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Linear Algebra: Kernel, Basis, Dimensions, injection, surjections

  1. Mar 29, 2014 #1
    1. The problem statement, all variables and given/known data
    SK85YSZ.jpg

    3. The attempt at a solution
    uP1S7hh.jpg

    Can someone please check my work?
     
  2. jcsd
  3. Mar 30, 2014 #2

    micromass

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    Some comments:

    a) You have not given the entire kernel. You have just produced a basis (or at least a spanning set) of the kernel.

    d) I don't see why you care about the range here.

    h) You did some calculations to show the map is invertible. OK. But there is a much easier way, do you see which one?
     
  4. Mar 30, 2014 #3
    For a), I don't know how to write out the full set of vectors. All I know is that the vectors are linear combinations of the basis of the kernel.

    I'm not sure what you mean by range for d). Injection refers to one-to-one and a theorem states that if the rank is equal to the number of columns, then it is injective or one-to-one. So if the rank is not equal to the numbers, it is not injective.

    For h), I can't think of any easier way or any other way for that matter. Does your method involve determinants? If so, I didn't learn those yet.
     
  5. Mar 30, 2014 #4

    micromass

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    Yes, just write that out:

    [tex]\textrm{Ker}(L) = \{(-2\alpha-\beta,3\alpha,\alpha,\beta)~\vert~\alpha,\beta\in \mathbb{R}\}[/tex]

    OK, this is right. I was thinking of ##L## being injective if and only if ##\textrm{dim}(\textrm{Ker}(L)) = 0##.

    If a function is invertible, then it must be both injective and surjective. That's not the case here, is it?
     
  6. Mar 30, 2014 #5
    Thanks a lot!
     
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