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Idempotent property

  1. Nov 29, 2013 #1
    How do you show that a linear transformation is idempotent?

    T:R^3 to R^3 T (x y z)^T = (0.5 (x-z) , y, 0.5 (z-x))


    I have no idea where to begin. I know a few facts about idempotent properties e.g such as their eigenvalues are either 0 or 1. How would I show that the above transformation has these eigenvalues. I know how to find them but the above form has thrown me off. Would I have to just prove a couple of idempotent to show it is idempotent?
    Thanks in advance
     
  2. jcsd
  3. Nov 29, 2013 #2

    LCKurtz

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    Why don't you just directly show that ##T^2=T##, either from the above definition or its matrix representation?
     
  4. Nov 29, 2013 #3
    ok so how would I square this matrix? if I write it in matrix form would it be 0.5x-0.5x
    y
    -0.5z+0.5z

    or do I not expand the bracket because the x variable and z variable cancel out?
     
  5. Nov 29, 2013 #4

    LCKurtz

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    You have to write it in matrix form if you want to get a matrix to square.$$
    T(x,y,z) = \begin{bmatrix} a&b&c\\d&e&f\\g&h&i\end{bmatrix}
    \begin{bmatrix}x\\y\\z\end{bmatrix}=
    \begin{bmatrix}\frac 12(x-z)\\y\\\frac 1 2(z-x)\end{bmatrix}$$Once you figure out that matrix ##A## you can square it to see if ##A^2=A##.

    Also, as I suggested before, you have the formula for ##T(x,y,z)##. You could just apply it twice.
     
  6. Nov 30, 2013 #5
    so if I multiply them I get ax+by+cz 0.5(x-z)
    dx+ey+fz = y
    gx+hy+iz 0.5(z-x)

    Is that what you meant? I am not sure where I am going with this.
     
  7. Nov 30, 2013 #6

    Erland

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    But it is obvious. If T(x,y,z)=(0.5(x-z), y, 0.5(z-x)) for all x,y,z, then what is T(0.5(x-z), y, 0.5(z-x))?
     
  8. Nov 30, 2013 #7
    I honestly can not see the answer. Could you give me an example if you do not mind? It does not have to be this specific problem. maybe I am looking at the problem from a different angle if it is meant to be that obvious.
    is it
    0.5 0 -0.5
    0 1 0
    -0.5 0 0.5
     
    Last edited: Nov 30, 2013
  9. Nov 30, 2013 #8

    Erland

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    You just have to calculate T(0.5(x-z), y, 0.5(z-x)) and see what the answer is.
     
  10. Nov 30, 2013 #9
    ahh ok so the matrix representation is what I said above and then you just check the properties of the idempotent apply
     
  11. Nov 30, 2013 #10

    LCKurtz

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    Yes, that is the matrix representation. What happens if you square that matrix, which represents ##T^2##?

    I am curious what course you are taking. Have you talked about matrix representations of linear transformations?
     
  12. Nov 30, 2013 #11

    LCKurtz

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    You are making this way too difficult. You have the formula for ##T##. Could you calculate ##T(1,2,3)##? ##T(a,b,c)##? What about ##T(0.5(x-z), y, 0.5(z-x))##? Just use the formula.
     
  13. Nov 30, 2013 #12
    I am doing maths.I find the course very well except for linear stuff. I can not picture things. Yeah if you square it you get itself again. we have only just touched on it. I am kind of going ahead of the course.
     
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