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Prove that dim(Ker(T)) is either n or n-1

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Having:

    [tex] \displaystyle T:V\to \mathbb{R}[/tex]
    [tex] \displaystyle \dim\left( V \right)=n[/tex]

    Prove that [tex] \displaystyle \dim\left( \ker \left( T \right) \right)=n\text{ or }n-1[/tex]


    2. Relevant equations

    3. The attempt at a solution

    I would start stating the theorem of the dimensions:

    [tex] \displaystyle \dim\left( V \right)=\dim\left( \ker \left( T \right) \right)+\dim\left( \operatorname{Im}\left( T \right) \right)[/tex]

    But I get nowhere with that theorem. Any clue?

    Thanks!
     
  2. jcsd
  3. Jun 19, 2012 #2
    What is dim(V)? What are possibilities for dim(im(T))?
     
  4. Jun 19, 2012 #3
    Hmm yes, I know how to determine the rank of a matrix. What do you mean by the dimension of T? Im(T)?

    Thanks
     
  5. Jun 19, 2012 #4
    Ahhhhmmm, dim(Im(T)) has to be less or equal than 1... or better: it either is 1 or zero Thanks!
     
  6. Jun 19, 2012 #5
    Alright, then I'm asked the following:

    Having:
    [tex] \displaystyle T:V\to \mathbb{R}[/tex]
    [tex] \displaystyle S:V\to \mathbb{R}[/tex]
    [tex] \displaystyle \dim\left( V \right)=n[/tex]

    Prove that ker(T)=ker(S) iff exists a real number a (different to zero) such that T=aS

    This looks obvious for me because if the transformations are equal, then not only their kernels but also their images and everything must be equal.. What is what I have to show then?

    Thanks
     
  7. Jun 19, 2012 #6

    micromass

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    Sure, if two transformations are equal then their kernels must be equal. But what does that have to do with the problem here??
     
  8. Jun 19, 2012 #7
    I don't see why that is obvious. One direction looks fairly clear, but you should still show the details, because I think it will familiarize you more with the elements of vector spaces and linear maps. (By the way, why do you say equal, we don't have T=S, we have T=aS.)

    The other direction looks less clear, if their kernels are equal, It's not obvious to me why they have to be scalar multiples of one another. It may require the use of the image being 1-d.



    You may ask, "why should I be so careful?" You could easily lead yourself to believing something false. We need to establish whether a given statement is true. That takes some of what's called mathematical maturity. You can come to recognize the care you must take. Your belief in the statement does not use the fact that a is nonzero. In fact, you seem to read it as S=T. Your belief in the statement does not use the fact that the codomain is ℝ, 1-d. It may require this fact. It seems if somebody adjusted the statement a bit so that it was false, you would not notice, and and that you would not be prepared to verify.

    Now these things come with lots of practice, but I just wanted to point out some possible goals. And let's keep trying to investigate the problem you stated above!
     
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