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Isomorphism proof

  1. May 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Let L:R->R be a linear operator with matrix C. Prove if the columns of C are linearly independent, then L is an isomorphism.


    2. Relevant equations



    3. The attempt at a solution

    Assume the columns of C are linearly independent. Then, the homogenous equation Cx=0 is the trivial solution. Need to show L is both 1-1 and onto. Assume that L(c1)=L(c2). Need to show that c1=c2. Well, since the matrix is linearly independent and Cx=0 is trivial solution, then each column represents its own solution. Therefore, if L(c1)=L(c2), then it must be that c1=c2. Now need to show L is onto. So for every d, there exists a c such that L(c)=d. Since C is linearly independent, no vector can be expressed as a linear combination of others. So, for each d, there will be a c where L(c)=d. Therefore, L is an isomorphism.
     
  2. jcsd
  3. May 10, 2013 #2

    mfb

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    I think this should be L:Rn->Rn?
    An equation is not the same as a solution. The equation has the trivial solution only.

    How do you mean that?
    Hint: Use linearity.

    "is onto"?

    No vector of your matrix, right.
    That does not follow from the previous statement.
     
  4. May 10, 2013 #3
    Okay so the equation has the trivial solution. So that means that the only solution is the trivial one, which is represented by the columns themselves, therefore 1-1. And since C is linearly independent, no vector in the matrix can be expressed as a linear combination of otheres. Therefore, each output has an input where L maps to that.
     
  5. May 11, 2013 #4

    mfb

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    The columns have no relation to the trivial solution. The trivial solution is x=0.


    Consider the linear operator F:R->R2, F(x)=(x,0). It satisfies everything you list in your posts, but it is not surjective: there is no x such that F(x)=(1,1). As long as you do not use that L maps Rn on itself, the argument cannot work.
     
  6. May 11, 2013 #5
    so how do you go about showing it is surjective
     
  7. May 12, 2013 #6
    Because the columns of A are linearly independent, each column will have a unique solutions such that there will exist an x such that L(x)=b which implies Ax=B
     
  8. May 12, 2013 #7

    mfb

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    Injective linear operators from R^n to R^n are always surjective. This follows from a more general result about the dimensions of the image and the kernel.
     
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