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How to prove a power series of matrix is onto

  1. Apr 17, 2009 #1
    1. The problem statement, all variables and given/known data
    How do I prove a power series is onto? Since I cannot calculate directly, especially I haven't learned Jodarn Normal form.




    2. Relevant equations



    3. The attempt at a solution
    By showing 1-1, I tried
    ∑(1/n!)[(M)^n-(N)^n]=0, what can I conclude from this step?
     
  2. jcsd
  3. Apr 18, 2009 #2

    matt grime

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    M is a square matrix, so is exp(M). Thus it is one-to-one if and only if it is onto which is if and only if it is invertible.
     
  4. Apr 18, 2009 #3

    matt grime

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    Please don't send me private messages instead of answering in the thread.

    From the pm, the question ought to have been written:

    show that the map

    exp: M_2(R) --> M_2(R)

    is not onto.

    This is a very different question from the one inferred from the first post.

    What have you attempted.
     
  5. Apr 18, 2009 #4
    Sorry about that. How can I show it is not onto, since I cannot find a B in Y, and calculate its A in X? Is there any other to prove it instead of using Jordan Normal Form
     
  6. Apr 18, 2009 #5

    matt grime

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    You are not aiming to calculate its image. Just to find one element not in its image, or some family of matrices not in its image.

    Just think about the exp map for real numbers. It maps R onto the strictly positive real numbers. Now, negative matrices don't make much sense immediately, so what about the zero matrix? Can you show that is not the image.

    If you don't like that, what else do you know? How about: what is exp(A)exp(-A)? Do you have any other results you can think of?
     
  7. Apr 18, 2009 #6
    I understand this part. But this is follow the assumption that we know
    exp(A)=[tex]\sum[/tex]1/n!(A)^n

    What if we don't know this theroem. All we know this is true for A is real number, not when A is matrix, then makes it hard to prove, right
     
  8. Apr 18, 2009 #7

    matt grime

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    What? That's not an assumption so much as the definition of what the expression exp(A) means.
     
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