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Kernel of eigenspace.

  1. Nov 30, 2013 #1
    Lets say you have a linear transformation P. The eigenvalues of the matrices are 0,1 and 2.
    How would you show that ker P belongs to the eigenspace corresponding to 0?

    So you have an eigenvalue 0. Let A be the 3X3 matrix.
    I was thinking of doing something like Ax=λx and substitute 0 for λ. And then show that x,y,z are equal to 0 and hence the eigenspace is 0. Would this be a good idea?
     
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  3. Nov 30, 2013 #2

    Dick

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    Thinking a little more about it would be the best idea. Isn't the definition of x being in ker P the same as the definition of x being a eigenvector with eigenvalue 0?
     
  4. Nov 30, 2013 #3
    yes so I assume the original suggestion was bad.
     
  5. Nov 30, 2013 #4

    Dick

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    It was certainly confusing. If you meant the vector x has the components (x,y,z) then Ax=0 doesn't necessarily mean x,y,z=0.
     
  6. Nov 30, 2013 #5
    Im sorry. I am really really bad/hopeless at linear mathematics. When I meant that A(x,y,z)=0
     
  7. Nov 30, 2013 #6

    Dick

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    If you are given a specific matrix A and you want to find ker A then that's what you do alright. But you don't have to find ker A to see that it's the same as the set of eigenvectors with eigenvalue 0.
     
  8. Nov 30, 2013 #7
    ok cool. I am starting to understand a little better. How do you know the kernel A is the same as the set of eigenvectors with eigenvalue 0? Where do I go from here?
     
  9. Nov 30, 2013 #8

    vela

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    What equation does a vector in the kernel of A satisfy? What equation does an eigenvector with ##\lambda=0## satisfy?
     
  10. Nov 30, 2013 #9
    what are you asking for the original matrix?
     
  11. Nov 30, 2013 #10

    vela

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    No. I'm asking you to tell us how to express the phrase "##\vec{x}## is in the kernel of a matrix A" mathematically. Similarly, how do you say "##\vec{x}## is an eigenvector of matrix A with eigenvalue 0" mathematically?
     
  12. Nov 30, 2013 #11
    for "vector x is in the kernel of A" ker(A)={x belongs to X: T(x)=0}
    I am not sure about the other one.
    Great question by the way, really forcing me to think and understand.
     
  13. Nov 30, 2013 #12

    vela

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    Your definition for the kernel of A isn't correct as there's no mention of A.

    Look up the definition of an eigenvector and eigenvalue. In math, you really should know the definitions.
     
  14. Nov 30, 2013 #13
    What should it be?
    to calculate eigenvalue you use Ax=lambda x
    I know it is a relatively new topic we have started and I can not stand it. But I try
     
  15. Nov 30, 2013 #14

    Dick

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    Try to take the think and understand challenge. What's wrong with saying
    for "vector x is in the kernel of A" ker(A)={x belongs to X: T(x)=0}? What does T have to do with it? And, yes, x is an eigenvector if Ax=lambda x. What value of lambda are you interested in?
     
  16. Dec 1, 2013 #15
    The value of lambda I am interested in is 0.

    AT the end should it be T(A)=0?
    Is this stuff needed to answer the question?
     
  17. Dec 1, 2013 #16

    HallsofIvy

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    No, it isn't! There was no "T" in your question and you cannot just throw one in without defining it! The kernel of A is the set of all x such that A(x)= 0, not "T(x)= 0" as you had before.

    Yes, knowing the definition of "Kernel" is needed to answer questions about the "kernel" of a linear transformation!
     
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