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It the following matrix isomorphic?

  1. Jan 21, 2009 #1
    Is the following matrix isomorphic?

    1. The problem statement, all variables and given/known data
    a) Is the following matrix isomorphic?
    [tex] L: \Re^{2} \rightarrow \Re^{2} [/tex]

    [tex] L(x) = (-x_{2}, 0), where x = (x_{1}, x_{2}) \in \Re^{2}.[/tex]

    b)
    Define Ker(L) and Im(L).

    3. The attempt at a solution
    a) I need to show that the L is bijection. I know that it is injection,
    because it can be presented as Ax = b. The mapping is also linear, so L is
    surjection. Thus, L is bijection and isomorphic.

    b)
    [tex]Ker(L) = \Re^{2}[/tex]
    [tex]Im(L) = \Re^{2}[/tex]

    I am not sure is this enough.
     
    Last edited: Jan 22, 2009
  2. jcsd
  3. Jan 21, 2009 #2
    This is incorrect. There are many examples of linear, non-surjective functions. For instance, the projective function. What is the definition of a surjective function? Moreover, what is the definition of an injective function?
     
  4. Jan 21, 2009 #3
    Injection: from one to one i.e. each element in the domain is determined once by the element at the codomain
    Surjection: all elements in the codomain are determined by those at the domain

    This seems to mean that L is not isomorphic. The reason is perhaps that L is not injection as x1 is not determined at the codomain by the domain. That is all elements of the codomain are not determined.
     
  5. Jan 21, 2009 #4

    Dick

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    L is neither an injection nor a surjection. Just cite an example for each. Find two elements in the domain that map to the same image point and find a point in the range that isn't hit by any domain point. Neither is at all hard.
     
  6. Jan 21, 2009 #5
    Let x2 = 0 and x1 = 0. This means that two points in the domain map to one codomain. So it is not an injection.

    The other one is harder. I first thought to let x2 = 1 and x1 = 1. Then, I started to wonder that it is not even a funtion as no domain points hit points at the codomain.

    Please, let me know whether I am at the right direction or not.
     
  7. Jan 21, 2009 #6

    Dick

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    You do mean (x1,x2)->(-x2,0), right? (1,1)->(-1,0), (0,1)->(-1,0), not injective. Nothing maps to (0,1), not surjective. Case closed.
     
  8. Jan 21, 2009 #7
    No, no. To prove that it's not an injection, you must find two points in the domain that map to the same point. The domain is [tex]\mathbb{R}^2[/tex]. Hence, you must find two points x=(x1,x2) and y=(y1,y2) such that [tex]L(x) = L(x_1,x_2) = L(y_1,y_2) = L(y)[/tex]. Since you seem to be stuck, I'll give you an example. Let x= (1,1) and y = (0,1). Then by definition of L, we have L(x) = (-x2,0) = (-1,0) = (1,0) and L(y) = (-y2,0) = (-1,0). This means that we have found two points in the domain of L that map to the same point. But this contradicts the definition of injectivity.
     
  9. Jan 21, 2009 #8
    Thank you!
    Your answer is excellent.
     
  10. Jan 21, 2009 #9

    Dick

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    You WILL have an easier time with the next one, right? Don't overthink it. It's not all that abstract.
     
  11. Feb 23, 2009 #10
    b)
    The image should be
    Im(L) = ([tex]x_{1}, x_{2}[/tex])

    The kernel should be got by solving the null space that is Ax = 0.
    I am wondering how to put the domain and codomain to Ax = 0.
    We should get two simple equations from the matrix.

    I am keen on saying that the kernel is
    Ker(L) = ([tex]x_{1} = 0, x_{2} = 0[/tex])
    It seems to be the only solution - not sure though.
     
  12. Feb 23, 2009 #11

    Dick

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    What linear map are you talking about now?
     
  13. Feb 24, 2009 #12
    I will correct the mistake.

    We get the kernel by solving the nullspace
    Ker(L) = {[tex](x_{1}, x_{2})^{T}: x_{1} = 0[/tex] and [tex] x_{2} = 0[/tex]}

    The image is the same as we apparently want it for the nullspace
    Im(L) = {[tex](x_{1}, x_{2})^{T}: x_{1} = 0[/tex] and [tex] x_{2} = 0[/tex]}

    It should now be correct.
     
  14. Feb 24, 2009 #13

    HallsofIvy

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    Please answer Dick's question! What linear map are you talking about? Are you still talking about L(x1,x2)= (-x2, 0)? If so, then, no, the kernel is NOT {(0,0)}.

    And it makes no sense to say of the image "apparently we want it for the nullspace". The image of L is set of all y such that L(x)= y for some x. It has little to do with the nullspace (= kernel).
     
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