choose Y in ℝ^n. Is there an X in ℝ^n such that A.X=Y? I'm not sure what to do here.
     
  5. Sep 23, 2012 #4

    Dick

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    You've shown A is injective, so you know the kernel of A is {0}. Use the rank-nullity theorem.
     
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